rotational dynamics

CHAPTER 9 ROTATIONAL DYNAMICS

PROBLEMS_________________________________________________________________________________________

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1. SSM REASONING AND SOLUTION Solving Equation 9.1 for the lever arm l , we obtain

l =

tF

= 3.0 N m12 N

= 0.25 m

____________________________________________________________________________________________

2. REASONING The torque is given by Equation 9.1, t = F l, where F is the magnitude of theapplied force and l is the lever arm. From the figure in the text, the lever arm is given by

l = ( 0.28 m) sin 50.0 . Since both t and l are known, Equation 9.1 can be solved for F.

SOLUTION Solving Equation 9.1 for F, we have

F =

tl

=45 N m

(0.28 m) sin 50.0 = 2.1 10 2 N

____________________________________________________________________________________________

3. REASONING AND SOLUTION The torque about the center of the cable car is t = FL,where L is the lever arm.

t = FL = 2(185 N)(4.60 m) = 1.70 10 3 N m_________________________________________________________________________________________

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4. REASONING To calculate the torques, we need to determine the lever arms for each of theforces. These lever arms are shown in the following drawings:

SOLUTIONa. Using Equation 9.1, we find that the magnitude of the torque due to the weight W is

2.50 m

Axis32.0

lT = (2.50 m) cos 32.0

T

2.50 m

Axis

32.0

WlW = (2.50 m) sin 32.0

320 ROTATIONAL DYNAMICS

t W W N m N m= = = Wl 10 200 2 5 32 13500b gb g. sin

b. Using Equation 9.1, we find that the magnitude of the torque due to the thrust T is

t T T N m N m= = = Tl 62 300 2 5 32 132 000b gb g. cos_________________________________________________________________________________________

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5. SSM REASONING The torque on either wheel is given by Equation 9.1, t = F l, where F isthe magnitude of the force and l is the lever arm. Regardless of how the force is applied, thelever arm will be proportional to the radius of the wheel.

SOLUTION The ratio of the torque produced by the force in the truck to the torque produced inthe car is

ttruck

tcar

=Fl

truck

Flcar

=Fr

truck

Frcar

=r

truck

rcar

=0.25 m

0.19 m= 1.3

_____________________________________________________________________________________________

6. REASONING To calculate the torque,we will need the lever arm of the 110 Nforce. The drawing identifies the leverarm l.

SOLUTION According to Equation9.1, the magnitude of the torque is

t = = = Wl 110 N 3 320 Nb gb g.0 m mcos 15_________________________________________________________________________________________

____

7. REASONING AND SOLUTION The torque due to F1 is

t1 = -F1L1 = -(20.0 N)(0.500 m) = -10.0 Nm (CW)

The torque due to F2 is

t2 = (F2 cos 30.0)L2 = (35.0 N)(1.10 m) cos 30.0 = 33.3 Nm (CCW)

The net torque is therefore,

3.0 m

W =110 N

l = (3.0 m)cos 15

15

Chapter 9 Problems 321

St = t1 + t2 = -10.0 Nm + 33.3 Nm = 23.3 N m, counterclockwise_________________________________________________________________________________________

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8. REASONING AND SOLUTION The torque produced by a force of magnitude F is given byEquation 9.1, t = F l, where l is the lever arm. In each case, the torque produced by the coupleis equal to the sum of the individual torques produced by each member of the couple.

a. When the axis passes through point A, the torque due to the force at A is zero. The lever armfor the force at C is L. Therefore, taking counterclockwise as the positive direction, we have

t = tA +tC = 0 +tC = FL

b. Each force produces a counterclockwise rotation. The magnitude of each force is F and eachforce has a lever arm of L / 2 . Taking counterclockwise as the positive direction, we have

t = tA +tC = F L

2

+ F

L

2

= FL

c. When the axis passes through point C, the torque due to the force at C is zero. The lever armfor the force at A is L. Therefore, taking counterclockwise as the positive direction, we have

t = tA +tC = tA + 0 = FL

Note that the value of the torque produced by the couple is the same in all three cases; in otherwords, when the couple acts on the tire wrench, the couple produces a torque that does notdepend on the location of the axis.

____________________________________________________________________________________________

9. SSM REASONING Since the meter stick does not move, it is in equilibrium. The forces andthe torques, therefore, each add to zero. We can determine the location of the 6.00 N force, byusing the condition that the sum of the torques must vectorially add to zero.

1l2

l

F1

F2

30.0

pinned

end

TOP VIEW

SOLUTION If we take counterclockwise torques as positive, then the torque of the first forceabout the pin is

t1 1 1= = F l (2.00 N)(1.00 m) = 2.00 N m


The torque due to the second force is

t 2 2 2 2 23= = F (sin 30.0 ) (6.00 N)(sin 30.0 ) = ( .00 N) l l l

The torque t 2 is negative because the force F2 tends to produce a clockwise rotation about thepinned end. Since the net torque is zero, we have

2 00 3 00 02. ( . ] N m +[ N) =lThus,

l 2 = =2.00 N m

3.00 N0.667 m

____________________________________________________________________________________________

10. REASONING AND SOLUTION The net torque about the axis in text drawing (a) is

St = t1 + t2 = F1b - F2a = 0.

Considering that F2 = 3F1, we have b 3a = 0. The net torque in drawing (b) is then

St = F1(1.00 m - a) - F2b = 0 or 1.00 m - a - 3b = 0.

Solving the first equation for b, substituting into the second equation and rearranging, gives

a = 0.100 m and b = 0.300 m

_____________________________________________________________________________________________

11. REASONING AND SOLUTIONa. Taking torques about an axis through the rear axle of the car gives

2Ff(2.54 m) - (1160 kg)(9.80 m/s2)(1.52 m) = 0 or Ff = 3.40 10

3 N

b. Taking torques about an axis through the front axle gives

(1160 kg)(9.80 m/s2)(1.02 m) - 2Fr(2.54 m) = 0 or Fr = 2.28 103 N

_____________________________________________________________________________________________

12. REASONING The drawingshows the forces acting on theperson. It also shows the leverarms for a rotational axis

Axis

FFEETFHANDS

W


perpendicular to the plane of thepaper at the place where thepersons toes touch the floor.Since the person is in equilibrium,the sum of the forces must bezero. Likewise, we know that thesum of the torques must be zero.

SOLUTION Taking upward to be the positive direction, we have

F F WFEET HANDS+ - = 0

Remembering that counterclockwise torques are positive and using the axis and the lever armsshown in the drawing, we find

W F

FW

l l

l

l

W HANDS HANDS

HANDSW

HANDS

N m

1.250 m N

- =

= = =

0

584 0 840392

b gb g.

Substituting this value into the balance-of-forces equation, we find

F W FFEET HANDS N N == - = -584 392 192 N

The force on each hand is half the value calculated above, or 196 N . Likewise, the force on

each foot is half the value calculated above, or 96 N ._________________________________________________________________________________________

____

13. SSM REASONING The minimum value for the coefficient of static friction between the ladderand the ground, so that the ladder does not slip, is given by Equation 4.7:

f FsMAX

s N= m

SOLUTION From Example 4, the magnitude of the force of static friction is Gx = 727 N. Themagnitude of the normal force applied to the ladder by the ground is Gy = 1230 N. The minimumvalue for the coefficient of static friction between the ladder and the ground is

ms

sMAX

N

727 N1230 N

0 591= = = =f

F

G

Gx

y

.


_____________________________________________________________________________________________

14. REASONING When the board just begins to tip, three forces act on the board. They are theweight W of the board, the weight WP of the person, and the force F exerted by the right support.

x

1.4 m

W = 225 N

W P = 450 N

F

Since the board will rotate around the right support, the lever arm for this force is zero, and thetorque exerted by the right support is zero. The lever arm for the weight of the board is equal toone-half the length of the board minus the overhang length: 2.5 m 1.1 m = 1.4 m .

The lever arm for the weight of the person is x. Therefore, taking counterclockwise torques aspositive, we have

- WP x + W (1.4 m) = 0

This expression can be solved for x.

SOLUTION Solving the expression above for x, we obtain

x =W (1.4 m)

WP

=(225 N)(1.4 m)

450 N= 0.70 m

_____________________________________________________________________________________________


15. SSM REASONING The figure at the right shows the doorand the forces that act upon it. Since the door is uniform, thecenter of gravity, and, thus, the location of the weight W, is atthe geometric center of the door.

Let U represent the force applied to the door by the upperhinge, and L the force applied to the door by the lower hinge.Taking forces that point to the right and forces that point up aspositive, we have

Fx = L x - U x = 0 or L x = U x (1)

Fy = Ly - W = 0 or L y = W = 140 N (2)

h = 2.1 m

d = 0.81 m

Ux

Ly

Lx

W

Taking torques about an axis perpendicular to the plane of the door and through the lower hinge,with counterclockwise torques being positive, gives

t = -Wd

2

+ U x h = 0 (3)

Equations (1), (2), and (3) can be used to find the force applied to the door by the hinges;Newton's third law can then be used to find the force applied by the door to the hinges.

SOLUTIONa. Solving Equation (3) for Ux gives

U x =

Wd

2 h= (140 N)(0.81 m)

2(2.1 m)= 27 N

From Equation (1), Lx = 27 N.

Therefore, the upper hinge exerts on the door a horizontal force of 27 N to the left .

b. From Equation (1), Lx = 27 N, we conclude that the bottom hinge exerts on the door ahorizontal force of 27 N to the right .

c. From Newton's third law, the door applies a force of 27 N on the upper hinge. This

force points horizontally to the right , according to Newton's third law.


d. Let D represent the force applied by the door to the lower hinge. Then, from Newton's thirdlaw, the force exerted on the lower hinge by the door has components Dx = 27 N andDy = 140 N.

From the Pythagorean theorem,

D D Dx y= + = - + - =2 2 ( 27 N) ( 140 N) 143 N2 2

The angle q is given by

q =FHG

IKJ=

-tan140

271 79

N

N

The force is directed 79 below the horizontal .

q

Dx

Dy

16. REASONING AND SOLUTION The net torque about an axis through the contact pointbetween the tray and the thumb is

St = F(0.0400 m) - (0.250 kg)(9.80 m/s2)(0.320 m) - (1.00 kg)(9.80 m/s2)(0.180 m)- (0.200 kg)(9.80 m/s2)(0.140 m) = 0

F = 70.6 N, up

Similarly, the net torque about an axis through the point of contact between the tray and the finger is

St = T(0.0400 m) - (0.250 kg)(9.80 m/s2)(0.280 m) - (1.00 kg)(9.80 m/s2)(0.140 m)- (0.200 kg)(9.80 m/s2)(0.100 m) = 0

T = 56.4 N, down _________________________________________________________________________________________

____

17. REASONING AND SOLUTION When the arm is in equilibrium, the sum of the torques iszero. Taking torques about the elbow joint with counterclockwise torques taken as positive, wehave

- FlF + M lM = 0Solving for M,

M = F lFl M

= (170 N)

0.30 m

0.051 m

= 1.0 10 3 N

The direction of the force is to the left .


_____________________________________________________________________________________________

18. REASONING Although this arrangement of body parts is vertical, we can apply Equation 9.3 tolocate the overall center of gravity by simply replacing the horizontal position x by the verticalposition y, as measured relative to the floor.

SOLUTION Using Equation 9.3, we have

yW y W y W y

W W Wcg

N m .760 m .250 m

N m

=+ ++ +

=+ +

+ +=

1 1 2 2 3 3

1 2 3

438 1 28 144 N 0 87 N 0

438 144 N 87 N1 03

b gb g b gb g b gb g..

_____________________________________________________________________________________________

19. REASONINGThe jet is in equilibrium, so the sum of the external forces is zero, and the sum of the externaltorques is zero. We can use these two conditions to evaluate the forces exerted on the wheels.

SOLUTIONa. Let Ff be the magnitude of the normal force that the ground exerts on the front wheel. Since thenet torque acting on the plane is zero, we have (using an axis through the points of contact betweenthe rear wheels and the ground)

St = -W lw + Ff lf = 0

where W is the weight of the plane, and lw and lf are the lever arms for the forces W and Ff ,

respectively. Thus,

St = -(1.00 106 N)(15.0 m - 12.6 m) + Ff (15.0 m) = 0

Solving for Ff gives Ff = 1 60 105. N .

b. Setting the sum of the vertical forces equal to zero yields

SFy = Ff + 2Fr - W = 0

where the factor of 2 arises because there are two rear wheels. Substituting in the data,

SFy = 1.60 105 N + 2Fr - 1.00 10

6 N = 0


Fr = 4.20 105 N

_____________________________________________________________________________________________

20. REASONING AND SOLUTIONa. Taking torques about an axis through the bolt we have

-(4.5 103 N)(2.50 m) - (12.0 103 N)(3.50 m) + T(5.00 m) sin 25.0 = 0

Solving for T yields

T = 2.52 10 4 N

b. Applying Newton's second law in the horizontal direction gives -T cos 25.0 + Fh = 0, so

Fh = 2.28 104 N

In the vertical direction Fv - WL - WB + T sin 25.0 = 0, and

Fv = 5.85 103 N

Now the force exerted on the beam by the bolt is

F = Fh2 + Fv

2 = 2.35 10 4 N_________________________________________________________________________________________

____

21. SSM WWW REASONING AND SOLUTION The figure below shows the massless boardand the forces that act on the board due to the person and the scales.

W

x

2.00 m

center of gravityof person

person's head person's feet

425 N 315 N

a. Applying Newton's second law to the vertical direction gives

315 N + 425 N W = 0 or W = 7.40 10 2 N, downward


b. Let x be the position of the center of gravity relative to the scale at the person's head. Takingtorques about an axis through the contact point between the scale at the person's head and theboard gives

(315 N)(2.00 m) (7.40 102 N)x = 0 or x = 0.851 m_________________________________________________________________________________________

____

22. REASONING AND SOLUTION The net torque about an axis through the knee joint is

St = M(sin 25.0)(0.100 m) - (44.5 N)(cos 30.0)(0.250 m) = 0

M = 228 N_________________________________________________________________________________________

____

23. SSM REASONING Since the man holds the ball motionless, the ball and the arm are inequilibrium. Therefore, the net force, as well as the net torque about any axis, must be zero.

SOLUTIONa. Using Equation 9.1, the net torque about an axis through the elbow joint is

St = M(0.0510 m) (22.0 N)(0.140 m) (178 N)(0.330 m) = 0

Solving this expression for M gives M = 1.21 10 3 N .

b. The net torque about an axis through the center of gravity is

St = (1210 N)(0.0890 m) + F(0.140 m) (178 N)(0.190 m) = 0

Solving this expression for F gives F = 1.01 10 3 N . Since the forces must add to give a net

force of zero, we know that the direction of F is downward .

_____________________________________________________________________________________________

24. REASONING If we assume that the system is in equilibrium, we know that the vector sum of allthe forces, as well as the vector sum of all the torques, that act on the system must be zero.

The figure below shows a free body diagram for the boom. Since the boom is assumed to beuniform, its weight W B is located at its center of gravity, which coincides with its geometricalcenter. There is a tension T in the cable that acts at an angle q to the horizontal, as shown. At thehinge pin P, there are two forces acting. The vertical force V that acts on the end of the boom


prevents the boom from falling down. The horizontal force H that also acts at the hinge pinprevents the boom from sliding to the left. The weight W L of the wrecking ball (the "load") acts atthe end of the boom.

f = 48P

q = 32

q = 32

W L

W B

T

V

H

f q

By applying the equilibrium conditions to the boom, we can determine the desired forces.

SOLUTION The directions upward and to the right will be taken as the positive directions. Inthe x direction we have

Fx = H - T cos q = 0 (1)

while in the y direction we have

Fy = V - T sin q - WL - WB = 0 (2)

Equations (1) and (2) give us two equations in three unknown. We must, therefore, find a thirdequation that can be used to determine one of the unknowns. We can get the third equation fromthe torque equation.

In order to write the torque equation, we must first pick an axis of rotation and determine the leverarms for the forces involved. Since both V and H are unknown, we can eliminate them from thetorque equation by picking the rotation axis through the point P (then both V and H have zero leverarms). If we let the boom have a length L, then the lever arm for W L is L cos f, while the leverarm for W B is ( L / 2) cos f. From the figure, we see that the lever arm for T is L sin(f q) . Ifwe take counterclockwise torques as positive, then the torque equation is

t = - WB L cos f

2

- WL L cos f+ TL sin f -q( ) = 0

Solving for T, we have

T =12 WB + WLsin( f q)

cos f (3)


a. From Equation (3) the tension in the support cable is

T =12

(3600 N) + 4800 Nsin(48 32 )

cos 48 = 1.6 10 4 N

b. The force exerted on the lower end of the hinge at the point P is the vector sum of the forces Hand V. According to Equation (1),

H = T cosq = 1.6 10 4 N( )cos 32 = 1.4 10 4 N

and, from Equation (2)

V = WL +W B + T sin q = 4800 N + 3600 N + 1.6 104 N( ) sin 32 = 1.7 10 4 N

Since the forces H and V are at right angles to each other, the magnitude of their vector sum canbe found from the Pythagorean theorem:

FP = H2 + V 2 = (1.4 10 4 N) 2 + (1.7 10 4 N) 2 = 2.2 10 4 N

_____________________________________________________________________________________________

25. REASONING AND SOLUTION Taking torques on the right leg about the apex of the Agives

St = -(120 N) L cos 60.0 - FL cos 30.0 + T(2L) cos 60.0 = 0

where F is the horizontal component of the force exerted by the crossbar and T is the tension in theright string. Due to symmetry, the tension in the left string is also T. Newton's second law applied tothe vertical gives

2T - 2W = 0 so T = 120 N

Substituting T = 120 N into the first equation yields F = 69 N ._________________________________________________________________________________________

____


26. REASONING The drawing shows the forces acting onthe board, which has a length L. Wall 2 exerts a normalforce P2 on the lower end of the board. The maximumforce of static friction that wall 2 can apply to the lowerend of the board is msP2 and is directed upward in thedrawing. The weight W acts downward at the board'scenter. Wall 1 applies a normal force P1 to the upper endof the board. We take upward and to the right as ourpositive directions.

q

P1

P2

P2m s

W

Wall 1 Wall 2

SOLUTION Then, since the horizontal forces balance to zero, we have

P1 P2 = 0 (1)

The vertical forces also balance to zero:

msP2 W = 0 (2)

Using an axis through the lower end of the board, we now balance the torques to zero:

WL

2

(cos q ) - P1 L (sin q ) = 0 (3)

Rearranging Equation (3) gives

tan q = W

2P1 (4)

But W = msP2 according to Equation (2), and P2 = P1 according to Equation (1). Therefore, W =msP1, which can be substituted in Equation (4) to show that

tan q = msP12P1

=m s2

=0.98

2

orq = tan1(0.49) = 26


From the drawing at the right,

cos q =d

L

Therefore, the longest board that can be propped between thetwo walls is

Ld 1.5m

cos26 1.7 m= =

=

cos q

d

q

L

_____________________________________________________________________________________________

27. REASONING AND SOLUTION The weight W of the left side of the ladder, the normal forceFN of the floor on the left leg of the ladder, the tension T in the crossbar, and the reaction force R

due to the right-hand side of the ladder, are shown in the figure below. In the vertical direction -W+ FN = 0, so that

FN = W = mg = (10.0 kg)(9.80 m/s2) = 98.0 N

In the horizontal direction it is clear that R = T.The net torque about the base of the ladder is

St = T [(1.00 m) sin 75.0o] W [(2.00 m) cos 75.0o] + R [(4.00 m) sin 75.0o] = 0

Substituting for W and using R = T, we obtain

75.0o

W

R

TF

N

T = ( N )(2.00 m) cos 75.0

(3.00 m) s in 75.0 =

98.0 17.5 N

_____________________________________________________________________________________________

28. REASONING The moment of inertia of the stool is the sum of the individual moments of inertiaof its parts. According to Table 9.1, a circular disk of radius R has a moment of inertia ofI M Rdisk disk= 12

2 with respect to an axis perpendicular to the disk center. Each thin rod isattached perpendicular to the disk at its outer edge. Therefore, each particle in a rod is located ata perpendicular distance from the axis that is equal to the radius of the disk. This means that each

of the rods has a moment of inertia of Irod = Mrod R2.

SOLUTION Remembering that the stool has three legs, we find that the its moment of inertia is


I I I M R M Rstool disk rod disk rod

2 kg m .15 kg m kg m

= + = +

= + =

3 3

1 2 0 16 3 0 0 16 0 027

12

2 2

12

2 2. . . .b gb g b gb g______________________________________________________________________________

29. SSM REASONING AND SOLUTION For a rigid body rotating about a fixed axis,Newton's second law for rotational motion is given by Equation 9.7, t = Ia, where I is themoment of inertia of the body and a is the angular acceleration expressed in rad/s2. Equation 9.7gives

I =t

a=

10 .0 N m8.00 rad/s 2

= 1.25 kg m 2

_____________________________________________________________________________________________

30. REASONING AND SOLUTION We know

St = Ia = (0.16 kgm2)(7.0 rad/s2) = 1.1 kgm2/s2 = 1.1 N m_________________________________________________________________________________________

____

31. REASONING AND SOLUTIONa. The net torque on the disk about the axle is

St = F1R - F2R = (0.314 m)(90.0 N - 125 N) = -11 N m

b. The angular acceleration is given by a = St/I. From Table 9.1, the moment of inertia of thedisk is

I = (1/2) MR2 = (1/2)(24.3 kg)(0.314 m)2 = 1.20 kgm2

a = (-11 Nm)/(1.20 kgm2) = 9.2 rad / s2

_____________________________________________________________________________________________

32. REASONING The rotational analog of Newton's second law is given by Equation 9.7,t = Ia. Since the person pushes on the outer edge of one pane of the door with a force F that

is directed perpendicular to the pane, the torque exerted on the door has a magnitude of FL, wherethe lever arm L is equal to the width of one pane of the door. Once the moment of inertia isknown, Equation 9.7 can be solved for the angular acceleration a .


The moment of inertia of the door relative to the rotation axis is I = 4 IP , where IP is the momentof inertia for one pane. According to Table 9.1, we find IP =

1

3ML2 , so that the rotational inertia

of the door is I = 43

ML 2 .

SOLUTION Solving Equation 9.7 for a , and using the expression for I determined above, wehave

a =FL

43

ML 2=

F43

ML=

68 N43

(85 kg)(1.2 m)= 0.50 rad/s 2

_____________________________________________________________________________________________

33. SSM REASONING The figure below shows eight particles, each one located at a differentcorner of an imaginary cube. As shown, if we consider an axis that lies along one edge of the cube,two of the particles lie on the axis, and for these particles r = 0. The next four particles closest tothe axis have r = l , where l is the length of one edge of the cube. The remaining two particleshave r = d , where d is the length of the diagonal along any one of the faces. From thePythagorean theorem, d = + =l l l2 2 2 .

l

ld

Axis

l

l

d

According to Equation 9.6, the moment of inertia of a system of particles is given by I = mr 2 .

SOLUTION Direct application of Equation 9.6 gives

I = mr2 = 4(ml2 ) + 2(md 2 ) = 4(ml2 ) + 2(2ml2 ) = 8ml2

or

I = 8( 0.12 kg)(0.25 m) 2 = 0.060 kg m 2

_____________________________________________________________________________________________

34. REASONING AND SOLUTION


a. We know that a = (w wo)/t. But w = (1/2)wo, so

a = -(1/2)wo/t = -(1/2)(88.0 rad/s)/(5.00 s) = -8.80 rad/s2

The magnitude of a is 8.80 rad/s 2 .

b. St = Ia and FR = Ia. The magnitude of the frictional force is

F = Ia/R = (0.850 kgm2)(8.80 rad/s2)/(0.0750 m) = 99.7 N_________________________________________________________________________________________

____

35. REASONING AND SOLUTION We know St = Ia, where

I = (1/2) mr2 = (1/2)(0.400 kg)(0.130 m)2 = 3.38 103 kgm2

Also, w = wo + at, so that a = (85.0 rad/s - 262 rad/s)/(18.0 s) = -9.83 rad/s2. Thus, the net

torque is

St = Ia = (3.38 103 kgm2)(-9.83 rad/s2) = -3.32 10 2 N m_________________________________________________________________________________________

____

36. REASONING The force applied to the baton creates a net torque that gives the rod an angularacceleration, according to Newtons second law as expressed for rotational motion (St = Ia).From the data given, we will calculate the moment of inertia I and the angular acceleration a.Newtons second law, then, will allow us to obtain the net torque St. From the net torque, we willbe able to determine the force, since the magnitude of the torque in this case is just the magnitude ofthe force times the lever arm of 2.0 cm.

SOLUTION The momentum of inertia of the baton is the sum of the individual moments of inertiaof its parts. For a thin rod of length L rotating about an axis perpendicular to the rod at its center,the moment of inertia is I M Lrod rod=

112

2 . For each ball (assumed to be very small compared to

the length of the rod), the moment of inertia is Iball = Mball (L/2)2. The total moment of inertia, then,

is

I I I M L M L M L M Lbaton rod ball rod ball rod ball= + = + = +2 21

122 1

2

2 112

2 12

2c h

Using Equation 8.4 from the equations of kinematics for rotational motion, we can determine theangular acceleration as


aw w

=- 0t

where w and w0 are, respectively, the final and initial angular velocities. Since only one force ofmagnitude F creates the torque and since Equation 9.1 gives the magnitude of the torque as themagnitude of the force times the lever arm l, Newtons second law becomes St = Fl= Ia.Substituting our expressions for the moment of inertia and angular acceleration of the baton into thesecond law gives

F I M L M Lt

l = = +-F

HGIKJbaton rod balla

w w1

122 1

22 0c h

In Newtons second law the angular acceleration a must be expressed in rad/s2. Therefore, thevalues for the angular speeds w and w0 must be expressed in rad/s. Recognizing that 2.0 rev/s is4.0 p rad/s and solving for F, we find that

FM L M L

t=

+ -FHG

IKJ

=+ -F

HGIKJ=

112

2 12

20

112

2 12

20 13 0 81 0 13 0 81

0 0204 0

78 N

rod ball

kg m kg m

m rad / s rad / s

0.40 s

c h

b gb g b gb gl

w w

p. . . ..

_____________________________________________________________________________________________

37. SSM REASONING AND SOLUTIONa. From Table 9.1 in the text, the moment of inertia of the rod relative to an axis that isperpendicular to the rod at one end is given by

I MLrod2 2 kg)(2.00 m) m= = = 13

2 13 2 00 2( . .67 kg

b. The moment of inertia of a point particle of mass m relative to a rotation axis located aperpendicular distance r from the particle is I mr= 2 . Suppose that all the mass of the rod werelocated at a single point located a perpendicular distance R from the axis in part (a). If this pointparticle has the same moment of inertia as the rod in part (a), then the distance R, the radius ofgyration, is given by

RIm

= =

=2.67 kg m

2.00 kg1.16 m

2

_____________________________________________________________________________________________


38. REASONING AND SOLUTION The final angular speed of the arm is w = vT/r, wherer = 0.28 m. The angular acceleration needed to produce this angular speed is a = (w - w0)/t. Thenet torque required is St = Ia. This torque is due solely to the force M, so that St = ML. Thus,

ML

It

L= =

-FHG

IKJt

w w0

Setting w0 = 0 and w = vT/r, the force becomes

M

Ir t

L

I

Lrt

kg m m / s

m m sN

T

T=

FHG

IKJ

= =

=

v

v 0 065 5 0

0 025 0 28 0 10460

2. .

. . .

c hb gb gb gb g

_____________________________________________________________________________________________

39. SSM REASONING The angular acceleration of the bicycle wheel can be calculated fromEquation 8.4. Once the angular acceleration is known, Equation 9.7 can be used to find the nettorque caused by the brake pads. The normal force can be calculated from the torque usingEquation 9.1.

SOLUTION The angular acceleration of the wheel is, according to Equation 8.4,

a =w -w0

t=

3.7 rad/s - 13 .1 rad/s3.0 s

= -3.1 rad/s 2

If we assume that all the mass of the wheel is concentrated in the rim, we may treat the wheel as ahollow cylinder. From Table 9.1, we know that the moment of inertia of a hollow cylinder of massm and radius r about an axis through its center is I = mr 2 . The net torque that acts on the wheeldue to the brake pads is, therefore,

t = Ia= (mr 2 )a (1)

From Equation 9.1, the net torque that acts on the wheel due to the action of the two brake pads is

t = 2 fk l (2)

where f k is the kinetic frictional force applied to the wheel by each brake pad, and l = 0.33 m isthe lever arm between the axle of the wheel and the brake pad (see the drawing in the text). Thefactor of 2 accounts for the fact that there are two brake pads. The minus sign arises because thenet torque must have the same sign as the angular acceleration. The kinetic frictional force can bewritten as (see Equation 4.8)

fk = mk FN (3)


where mk is the coefficient of kinetic friction and FN is the magnitude of the normal force appliedto the wheel by each brake pad.

Combining Equations (1), (2), and (3) gives

2(mk FN )l = (mr2 )a

FN = mr 2a2mk l

=(1.3 kg)(0.33 m) 2 (3.1 rad/s 2 )

2(0.85) ( 0.33 m)= 0.78 N

_____________________________________________________________________________________________

40. REASONING AND SOLUTION The moment of inertia of a solid cylinder about an axis

coinciding with the cylinder axis which contains the center of mass is Icm = (1/2) MR2. The parallel

axis theorem applied to an axis on the surface (h = R) gives I = Icm + MR2, so that

I =3

2MR 2

_____________________________________________________________________________________________

41. REASONING AND SOLUTION For door A, q = (1/2) aAtA2. For door B, q = (1/2) aBtB

2.Equating gives

t tB A

A

B

=aa

We need the angular accelerations. Applying Newton's second law for rotation to door A

FL = (1/3)ML2aA and to door B F(1/2)L = (1/12)ML2aB

Division of the previous two equations yields aA/aB = 1/2. Now

tB = tA/ 2 = (3.00 s)/(1.41) = 2.12 s

_____________________________________________________________________________________________

42. REASONING The drawing shows the drum, pulley, and the crate, as well as the tensions in thecord


r 1

r 2

D r u m

m 1 = 1 5 0 k g

r 1 = 0 . 7 6 m

P u l l e y

m 2 = 1 3 0 k g

C r a t e

m 3 = 1 8 0 k g

T 1

T 2T 1

T 2

Let T1 represent the magnitude of the tension in the cord between the drum and the pulley. Then,

the net torque exerted on the drum must be, according to Equation 9.7, St = I1a1, where I1 is themoment of inertia of the drum, and a1 is its angular acceleration. If we assume that the cabledoes not slip, then Equation 9.7 can be written as

- +

=FHG

IKJT r m r

I

ar1 1 1 1

2

1

1

1

t

t a

1 24 34 123123

c h (1)

where t is the counterclockwise torque provided by the motor, and a is the acceleration of thecord (a = 1.2 m/s2). This equation cannot be solved for t directly, because the tension T1 is notknown.

We next apply Newtons second law for rotational motion to the pulley in the drawing:

+ -

=FHG

IKJT r T r m r

I

ar1 2 2 2

12 2 2

2

2

2

2

t a

1 244 344 1 24 34123

c h (2)

where T2 is the magnitude of the tension in the cord between the pulley and the crate, and I2 is themoment of inertial of the pulley.

Finally, Newtons second law for translational motion (SFy = m a) is applied to the crate, yielding

+ -

=T m g

F

m a

y

2 3 31 244 344 (3)


SOLUTION Solving Equation (1) for T1 and substituting the result into Equation (2), then solvingEquation (2) for T2 and substituting the result into Equation (3), results in the following value for thetorque

t = + + +

= + =

r a m m m m g1 112 2 3 3

0 76 150 130 180 1700

c hc h( . ) ( )m) (1.2 m / s kg + kg + 180 kg kg)(9.80 m / s N m2 12 2

_____________________________________________________________________________________________

43. SSM REASONING The kinetic energy of the flywheel is given by Equation 9.9. The momentof inertia of the flywheel is the same as that of a solid disk, and, according to Table 9.1 in the text,is given by I MR= 12

2 . Once the moment of inertia of the flywheel is known, Equation 9.9 can besolved for the angular speed w in rad/s. This quantity can then be converted to rev/min.

SOLUTION Solving Equation 9.9 for w, we obtain,

w = = = = 2 2

12

2

( ) ( )KE KE 4(1.2 10 J)(13kg)(0.30 m)

6.4 10 rad / sR R9

24

I MR

Converting this answer into rev/min, we find that

wp

= FHGIKJFHG

IKJ= 6.4 10 rad / s

1 rev2 rad

60 s

1 min6.1 10 rev / min4 5c h

_____________________________________________________________________________________________

44. REASONING The rotational kinetic energy of each object is given by Equation 9.9,KE R =

12

2Iw . To solve this problem, we need only set the two kinetic energies equal, being sureto use the proper moment of inertia for each object, as given in Table 9.1.

SOLUTION According to Table 9.1 the moment of inertia of the solid sphere isI MRsolid =

25

2 , while the moment of inertia of the shell is I MRshell =23

2 . Using these

expressions in Equation 9.9 for the rotational kinetic energy, we obtain


12

12

12

25

2 12

23

2

25

23

53

53 7 0 9 0

I I

MR MR

solid solid2

Rotational KEof solid sphere

shell shell2

Rotational KEof spherical shell

solid2

shell2

solid2

shell2

solid shell rad / s rad / s

w w

w w

w w

w w

1 24 34 1 24 34=

=

=

= = =

c h c h

b g. ._________________________________________________________________________________________

____

45. SSM REASONING AND SOLUTIONa. The tangential speed of each object is given by Equation 8.9, v rT = w . Therefore,

For object 1: vT1 = (2.00 m)(6.00 rad/s) = 12.0 m / s



b. The total kinetic energy of this system can be calculated by computing the sum of the kineticenergies of each object in the system. Therefore,

KE = + +12 1 12 1

2 2 22 1

2 3 32m v m v m v

KE kg)(12.0 m / s) kg)(9.00 m / s) kg)(18.0 m / s) 1.08 2 2 2= + + = 12 6 00 4 00 3 00 10( . ( . ( .3 J

c. The total moment of inertia of this system can be calculated by computing the sum of themoments of inertia of each object in the system. Therefore,

I mr m r m r m r= = + + 2 1 12

2 22

3 32

I = + + = ( . ( . ( . .6 00 4 00 3 00 60 0 kg)(2.00 m) kg)(1.50 m) kg)(3.00 m) kg m2 2 2 2

d. The rotational kinetic energy of the system is, according to Equation 9.9,

KE (60.0 kg m )(6.00 rad / s) 1.08R2 2= = = 12

2 12 10Iw

3 J

This agrees, as it should, with the result for part (b).


_____________________________________________________________________________________________

46. REASONINGa. The kinetic energy is given by Equation 9.9 as KE R =

12

2Iw . Assuming the earth to be auniform solid sphere, we find from Table 9.1 that the moment of inertia is I MR= 25

2 . The mass

and radius of the earth is M = 5.98 1024 kg and R = 6.38 106 m (see the inside of the textsfront cover). The angular speed w must be expressed in rad/s, and we note that the earth turnsonce around its axis each day, which corresponds to 2p rad/day.

b. The kinetic energy for the earths motion around the sun can be obtained from Equation 9.9 asKE R =

12

2Iw . Since the earths radius is small compared to the radius of the earths orbit

(Rorbit = 1.50 1011 m, see the inside of the texts front cover), the moment of inertia in this case

is just I MR= orbit2 . The angular speed w of the earth as it goes around the sun can be obtained

from the fact that it makes one revolution each year, which corresponds to 2p rad/year.

SOLUTIONa. According to Equation 9.9, we have

KE

kg m rad

day day24 h

h3600 s

J

R = =

= FHG

IKJFHG

IKJFHG

IKJ

LNM

OQP

=

12

2 12

25

2 2

12

25

24 6 22

29

5 98 10 6 38 1021

1 1

2 57 10

I MRw w

p

c h

c hc h. .

.

b. According to Equation 9.9, we have

KE

kg m rad yr

yr365 day

day24 h

h3600 s

J

R orbit2= =

= FHGIKJFHG

IKJFHG

IKJFHG

IKJ

LNM

OQP

=

12

2 12

2

12

24 11 22

33

5 98 10 1 50 1021

1 1 1

2 67 10

I MRw w

p

c h

c hc h. .

._________________________________________________________________________________________

____

47. REASONING AND SOLUTION The conservation of energy gives for the cube (1/2) mvc2 =

mgh, so vc = 2gh . The conservation of energy gives for the marble (1/2) mvm2 + (1/2) Iw2 =

mgh. Since the marble rolls without slipping: w = v/R. For a solid sphere we know I = (2/5) mR2.Combining the last three equations gives


vgh

m =10

7Thus,

v

vc

m

= =75

118.

_____________________________________________________________________________________________

48. REASONING AND SOLUTION The rotational kinetic energy of the shell is KEr = (1/2) Iw2

and the translational kinetic energy is KEt = (1/2) Mv2. Now, v = Rw since the sphere rolls

without slipping, so

KEt = (1/2) MR2w2

The desired fraction is

KEr/KE = I/(I + MR2) = 2/5

_____________________________________________________________________________________________

49. SSM REASONING AND SOLUTION The only force that does work on the cylinders asthey move up the incline is the conservative force of gravity; hence, the total mechanical energy isconserved as the cylinders ascend the incline. We will let h = 0 on the horizontal plane at thebottom of the incline. Applying the principle of conservation of mechanical energy to the solidcylinder, we have

mgh s =1

2mv 0

2 + 12

Is w02 (1)

where, from Table 9.1, Is =1

2mr 2 . In this expression, v0 and w0 are the initial translational and

rotational speeds, and hs is the final height attained by the solid cylinder. Since the cylinder rollswithout slipping, the rotational speed w0 and the translational speed v0 are related according toEquation 8.12, w0 = v 0 / r . Then, solving Equation (1) for hs , we obtain

hs =3 v 0

2

4 g

Repeating the above for the hollow cylinder and using Ih = mr2 we have

hh =v0

2

g

The height h attained by each cylinder is related to the distance s traveled along the incline and theangle q of the incline by


s s =hs

sin q and s h =

hhsin q

Dividing these givessssh

= 3/4

_____________________________________________________________________________________________

50. REASONING AND SOLUTION The conservation of energy gives

mgh + (1/2) mv2 + (1/2) Iw2 = (1/2) mvo2 + (1/2) Iwo

2

If the ball rolls without slipping, w = v/R and wo = vo/R. We also know I = (2/5) mR2.

Substitution of the last two equations into the first and rearrangement gives

v v gh m / s m / s m 1.3 m / s02 10

7107

2= - = - =3 50 9 80 0 7602. . .b g c hb g_________________________________________________________________________________________

____

51. REASONING We first find the speed v0 of the ball when it becomes airborne using theconservation of mechanical energy. Once v0 is known, we can use the equations of kinematics tofind its range x.

SOLUTION When the tennis ball starts from rest, its total mechanical energy is in the form ofgravitational potential energy. The gravitational potential energy is equal to mgh if we take h = 0at the height where the ball becomes airborne. Just before the ball becomes airborne, itsmechanical energy is in the form of rotational kinetic energy and translational kinetic energy. Atthis instant its total energy is

1

2mv 0

2 + 12

Iw2 . If we treat the tennis ball as a thin-walled sphericalshell of mass m and radius r, and take into account that the ball rolls down the hill without slipping,its total kinetic energy can be written as

1

2mv 0

2 + 12

Iw2 = 12

mv 02 + 1

2(

2

3mr

2)

v0r

2

= 56

mv 02

Therefore, from conservation of mechanical energy, he have

mgh = 56

mv 02 or v 0 =

6 gh

5

The range of the tennis ball is given by x = v0 x t = v0 (cos q) t , where t is the flight time of the ball.From Equation 3.3b, we find that the flight time t is given by


t =v - v0 y

ay=

(-v0 y ) - v 0 yay

= 2v0 sin q

ay

Therefore, the range of the tennis ball is

x = v0 x t = v0 (cos q) 2v0 sin q

ay

If we take upward as the positive direction, then using the fact that a y = g and the expression forv0 given above, we find

x = 2 cos q sin qg

v 0

2 = 2 cos q sin qg

6 gh

5

2

= 125

h cos q sin q

= 125

(1.8 m) (cos 35)(sin 35) = 2.0 m

_____________________________________________________________________________________________

52. REASONING AND SOLUTION The angular momentum of the satellite is conserved if only thegravitational force of the earth is acting on it. We have LA = LP or mvArA = mvPrP. Now

rP = (vA/vP)rA = (1/1.20)(1.30 107 m) = 1.08 10 7 m

_____________________________________________________________________________________________

53. SSM WWW REASONING Let the two disks constitute the system. Since there areno external torques acting on the system, the principle of conservation of angular momentumapplies. Therefore we have L Linitial final= , or

I I I IA A B B A B finalw w w+ = +( )

This expression can be solved for the moment of inertia of disk B.

SOLUTION Solving the above expression for I B , we obtain

I IB A

final A

B final

2( 4 kg m2.4 rad / s 7.2 rad / s

9.8 rad / s (2.4 rad / s)4.4 kg m=

FHG

IKJ=

LNM

OQP=

w ww w

. )3 2

_____________________________________________________________________________________________

54. REASONING AND SOLUTION Angular momentum is conserved, Iw = I0w0, so


w w=FHG

IKJ =

FHG

IKJ =

I

I0

0

5.40 kg m

3.80 kg m5.00 rad / s 7 11 rad s

2

2 b g . /_________________________________________________________________________________________

____

55. REASONING AND SOLUTIONa. Angular momentum is conserved, so that Iw = Iowo, where

I = Io + 10mbRb2 = 2100 kgm2

w = (Io/I)wo = [(1500 kgm2)/(2100 kgm2)](0.20 rad/s) = 0.14 rad/s

b. A net external torque must be applied in a direction that is opposite to the angular decelerationcaused by the baggage dropping onto the carousel.

_____________________________________________________________________________________________

56. REASONING AND SOLUTIONThe conservation of angular momentum applies about the indicated axis

Idwd + MpRp2wp = 0 where wp = vp/Rp

Now Id = (1/2) MdRd2. Combining and solving yields

wd = -2(Mp/Md)(Rp/Rd2)vp = -

FHG

IKJLNMM

OQPP2

40 010

1 252 00

2 2

. ..

kg1.00 kg

m

2.00 m m / sb g b g = -0.500 rad/s

The magnitude of wd is 0.500 rad/s . The negative sign indicates that the disk rotates in a

direction opposite to the motion of the person._________________________________________________________________________________________

____

57. SSM REASONING Let the space station and the people within it constitute the system. Thenas the people move radially from the outer surface of the cylinder toward the axis, any torques thatoccur are internal torques. Since there are no external torques acting on the system, the principleof conservation of angular momentum can be employed.

SOLUTION Since angular momentum is conserved,

I Ifinal final 0w w= 0


Before the people move from the outer rim, the moment of inertia is

I I m r0

500= +station person person

2

orI 0

93 00 10 10= + = . ( kg m kg m2 9 2500)(70.0 kg)(82.5 m) 3.242

If the people all move to the center of the space station, the total moment of inertia is

I Ifinal station= = 3 00 10.9 2 kg m

Therefore,ww

final

0

0

final

3.24= =

=I

I1010

1 089

9 2

2

kg m 3.00 kg m

.

This fraction represents a percentage increase of 8 percent .

_____________________________________________________________________________________________

58. REASONING When the modules pull together, they do so by means of forces that are internal.These pulling forces, therefore, do not create a net external torque, and the angular momentum ofthe system is conserved. In other words, it remains constant. We will use the conservation ofangular momentum to obtain a relationship between the initial and final angular speeds. Then, wewill use Equation 8.9 (v = rw) to relate the angular speeds w0 and wf to the tangential speeds v0and vf.

SOLUTION Let L be the initial length of the cable between the modules and w0 be the initialangular speed. Relative to the center-of-mass axis, the initial momentum of inertia of the two-

module system is I0 = 2M(L/2)2, according to Equation 9.6. After the modules pull together, the

length of the cable is L/2, the final angular speed is wf, and the momentum of inertia is

If = 2M(L/4)2. The conservation of angular momentum indicates that

I I

ML

ML

f f

Final angularmomentum

0 0

Initial angularmomentum

f 0

f 0

w w

w w

w w

123 123=

FHG

IKJ

LNMM

OQPP

= FHGIKJ

LNMM

OQPP

=

24

22

4

2 2

According to Equation 8.9, wf = vf/(L/4) and w0 = v0/(L/2). With these substitutions, the resultthat wf = 4w0 becomes


v

L

v

Lv vf f or m / s m / s/ /4

42

2 2 17 340 0=FHG

IKJ = = =b g

_____________________________________________________________________________________________

59. REASONING AND SOLUTION After the mass has moved inward to its final path thecentripetal force acting on it is T = 105 N.

10 5 N

r

Its centripetal acceleration is

ac = v2/R = T/m

Now

v = wR so R = T/(w2m)

The centripetal force is parallel to the line of action (the string), so the force produces no torque onthe object. Hence, angular momentum is conserved.

Iw = Iowo so that w = (Io/I)wo = (Ro2/R2)wo

Substituting and simplifying

R3 = (mRo4 wo

2)/T, so that R = 0.573 m

_____________________________________________________________________________________________

60. REASONING AND SOLUTION The block will just start to move when the centripetal force onthe block just exceeds f s

max . Thus, if rf is the smallest distance from the axis at which the blockstays at rest when the angular speed of the block is wf, then ms FN = mrf wf2, or

ms mg = mrf wf2

ms g = rf wf2 (1)


Since there are no external torques acting on the system, angular momentum will be conserved.I0 w0 = If wf

where I0 = mr02, and If = mrf2. Making these substitutions yields

r02 w0 = rf2 wf (2)

Solving Equation (2) for wf and substituting into Equation (1) yields:

ms

g = rfw

0

2r

0

4

rf

4

Solving for rf gives

r f =w

02 r0

4

ms g

1/3

= (2.2 rad/s)2 (0.30 m) 4

(0.75)(9.80 m/s 2 )

1/3

= 0.17 m

_____________________________________________________________________________________________

61. SSM WWW REASONING The maximum torque will occurwhen the force is applied perpendicular to the diagonal of the square asshown. The lever arm l is half the length of the diagonal. From thePythagorean theorem, the lever arm is, therefore,

l = + =122 2( ) ( )0.40 m 0.40 m 0.28 m

Since the lever arm is now known, we can use Equation 9.1 to obtainthe desired result directly.

SOLUTION Equation 9.1 gives

t = = Fl (15 N)(0.28 m) = 4.2 N m

F

axis

0.40

m0.40 m

____________________________________________________________________________________________

62. REASONING AND SOLUTION For a solid disk we know I = (1/2)MR2 so M = 2I/R2.

Newton's second law for rotation yields I = (St)/a, so M = 2(St)/(aR2).

a = (w - wo)/t = (-3.49 rad/s)/(15.0 s) = -0.233 rad/s2, so that


M =2 -6.20 10 -3 N m( )

( -0.233 rad/s 2 )( 0.150 m ) 2= 2.37 kg

_____________________________________________________________________________________________

163. REASONING AND SOLUTION The sum ofthe forces must be zero:

SFy = FW + FC mg = 0

Therefore,

FW + FC = mg

= (71.0 kg)(9.80 m/s2) = 696 N

1.40 m 0.60 m

FW

FC

mg

The sum of the torques must also be zero. Taking the axis for torques at the point just below theboulder, we have

St = FC (0.60 m) FW (1.40 m) = 0.

a. Substituting for FW yields

FC(0.60 m) (696 FC)(1.40 m) = 0

Solving for FC gives FC = 487 N .

b. Also, FW = 696 N FC = 209 N

_____________________________________________________________________________________________

64. REASONING AND SOLUTION Use conservation of angular momentum, Iowo = (Io + I) w,where

Io = 1.2 103 kgm2 and wo = 0.40 rad/s

The moment of inertia of the bug is

I = mL2 = (5.0 103 kg)(0.20 m)2 = 2.0 104 kgm2

The final angular velocity isw = Iowo/(Io + I) = 0.34 rad/s


_____________________________________________________________________________________________

65. SSM REASONING AND SOLUTIONa. The rim of the bicycle wheel can be treated as a hoop. Using the expression given in Table 9.1in the text, we have

Ihoop = MR2 = (1.20 kg)(0.330 m) 2 = 0.131 kg m 2

b. Any one of the spokes may be treated as a long, thin rod that can rotate about one end. Theexpression in Table 9.1 gives

Irod =1

3ML2 = 1

3(0.010 kg)(0.330 m) 2 = 3.6 10 4 kg m2

c. The total moment of inertia of the bicycle wheel is the sum of the moments of inertia of eachconstituent part. Therefore, we have

I = Ihoop + 50 I rod = 0.131 kg m2 + 50(3.6 10 4 kg m 2 ) = 0.149 kg m 2

_____________________________________________________________________________________________

66. REASONING AND SOLUTION The net torque about an axis through the elbow joint is

St = (111 N)(0.300 m) - (22.0 N)(0.150 m) - (0.0250 m)M = 0 or M = 1.20 10 3 N_________________________________________________________________________________________

____

67. SSM REASONING The drawing shows the forces actingon the board, which has a length L. The ground exerts thevertical normal force V on the lower end of the board. Themaximum force of static friction has a magnitude of msV andacts horizontally on the lower end of the board. The weight Wacts downward at the board's center. The vertical wall appliesa force P to the upper end of the board, this force beingperpendicular to the wall since the wall is smooth (i.e., there isno friction along the wall). We take upward and to the right asour positive directions. Then, since the horizontal forcesbalance to zero, we have

V

q

ms V

P

W

msV P = 0 (1)

The vertical forces also balance to zero giving

V W = 0 (2)


Using an axis through the lower end of the board, we express the fact that the torques balance tozero as

PL WL

sin = 0q q- FHGIKJ2 cos (3)

Equations (1), (2), and (3) may then be combined to yield an expression for q .

SOLUTION Rearranging Equation (3) gives

tan = q WP2

(4)

But, P = msV according to Equation (1), and W = V according to Equation (2). Substituting theseresults into Equation (4) gives

tan = 2

12s s

qm mV

V=

Therefore,

qm

=FHG

IKJ=

LNM

OQP=

- -tan tan(

1 112

12 0.650)s

37.6

_____________________________________________________________________________________________

68. REASONING AND SOLUTIONa. The net torque about an axis through the point of contact between the floor and her shoes is

St = - (5.00 102 N)(1.10 m)sin 30.0 + FN(cos 30.0)(1.50 m) = 0

FN = 212 N

b. Newton's second law applied in the horizontal direction gives Fh - FN = 0, so Fh = 212 N

c. Newton's second law in the vertical direction gives Fv - W = 0, so Fv = 5.00 102 N

_____________________________________________________________________________________________

69. REASONING AND SOLUTION Since the change occurs without the aid of external torques,the angular momentum of the system is conserved: Ifwf = I0w0. Solving for wf gives

w f = w0I0I f


where for a rod of mass M and length L, I 0 =1

12ML 2 . To determine If, we will treat the arms of

the "u" as point masses with mass M/4 a distance L/4 from the rotation axis. Thus,

I f =1

12

M

2

L

2

2

+ 2

M

4

L

4

2

=

1

24ML 2

and

wf

=w0

1

12ML 2

1

24ML 2

= (7.0 rad/s)(2) = 14 rad/s

_____________________________________________________________________________________________

70. REASONING AND SOLUTION The conservation of energy gives (1/2) Iw2 = mg[(1/2) L].Now the tip of the rod has, as it hits the ground, a speed of v = Lw, so w = v/L. The moment ofinertia of the rod is given by I = (1/3) mL2. Substituting the last two equations into the first equationand simplifying yields

v 3gL 3 9.8 m / s 2.00 m 7.67 m / s2= = =c hb g_________________________________________________________________________________________

____

71. SSM WWW REASONING AND SOLUTION Consider the left board, which has a length Land a weight of (356 N)/2 = 178 N. Let Fv be the upward normal force exerted by the ground on

the board. This force balances the weight, so Fv = 178 N. Let fs be the force of static friction,

which acts horizontally on the end of the board in contact with the ground. fs points to the right.

Since the board is in equilibrium, the net torque acting on the board through any axis must be zero.Measuring the torques with respect to an axis through the apex of the triangle formed by theboards, we have

+ (178 N)(sin 30.0)L

2

+ fs(L cos 30.0) Fv (L sin 30.0) = 0

or44.5 N + fs cos 30 .0 FV sin 30.0 = 0

so that

fs =(178 N)(sin 30.0 ) 44.5 N

cos 30.0 = 51.4 N

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72. REASONING AND SOLUTION Newton's law applied to the 11.0-kg object gives


T2 - (11.0 kg)(9.80 m/s2) = (11.0 kg)(4.90 m/s2) or T2 = 162 N

A similar treatment for the 44.0-kg object yields

T1 - (44.0 kg)(9.80 m/s2) = (44.0 kg)(-4.90 m/s2) or T1 = 216 N

For an axis about the center of the pulley

T2r - T1r = I(-a) = (1/2) Mr2 (-a/r)

Solving for the mass M we obtain

M = (-2/a)(T2 - T1) = [-2/(4.90 m/s2)](162 N - 216 N) = 22.0 kg

_____________________________________________________________________________________________

73. CONCEPT QUESTIONS a. The magnitude of a torque is the magnitude of the force times thelever arm of the force, according to Equation 9.1. The lever arm is the perpendicular distancebetween the line of action of the force and the axis. For the force at corner B, the lever arm is halfthe length of the short side of the rectangle. For the force at corner D, the lever arm is half thelength of the long side of the rectangle. Therefore, the lever arm for the force at corner D is greater.Since the magnitudes of the two forces are the same, this means that the force at corner D createsthe greater torque.

b. As we have seen, the torque at corner D (clockwise) has a greater magnitude than the torque atcorner B (counterclockwise). Therefore, the net torque from these two contributions is clockwise.This means that the force at corner A must produce a counterclockwise torque, if the net torqueproduced by the three forces is to be zero. The force at corner A, then, must point toward cornerB.

SOLUTION Let L be the length of the short side of the rectangle, so that the length of the longside is 2L. We can use Equation 9.1 for the magnitude of each torque. Remembering thatcounterclockwise torques are positive and that the torque at corner A is counterclockwise, we canwrite the zero net torque as follows:

St = + - =

+ - =

F F F

F L L L

A A B B D D

A N N

l l l 0

12 12 012b gc h b g

The length L can be eliminated algebraically from this result, which can then be solved for FA:

FA N N N (pointing toward corner B)= - + =12 12 6 012b gch b g .


_____________________________________________________________________________________________

74. CONCEPT QUESTIONS a. In the left design, both the 60.0 and the 525-N forces createclockwise torques with respect to the rotational axis at the point where the tire contacts the ground.In the right design, only the 60.0-N force creates a torque, because the line of action of the 525-Nforce passes through the axis. Thus, the total torque from the two forces is greater for the leftdesign.

b. The man is supporting the wheelbarrow in equilibrium. Therefore, his force must create a torquethat balances the total torque from the other two forces. To do this, his torque must becounterclockwise and have the same magnitude as the total torque from the other two forces. Sincethe two forces create more torque in the left design, the magnitude of the mans torque must begreater for that design.

c. The magnitude of a torque is the magnitude of the force times the lever arm of the force,according to Equation 9.1. The drawing shows that the lever arms of the mans forces in the twodesigns is the same (1.300 m). Thus, to create the greater torque necessitated by the left design, theman must apply a greater force for that design.

SOLUTION The lever arms for the forces can be obtained from the distances shown in thedrawing for each design. We can use Equation 9.1 for the magnitude of each torque.Remembering that counterclockwise torques are positive, we can write an expression for the zeronet torque for each design. These expressions can be solved for the mans force F in each case:

Left design

Right design

S

S

t

t

= - - + =

=+

=

= - + =

= =

525 0 400 60.0 N 0 600 1 300 0

525 0 400 60.0 N 0 600

1 300189

60.0 N 0 600 1 300 0

60.0 N 0 600

1 30027.7 N

N m m m

N m m

m N

m m

m

m

b gb g b gb g b gb gb g b gb g

b gb g b gb gb g

. . .

. .

.

. .

.

.

F

F

F

F

_____________________________________________________________________________________________

75. CONCEPT QUESTIONS a. The forces in part b of the drawing cannot keep the beam inequilibrium (no acceleration), because neither V nor W, both being vertical, can balance thehorizontal component of P. This unbalanced component would accelerate the beam to the right.

b. The forces in part c of the drawing cannot keep the beam in equilibrium (no acceleration). Tosee why, consider a rotational axis perpendicular to the plane of the paper and passing through thepin. Neither P nor H create torques relative to this axis, because their lines of action pass directlythrough it. However, W does create a torque, which is unbalanced. This torque would cause thebeam to have a counterclockwise angular acceleration.


c. The forces in part d are sufficient to eliminate the accelerations discussed above, so they cankeep the beam in equilibrium.

SOLUTION Assuming that upward and to the right are the positive directions, we obtain thefollowing expressions by setting the sum of the vertical and the sum of the horizontal forces equal tozero:

Horizontal forces

Vertical forces

P H

P V W

cos ( )

sin ( )

q

q

- =

+ - =

0 1

0 2

Using a rotational axis perpendicular to the plane of the paper and passing through the pin andremembering that counterclockwise torques are positive, we also set the sum of the torques equal tozero. In doing so, we use L to denote the length of the beam and note that the lever arms for Wand V are L/2 and L, respectively. The forces P and H create no torques relative to this axis,because their lines of action pass directly through it.

Torques W L VL12 0 3c h- = ( )Since L can be eliminated algebraically, Equation (3) may be solved immediately for V:

V W= = =1212 340 170 N Nb g

Substituting this result into Equation (2) gives

P W W

PW

sin

sin sin

q

q

+ - =

= =

=

12 0

2340

2 39270

N N

Substituting this result into Equation (1) gives

WH

HW

20

2340

39210

sincos

tan tan

qq

q

FHG

IKJ - =

= =

= N2

N

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76. CONCEPT QUESTIONS a. Equation 8.7 from the equations of rotational kinematics indicatesthat the angular displacement q is q w a= +0

12

2t t , where w0 is the initial angular velocity, t is the


time, and a is the angular acceleration. Since both wheels start from rest, w0 = 0 rad/s for each.Furthermore, each wheel makes the same number of revolutions in the same time, so q and t arealso the same for each. Therefore, the angular acceleration a must be the same for each.

b. For the hoop, all of the mass is located at a distance from the axis equal to the radius. For thedisk, some of the mass is located closer to the axis. According to Equation 9.6, the moment ofinertia is greater when the mass is located farther from the axis. Therefore, the moment of inertia ofthe hoop is greater. Table 9.1 indicates that the moment of inertia of a hoop is I MR

hoop= 2 , while

the moment of inertia of a disk is I MRdisk =12

2 .

c. Newtons second law for rotational motion indicates that the net external torque is equal to themoment of inertia times the angular acceleration. Both disks have the same angular acceleration, butthe moment of inertia of the hoop is greater. Thus, the net external torque must be greater for thehoop.

SOLUTION Using Equation 8.7, we can express the angular acceleration as follows:

q w a aq w

= + =-

012

2 02

2t t

t

t or

c h

This expression for the acceleration can now be used in Newtons second law for rotational motion.Accordingly, the net external torque is

Hoop

kg m rad rad / s s

s N m

hoopSt aq w

= =-L

NMM

OQPP

=-L

NMM

OQPP=

I MRt

t2 0

2

2

2

2

4 0 0 352 13 2 0 8 0

8 00 20

c h

b gb g b g b gb gb g. .

.

..

Disk

kg m rad rad / s s

s N m

diskSt aq w

= =-L

NMM

OQPP

=-L

NMM

OQPP=

I MRt

t12

2 0

2

12

2

2

2

4 0 0 352 13 2 0 8 0

8 00 10

c h

b gb g b g b gb gb g. .

.

..

_____________________________________________________________________________________________

77. CONCEPT QUESTIONS a. According to Equation 9.6, the moment of inertia for rod A is justthat of the attached particle, since the rod itself is massless. For rod A with its attached particle,then, the moment of inertia is I MLA =

2 . According to Table 9.1, the moment of inertia for rod B

is I MLB =13

2 . The moment of inertia for rod A with its attached particle is greater.


b. Since the moment of inertia for rod A is greater, its kinetic energy is also greater, according toEquation 9.9 KE R =

12

2Iw .

SOLUTION Using Equation 9.9 to calculate the kinetic energy, we find

Rod A KE

kg m rad / s J

R A= =

= =

12

2 12

2 2

12

2 20 66 0 75 4 2 3 3

I MLw wc hb gb gb g. . . .

Rod B KE

kg m rad / s .1 J

R B= =

= =

12

2 12

13

2 2

16

2 20 66 0 75 4 2 1

I MLw wc hb gb gb g. . .

_____________________________________________________________________________________________

78. CONCEPT QUESTIONS a. The force is applied to the person in the counterclockwisedirection by the bar that he grabs when climbing aboard.

b. According to Newtons action-reaction law, the person must apply a force of equal magnitudeand opposite direction to the bar and the carousel. This force acts on the carousel in the clockwisedirection and, therefore, creates a clockwise torque.

c. Since the carousel is moving counterclockwise, the clockwise torque applied to it when theperson climbs aboard decreases its angular speed.

SOLUTION We consider a system consisting of the person and the carousel. Since the carouselrotates on frictionless bearings, no net external torque acts on this system. Therefore, angularmomentum is conserved, and we can set the total angular momentum of the system before theperson hops on equal to the total angular momentum afterwards. Afterwards, the angularmomentum includes a contribution from the person. According to Equation 9.6, his moment ofinertia is I MR

person= 2 , since he is at the outer edge of the carousel.

I I I

I

I I

I

I MR

carousel f person f

Final totalangular mo mentum

carousel

Initial totalangular momentum

fcarousel

carousel person

carousel

carousel

2

2

kg m rad / s

kg m kg m rad / s

w w w

ww w

+ =

=+

=+

=

+=

1 2444 3444 1 24 340

0 02

2

125 3 14

125 40 0 1 501 83

c hb gb gb g

.

. ..


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rotational dynamics

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