dynamics ii: otion in a lane - santa rosa junior...

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8-1

Conceptual Questions

8.1. For an object in uniform circular motion the speed is constant (that’s what uniform means), the angular velocity is constant, and the magnitude of the net force is constant. The velocity and centripetal acceleration are both vectors whose magnitude is constant but whose direction changes.

8.2. The free-body diagram (a) is correct. The forces acting on the car at the bottom of the hill are the downward gravitational force and an upward normal force. The car can be considered to be in circular motion about a point above the bottom center of the valley, which requires a net force toward the center of the circle. In this case, the circle center is above the car, so the normal force is greater than the gravitational force.

8.3. c a d bT T T T> = > . Use 2mvT

r= . For (a),

2

amvT

r= . For (b),

2

b a1

2 2mvT T

r= = . For (c),

2

c a(2 ) 2m vT T

r= = .

For (d), 2

d a(2 )

2m vT T

r= = .

8.4. The tension in the vine at the lowest part of Tarzan’s swing is greater than the gravitational force on Tarzan. If Tarzan is at rest on the vine, just hanging, the tension in the vine is equal to the gravitational force on Tarzan. But when Tarzan is swinging he is in circular motion, with the center of the circle at the top end of the vine, and the vine must provide the additional centripetal force necessary to move him in a circle.

8.5. (a) The difference in the tension between A and B is due to the centripetal force 2

net( )rmvF

r= . Since the

velocity v is the same for both, the greater radius for B means that the tension in case A is greater than for case B. (b) In this case, we use 2( )r netF m rω= . The angular velocity ω is the same for both A and B, so the larger radius for B means that the tension is case A is less than the tension for case B.

8.6. Neither Sally nor Raymond is completely correct. Both have partially correct descriptions, but are missing key points. In order to speed up, there must be a nonzero acceleration parallel to the track. In order to move in a circle, there must be a nonzero centripetal acceleration. Since both of these are required, the net force points somewhere between the forward direction (parallel to the track) and the center of the circle.

8.7. (a) The plane is in dynamic equilibrium, so the net force on the plane is zero. (b) The vertical forces cancel, and so do the horizontal forces, so the net force is zero. The plane is traveling in the positive x direction.

DYNAMICS II: MOTION IN A PLANE

8

8-2 Chapter 8


(c) As seen from behind, with the velocity and positive x direction into the page.

(d) The net force must be toward the center of the circle during a turn at constant speed and altitude. Note that the radial component of the lift force provides the centripetal force while the vertical component balances the gravitational force. The velocity is into the page.

8.8. Yes, the bug is weightless because it, like the projectile it is riding in, is in free fall around the planet.

8.9. When the gravitational force on the ball is greater than the required centripetal force 2

,mvr

the ball is no longer

in circular motion. As the figure shows, at the top of the circle the net force on the ball is 2

net G( )rmvF F T

r= + = .

When the string goes slack, 0,T = leaving 2

GmvF

r= . If the velocity is not high enough to make this equality true,

the equation above becomes an inequality, 2

G ,mvFr

> and the ball begins to fall downward since the net force

downward is greater than the centripetal force required for circular motion.

8.10.

The golfer is swinging the club in circular motion. The club is speeding up as it swings. This motion requires a linear acceleration in the direction of motion of the club to speed it up and a centripetal acceleration to maintain circular motion. The vector sum yields a total acceleration pointing approximately toward the golfer’s feet (c), as shown in the figure.

Dynamics II: Motion in a Plane 8-3


Exercises and Problems

Section 8.1 Dynamics in Two Dimensions

8.1. Model: The model rocket and the target will be treated as particles. The kinematics equations in two dimensions apply. Visualize:

Solve: For the rocket, Newton’s second law along the y-direction is

net R R

2 2R R

( )

1 1( ) [15 N (0 8 kg)(9 8 m/s )] 8 95 m/s0 8 kg

yF F mg ma

a F mgm

= − =

⇒ = − = − . . = ..

Using the kinematic equation 211R 0R 0R 1R 0R R 1R 0R2( ) ( ) ( ) ,yy y v t t a t t= + − + −

2 211R230 m 0 m 0 m (8 95 m/s )( 0 s)t= + + . − 1R 2 589 st⇒ = .

For the target 1T 1R(noting ),t t=

211T 0T 0T 1T 0T T 1T 0T2( ) ( ) ( ) 0 m (15 m/s)(2.589 s s) 0 m 39 mxx x v t t a t t= + − + − = + − 0 + =

You should launch when the target is 39 m away. Assess: The rocket is to be fired when the target is at 0Tx . For a net acceleration of approximately 29 m/s in the vertical direction and a time of 2.6 s to cover a vertical distance of 30 m, a horizontal distance of 39 m is reasonable.

8.2. Model: The model rocket will be treated as a particle. Kinematic equations in two dimensions apply. Air resistance is neglected.

8-4 Chapter 8


Visualize:

The horizontal velocity of the rocket is equal to the speed of the car, which is 3.0 m/s. Solve: For the rocket, Newton’s second law along the y-direction is:

net R R( )yF F mg ma= − = 21 [(8 0 N) (0 5 kg)(9 8 m/s )]0 5 kgya⇒ = . − . ..

26 2 m/s= .

Thus using 211 0 0 1 0 1 02( ) ( ) ( ) ,y yy y v t t a t t= + − + −

2 211R2(20 m) 0 m 0 m (6 2 m/s )( 0 s)t= + + . − 2 2

1(20 m) (3 1 m/s )t⇒ = . 1 2 54 st⇒ = .

Since 1t is also the time for the rocket to move horizontally up to the hoop, 21

1 0 0 1 0 1 02( ) ( ) ( )x xx x v t t a t t= + − + − 0 m (3 0 m/s)(2 54 0 s) 0 m 7 6 ms= + . . − + = .

Assess: In view of the rocket’s horizontal speed of 3.0 m/s and its vertical thrust of 8.0 N, the above-obtained value for the horizontal distance is reasonable.

8.3. Model: The asteroid and the giant rocket will be treated as particles undergoing motion according to the constant-acceleration equations of kinematics. Visualize:

Solve: (a) The time it will take the asteroid to reach the earth is 6

5displacement 4.0 10 km 2.0 10 s 56 hvelocity 20 km/s

×= = × =

(b) The angle of a line that just misses the earth is

1 16

0 0

R R 6400 kmtan tan tan 0.0924.0 10 kmy y

θ θ − − = ⇒ = = = ° ×



(c) When the rocket is fired, the horizontal acceleration of the asteroid is 9

210

5 0 10 N 0 125 m/s4 0 10 kgxa . ×

= = .. ×

(Note that the mass of the rocket is much smaller than the mass of the asteroid and can therefore be ignored completely.) The velocity of the asteroid after the rocket has been fired for 300 s is

20 0( ) 0 m/s (0 125 m/s )(300 s 0 s) 37 5 m/sx x xv v a t t= + − = + . − = .

After 300 s, the vertical velocity is 42 10 m/syv = × and the horizontal velocity is 37 5 m/sxv = . . The deflection due to this horizontal velocity is

14

37 5 m/stan tan 0 1072 10 m/s

x

y

vv

θ θ . = ⇒ = = . ° × 2

That is, the earth is saved.

Section 8.2 Uniform Circular Motion

8.4. Model: We are using the particle model for the car in uniform circular motion on a flat circular track. There must be friction between the tires and the road for the car to move in a circle. Visualize:

Solve: The centripetal acceleration is 2 2

2(25 m/s) 6 25 m/s100 mr

var

= = = .

The acceleration points to the center of the circle, so the net force is 2(1500 kg)(6 25 m/s , toward center)

(9380 N, toward center) (9400 N, toward center)rF ma= = .

= ≈

This force is provided by static friction s 9 4 kNrf F= = .

8-6 Chapter 8


8.5. Model: We will use the particle model for the car which is in uniform circular motion. Visualize:

Solve: The centripetal acceleration of the car is 2 2

2(15 m/s) 4 5 m/s50 mr

var

= = = .

The acceleration is due to the force of static friction. The force of friction is 2s (1500 kg)(4.5 m/s )rf ma= = =

6750 N 6.8 kN.= Assess: The model of static friction is s max s s( ) 15,000 Nf n mg mgµ µ= = ≈ ≈ since s 1µ ≈ for a dry road surface. We see that s s max( ) ,f f< which is reasonable.

8.6. Model: Treat the block as a particle attached to a massless string that is swinging in a circle on a frictionless table. Visualize:

Solve: (a) The angular velocity and speed are rev 2 rad75 471 2 rad/minmin 1 rev

πω = × = .1min(0 50 m)(471 2 rad min) 3 93 m/s60 stv r /ω= = . . × = .

The tangential velocity is 3.9 m/s. (b) The radial component of Newton’s second law is

2

rmvF T

r∑ = =

Thus 2(3 93 m/s)(0 20 kg) 6 2 N

0 50 mT .

= . = ..



8.7. Solve: Newton’s second law is 2r rF ma mrω= = . Substituting into this equation yields:

8

31 118 2 10 N

(9 1 10 kg)(5 3 10 m)rF

mrω . ×

= =. × . ×

2

2 2

16 16 15rad 1 rev4 37 10 rad/s 4 37 10 6 6 10 rev/ss 2 radπ

= . × = . × × = . ×

Assess: This is a very high number of revolutions per second.

8.8. Model: The vehicle is to be treated as a particle in uniform circular motion. Visualize:

On a banked road, the normal force on a vehicle has a horizontal component that provides the necessary centripetal acceleration. The vertical component of the normal force balances the gravitational force. Solve: From the physical representation of the forces in the r-z plane, Newton’s second law can be written

2sinr

mvF nr

θ∑ = = cos 0 coszF n mg n mgθ θ∑ = − = ⇒ =

Dividing the two equations and making the conversion 90 km h 25 m/s/ = yields: 2 2

2(25 m/s)tan 0 128 7 3

(9 8 m/s )500 mvrg

θ θ= = = . ⇒ = . °.

Assess: Such a banking angle for a speed of approximately 55 mph is clearly reasonable and within our experience as well.

8.9. Model: The motion of the moon around the earth will be treated through the particle model. The circular motion is uniform. Visualize:

8-8 Chapter 8


Solve: The tension in the cable provides the centripetal acceleration. Newton’s second law is 2

2

moon

2rF T mr mr

Tπω

∑ = = =

222 8 202 1 day 1 h(7 36 10 kg)(3 84 10 m) 2 01 10 N

27 3 days 24 h 3600 sπ

= . × . × × × = . × .

Assess: This is a tremendous tension, but clearly understandable in view of the moon’s large mass and the large radius of circular motion around the earth. This is the same answer we’ll get later with Newton’s law of universal gravitation.

8.10. Model: Model the person as a particle in uniform circular motion. Visualize:

Solve: The only force acting on the passengers is the normal force of the wall. Newton’s second law along the r-axis is:

2rF n mrω∑ = =

To create “normal” gravity, the normal force by the inside surface of the space station equals mg. Therefore,

22

2 500 m2 2 45 s9 8 m/s

g rmg mr TT r gπω ω π π= ⇒ = = ⇒ = = =

.

Assess: This is a fast rotation. The tangential speed is 2 2 (500 m) 70 m/s 140 mph

45 srv

Tπ π

= = = ≈

Section 8.3 Circular Orbits

8.11. Model: The satellite is considered to be a particle in uniform circular motion around the moon. Visualize:

Solve: The radius of the moon is 61 738 10 m. × and the satellite’s distance from the center of the moon is the same quantity. The angular velocity of the satellite is

42 2 rad 1min 9 52 10 rad/s110 min 60 sT

π πω −= = × = . ×



and the centripetal acceleration is 2 6 4 2 2(1 738 10 m)(9 52 10 rad/s) 1 58 m/sra rω −= = . × . × = .

The acceleration of a body in orbit is the local “g” experienced by that body.

8.12. Model: The earth is considered to be a particle in uniform circular motion around the sun. Solve: The earth orbits the sun in 365 days and is 111 5 10 m. × from the sun. The angular velocity and centripetal acceleration are

72 rad 1 day 1h 2 0 10 rad/s365 days 24 h 3600 s

πω −= × × = . ×

2 11 7 2 3 2(1 5 10 m)(2 0 10 rad/s) 6 0 10 m/sra g rω − −= = = . × . × = . × Assess: The smallness of this acceleration due to gravity is essentially due to the large earth-sun distance.

Section 8.4 Fictitious Forces

8.13. Model: Use the particle model for the car which is undergoing circular motion. Visualize:

Solve: The car is in circular motion with the center of the circle below the car. Newton’s second law at the top of the hill is

2

G( )r r r rmvF F n mg n ma

r∑ = − = − = = 2 nv r g

m ⇒ = −

Maximum speed is reached when 0n = and the car is beginning to lose contact with the road. 2

max (50 m)(9 8 m/s ) 22 m/sv rg= = . = Assess: A speed of 22 m/s is equivalent to 49 mph, which seems like a reasonable value.

8.14. Model: The passengers are particles in circular motion. Visualize:

Solve: The center of the circle of motion of the passengers is directly above them. There must be a net force pointing up that provides the needed centripetal acceleration. The normal force on the passengers is their weight. Ordinarily their weight is G ,F so if their weight increases by 50%, G1 5n F= . . Newton’s second law at the bottom of the dip is

2

G G

2

(1 5 1) 0 5

0 5 0 5(9 8 m/s )(30 m) 12 m/s

rmvF n F F mg

r

v gr

∑ = − = . − = . =

⇒ = . = . . =

Assess: A speed of 12.1 m/s is 27 mph, which seems very reasonable.

8-10 Chapter 8


8.15. Model: Model the roller coaster car as a particle at the top of a circular loop-the-loop undergoing uniform circular motion. Visualize:

Notice that the r-axis points downward, toward the center of the circle. Solve: The critical speed occurs when n goes to zero and GF

provides all the centripetal force pulling the car in the

vertical circle. At the critical speed 2c / ,mg mv r= therefore cv rg= . Since the car’s speed is twice the critical speed,

c2tv v= and the centripetal force is 2 2

cG

(4 ) (4 ) 4rmv m v m rgF n F mg

r r r∑ = + = = = =

Thus the normal force is 3n mg= . Consequently, G 3n/F = .

8.16. Model: Model the roller coaster car as a particle undergoing uniform circular motion along a loop. Visualize:

Notice that the r-axis points downward, toward the center of the circle. Solve: In this problem the normal force is equal to the gravitational force: Gn F mg= = . We have

22

G 2 2(20 m)(9 8 m/s ) 19 8 m/s 20 m/srmvF n F mg mg v rg

r∑ = + = = + ⇒ = = . = . ≈



8.17. Model: Model the bucket of water as a particle in uniform circular motion. Visualize:

Solve: Let us say the distance from the bucket handle to the top of the water in the bucket is 35 cm. This makes the shoulder to water distance 65 cm 35 cm 1 00 m+ = . . The minimum angular velocity for swinging a bucket of water in a vertical circle without spilling any water corresponds to the case when the speed of the bucket is critical. In this case, 0 Nn = when the bucket is in the top position of the circular motion. We get

22c

G c

2

c

0 N

9 8 m/s 1 rev 60 s/ 3 13 rad/s 3 13 rad/s 30 rpm1 00 m 2 rad 1 min

rmvF n F mg mr

r

g r

ω

ωπ

∑ = + = + = =

.⇒ = = = . = . × × =

.

8.18. Model: Use the particle model for yourself while in uniform circular motion. Visualize:

Solve: (a) The speed and acceleration are 2 2

22 2 (15 m) (3 77 m/s)3 77 m/s 0 95 m/s25 s 15 mr

r vv aT rπ π .

= = = . = = = .

So the speed is 3.8 m/s and the centripetal acceleration is 20 95 m/s. . (b) The weight ,w m= the normal force. On the ground, your weight is the same as the gravitational force GF . Newton’s second law at the top is

2

Gr rmvF F n ma

r∑ = − = =

2 22 2(3 77 m/s)9 8 m/s (8 85 m/s )

15 mvn w m g m mr

.⇒ = = − = . − = .

2

2G

8 85 m/s 0 909 8 m/s

wF

.⇒ = = .

.

8-12 Chapter 8


(c) Newton’s second law at the bottom is 2

Gr rmvF n F ma

r∑ = − = =

2 22 2(3 77 m/s )9 8 m/s (10 75 m/s )

15 mvn w m g m mr

.⇒ = = + = . + = .

2

2G

10 75 m/s 1 19 8 m/s

wF

.⇒ = = .

.

Section 8.5 Nonuniform Circular Motion

8.19. Model: Use the particle model for the car, which is undergoing nonuniform circular motion. Visualize:

Solve: The car is in circular motion with radius 100 m2dr = = . We require

2 22 2 11 5 m/s 1 5 m/s1 5 m/s 0 122 s

100 mra rr

ω ω −. .= = . ⇒ = = = .

The definition of the angular velocity can be used to determine the time t∆ using the angular acceleration 2

2 21 5 m/s 1 5 10 s100 m

tar

α − −.= = = . × .

i tω ω α= + ∆

1 1i

20 122 s 0 s 8 2 s

0 015 t

s

ω ωα

− −

−

− . −⇒ ∆ = = = .

.

8.20. Model: The train is a particle undergoing nonuniform circular motion. Visualize:

Solve: (a) Newton’s second law in the vertical direction is net G( ) 0yF n F= − =



from which n mg= . The rolling friction is R R Rf n mgµ µ= = . This force provides the tangential acceleration

RRt

fa gm

µ= − = −

The angular acceleration is 2

2R (0 10)(9 8 m/s ) 1 96 rad/s0 50 m

ta gr r

µα − − . .= = = = − .

.

The magnitude is 22.0 rad/s .

(b) The initial angular velocity is rev 1 min 2 rad30 3 14 rad/smin 60 sec rev

π = . .

The time to come to a stop due to the

rolling friction is f i

20 3 14 rad/s 1 6 s

1 96 rad/st ω ω

α− − .

∆ = = = .− .

Assess: The original angular speed of rad/sπ means the train goes around the track one time every 2 seconds, so a stopping time of less than 2 s is reasonable.

Problems

8.21. Model: The object is treated as a particle in the model of kinetic friction with its motion governed by constant-acceleration kinematics. Visualize:

Solve: The velocity 1xv as the object sails off the edge is related to the initial velocity 0xv by 2 21 0 1 02 ( )x x xv v a x x= + − .

Using Newton’s second law to determine xa while sliding gives

k 0 Nx x yF f ma F n mg n mg∑ = − = ⇒ ∑ = − = ⇒ =

Using this result and the model of kinetic friction k k( ),f nµ= the x-component equation can be written as

k xmg maµ− = . This implies 2 2

k (0 50)(9 8 m/s ) 4 9 m/sxa gµ= − = − . . = − . Kinematic equations for the object’s free fall can be used to determine 1 :xv

2 212 1 1 2 1 2 1 2 1 2 12( ) ( )( ) 0 m 1 0 m 0 m ( ) ( ) 0 4518 s

2ygy y v t t g t t t t t t= + − + − − ⇒ = . + − − ⇒ − = .

2 1 1 2 1 1 1( ) 2 30 m 2 0 m (0 4518 s) 0 664 m/sx x xx x v t t v v= + − = . = . + . ⇒ = .

Having determined 1xv and ,xa we can go back to the velocity equation 2 21 0 1 02 ( ):x x xv v a x x= + −

2 2 20 0(0 664 m/s) 2( 4 9 m/s )(2 0 m) 4 5 m/sx xv v. = + − . . ⇒ = .

Assess: 0 4 5 m/sxv = . is about 10 mph and is a reasonable speed.

8-14 Chapter 8


8.22. Model: Treat the motorcycle and rider as a particle. Visualize: This is a two-part problem. Use an s-axis parallel to the slope for the first part, regular xy-coordinates for the second. The motorcycle’s final velocity at the top of the ramp is its initial velocity as it becomes airborne.

Solve: The motorcycle’s acceleration on the ramp is given by Newton’s second law: net r r r 0

2 20 r

( ) sin20 sin20 cos20 sin20

( cos20 sin20 ) (9.8 m/s )((0.02)cos20 sin20 ) 3.536 m/ssF f mg n mg mg mg ma

a g

µ µ

µ

= − − ° = − − ° = − ° − ° =

= − ° + ° = − ° + ° = −

The length of the ramp is 1 (2.0 m)/sin 20 5.85 m.s = ° = We can use kinematics to find its speed at the top of the ramp:

2 2 21 0 0 1 0 0 0 1

2 21

2 ( ) 2

(11.0 m/s) 2( 3.536 m/s )(5.85 m) 8.92 m/s

v v a s s v a s

v

= + − = +

⇒ = + − =

This is the motorcycle’s initial speed into the air, with velocity components 1 1cos20 8.38 m/sxv v= ° = and

1 1sin20 3.05 m/s.yv v= ° = We can use the y-equation of projectile motion to find the time in the air: 2 2 21

2 1 1 2 1 2 2 220 m 2 0 m (3 05 m/s) (4 90 m/s )y yy y v t a t t t= = + + = . + . − .

This quadratic equation has roots 2 0 399 st = − . (unphysical) and 2 1 021 st = . . The x-equation of motion is thus

2 1 1 2 20 m (8 38 m/s) 8 56 mxx x v t t= + = + . = . 8 56 m 10 0 m,. < . so it looks like crocodile food.

8.23. Model: Treat Sam as a particle. Visualize: This is a two-part problem. Use an s-axis parallel to the slope for the first part, regular xy-coordinates for the second. Sam’s final velocity at the top of the slope is his initial velocity as he becomes airborne.



Solve: Sam’s acceleration up the slope is given by Newton’s second law: net 0

2 20

( ) sin10200 Nsin10 (9 8 m/s )sin10 0 965 m/s75 kg

sF F mg maFa gm

= − ° =

= − ° = − . ° = .

The length of the slope is 1 (50 m)/ sin10 288 ms = ° = . His velocity at the top of the slope is 2 2 21 0 0 1 0 0 1 12 ( ) 2 2(0 965 m/s )(288 m) 23 6 m/sv v a s s a s v= + − = ⇒ = . = .

This is Sam’s initial speed into the air, giving him velocity components 1 1 cos10 23 2 m/sxv v= ° = . and

1 1sin10 410 m/syv v= ° = . This is not projectile motion because Sam experiences both the force of gravity and the thrust of his skis. Newton’s second law for Sam’s acceleration is

2net1

2net 2

1

( ) (200 N)cos10 2 63 m/s75 kg

( ) (200 N)sin10 (75 kg)(9 80 m/s ) 9 34 m/s75 kg

xx

yy

Fam

Fa

m

°= = = .

° − .= = = − .

The y-equation of motion allows us to find out how long it takes Sam to reach the ground: 2 2 21

2 1 1 2 1 2 2 220 m 50 m (4 10 m/s) (4 67 m/s )y yy y v t a t t t= = + + = + . − .

This quadratic equation has roots 2 2 86 st = − . (unphysical) and 2 3 74 st = . . The x-equation of motion—this time with an acceleration—is

2 2 21 12 1 1 2 1 2 2 22 20 m (23 2 m/s) (2 63 m/s ) 105 mx xx x v t a t t t= + + = + . − . =

Sam lands 105 m from the base of the cliff.

8.24. Visualize:

8-16 Chapter 8


Solve: From Chapter 6 the drag on a projectile is 21 , direction opposite to motion ,4

D Av =

where A is the cross-

sectional area. Using the free-body diagram above, apply Newton’s second law to each direction. In the x-direction,

21net 4

21net G 4

cos( ) cos

( ) sinsin

xx

yy

AvF Dam n m

F AvD Fa gm m m

θθ

θθ

= = =

−= = = −

2 2

2 2

Since 2 2 ,x yv v v= + cos ,xv v θ= and sin ,yv v θ= we can rewrite these as

2 2

2 2

( cos )4 m 4 m

( sin )4 m 4 m

x x yx

y x yy

Av v vA v va

Av v vA v va g g

θ

θ

+= =

+= − = − −

2 2

2

8.25. Model: Use the particle model and the constant-acceleration equations of kinematics for the rocket. Solve: (a) The acceleration of the rocket in the launch direction is obtained from Newton’s second law :F ma=

2140,700 N (5000 kg) 28 14 m/sa a= ⇒ = .

Therefore, 2cos44 7 20 0 m/sxa a= . ° = . and

2sin 44 7 19 8 m/sya a= . ° = . . The net acceleration in the y-direction is thus 2 2

net( ) (19 8 9 8) m/s 10 0 m/sy ya a g= − = . − . = . With this acceleration, we can write the equations for the x- and y-motions of the rocket.

2 2 2 2 21 10 0 0 net 02 2( ) ( ) ( ) 0 m 0 m (10 0 m/s ) (5 00 m/s )y yy y v t t a t t t t= + − + − = + + . = .

2 2 2 2 21 10 0 0 net 02 2( ) ( ) ( ) 0 m 0 m (20 0 m/s ) (10 0 m/s )x xx x v t t a t t t t= + − + − = + + . = .

From these two equations, 2 2

2 2(10 0 m/s ) 2(5 00 m/s )

x ty t

.= =

.

The equation that describes the rocket’s trajectory is 12y x= .

(b) It is a straight line with a slope of 12 .

(c) In general, 2

0 net 1 0 1

20 net 1 0 1

( ) ( ) 0 (10.0 m/s )

( ) ( ) 0 (20.0 m/s )

y y y

x x x

v v a t t t

v v a t t t

= + − = +

= + − = +

2 2 2 2 2 2 21 1 1(10 0 m/s ) (20 0 m/s ) (22 36 m/s )v t t t= . + . = .

The time required to reach the speed of sound is calculated as follows: 2

1 1330 m/s (22 36 m/s ) 14 76 st t= . ⇒ = . We can now obtain the elevation of the rocket. From the y-equation,

2 2 2 21(5 00 m/s ) (5 00 m/s )(14 76 s) 1090 my t= . = . . =

8.26. Model: The hockey puck will be treated as a particle whose motion is determined by constant-acceleration kinematic equations. We break this problem in two parts, the first pertaining to motion on the table and the second to free fall.



Visualize:

Solve: Newton’s second law is:

x xF ma= 22 0 N 2 0 m/s1 0 kg

xx

Fam

.⇒ = = = .

.

The kinematic equation 2 21 0 1 02 ( )x x xv v a x x= + − yields:

2 2 2 21 0 m /s 2(2 0 m/s )(4 0 m)xv = + . . 1 4 0 m/sxv⇒ = .

Let us now find the time of free fall 2 1( ):t t− 21

2 1 1 2 1 1 2 12( ) ( )y yy y v t t a t t= + − + − 2 21

2 120 m 2 0 m 0 m ( 9 8 m/s )( )t t⇒ = . + + − . − 2 1( ) 0 639 st t⇒ − = .

Having obtained 1xv and 2 1( ),t t− we can now find 2 1( )x x− as follows: 21

2 1 1 2 1 2 12( ) ( )x xx x v t t a t t= + − + − 2 21

2 1 2(4 0 m/s)(0 639 s) (2 0 m/s )(0 639 s) 3 0 mx x⇒ − = . . + . . = .

Assess: For a modest horizontal thrust of 2.0 N, a landing distance of 3.0 m is reasonable.

8.27. Model: The model rocket is treated as a particle and its motion is determined by constant-acceleration kinematic equations. Visualize:

Solve: As the rocket is accidentally bumped 0 0 5 m/sxv = . and 0 0 m/syv = . On the other hand, when the engine is fired

220 N 40 m/s0 500 kg

xx x x

FF ma am

= ⇒ = = =.

8-18 Chapter 8


(a) Using 211 0 0 1 0 1 02( ) ( ) ,y yy y v t t a t t= + − + −

2 211 120 m 40 m 0 m ( 9 8 m/s ) 2 857 st t= + + − . ⇒ = .

The distance from the base of the wall is 21

1 0 0 1 0 1 02( ) ( )x xx x v t t a t t= + − + − 2 2120 m (0 5 m/s)(2 857s) (40 m/s )(2 857 s) 165 m= + . . + . =

(b) The x- and y-equations are 2 21

0 0 0 022 21

0 0 0 02

( ) ( ) 40 4 9

( ) ( ) 0 5 20

y y

x x

y y v t t a t t t

x x v t t a t t t t

= + − + − = − .

= + − + − = . +

Except for a brief interval near 20, 20 0 5t t t= . . Thus

220 ,x t≈ or 2 /20t x= . Substituting this into the y-equation gives 40 0 245y x= − .

This is the equation of a straight line, so the rocket follows a linear trajectory to the ground.

8.28. Model: Model the plane as a particle with constant xa and constant .ya Visualize: The plane is taking off toward the northwest as we can see by plotting the -x y data.

But we analyze the data in each direction separately and then apply 2 2 .x ya a a a= = +

Solve: In each direction we apply the kinematic equation 210 0 2( ) ( ) .s ss t s v t a t= + + With 0 0v = we can graph s vs.

2t and expect to get a straight line whose slope is 12 sa and whose intercept is 0.s

From the x vs. 2t graph we see that 2 212 2 90 m/s 5 8 m/s .x xa a= − . ⇒ = − .

From the y vs. 2t graph we see that 2 212 4 00 m/s 8 0 m/s .y ya a= . ⇒ = .



2 2 2 2 2 2 2( 5 8 m/s ) (8 0 m/s ) 9 9 m/sx ya a a a= = + = − . + . = .

Assess: The intercepts of the best-fit lines are close to what the data table has.

8.29. Model: Assume the particle model for the satellite in circular motion. Visualize:

To be in a geosynchronous orbit means rotating at the same rate as the earth, which is 24 hours for one complete rotation. Because the altitude of the satellite is

7 7 7 6 7e3 58 10 m, 3 58 10 m, 3 58 10 m 6 37 10 m 4 22 10 mr r. × = . × = . × + . × = . × .

Solve: (a) The period (T) of the satellite is 24.0 hours. (b) The acceleration due to gravity is

2 22 7 22 2 1 hr(4 22 10 m) 0 223 m/s

24 0 hr 3600 srg a r rTπ πω = = = = . × × = . .

(c) There is no normal force on a satellite, so the weight is zero. It is in free fall.

8.30. Model: Treat the man as a particle. The man at the equator undergoes uniform circular motion as the earth rotates. Visualize:

Solve: The scale reads the man’s weight G ,F n= the force of the scale pushing up against his feet. At the north pole, where the man is in static equilibrium,

P G 735 Nn F mg= = = At the equator, there must be a net force toward the center of the earth to keep the man moving in a circle. The r-axis points toward the center, so

2 2 2G E E PrF F n m r n mg m r n m rω ω ω∑ = − = ⇒ = − = −

The equator scale reads less than the north pole scale by the amount 2m rω . The man’s angular velocity is that of the equator, or

52 2 rad 7 27 10 rad/s24 hours (3600 s/1 h)T

π πω −= = = . ××

8-20 Chapter 8


Thus the north pole scale reads more than the equator scale by 5 2 6(75 kg)(7 27 10 rad/s) (6 37 10 m) 2 5 Nw −∆ = . × . × = .

Assess: The man at the equator appears to have lost / 0 25 kg,m w g∆ = ∆ ≈ . or the equivalent of 12 lb≈ .

8.31. Model: Model the ball as a particle which is in a vertical circular motion. Visualize:

Solve: At the bottom of the circle, 2 2

2G

(0 500 kg)(15 N) (0 500 kg)(9 8 m/s ) 5 5 m/s(1 5 m)r

mv vF T F vr

.∑ = − = ⇒ − . . = ⇒ = .

.

8.32. Model: We will use the particle model for the car, which is undergoing uniform circular motion on a banked highway, and the model of static friction. Visualize:

Note that we need to use the coefficient of static friction s,µ which is 1.0 for rubber on concrete. Solve: Newton’s second law for the car is

2

s cos sinrmvF f n

rθ θ∑ = + = s Gcos sin 0zF n f Fθ θ∑ = − − = N

Maximum speed is when the static friction force reaches its maximum value s max s( )f nµ= . Then 2

s( cos15 sin15 ) mvnr

µ ° + ° = s(cos15 sin15 )n mgµ° − ° =



Dividing these two equations and simplifying, we get 2

s s

s s

tan15 tan151 tan15 1 tan15

v v grgr

µ µµ µ+ ° + °

= ⇒ =− ° − °

2 (1 0 0 268)(9 80 m/s )(70 m) 34 m/s(1 0 268). + .

= . =− .

Assess: The above value of 34 m/s 70 mph≈ is reasonable.

8.33. Model: Use the particle model for the rock, which is undergoing uniform circular motion. Visualize: L is the hypotenuse of the right triangle. The radius of the circular motion is cosr L θ= .

Solve: (a) Apply Newton’s second law in the -z and -directions.r

sin 0sinzmgF T mg Tθ

θ∑ = − = ⇒ =

2 2 2cos ( cos )rF T m r m L T m Lθ ω ω θ ω∑ = = = ⇒ = Set the two expressions for T equal to each other and solve for .ω

2

sin sinmg gm L

Lω ω

θ θ= ⇒ =

(b) Insert 1 0 mL = . and 10 .θ = ° 29 8 m/s 1 rev 60 s7 51 rad/s 72 rpm

sin (1 0 m)sin10 2 rad 1 ming

Lω

θ π .

= = = . = . °

Assess: Notice that the mass canceled out of the equation so the 500 g was unnecessary information. In other words, the answer, 72 rpm, would be the same regardless of the mass. The dependencies of ω on g, L, andθ seem to be in the right directions.

8.34. Model: Use the particle model and static friction model for the coin, which is undergoing circular motion. Visualize:

8-22 Chapter 8


Solve: The force of static friction is s s sf n mgµ µ= = . This force is equivalent to the maximum centripetal force that can be applied without sliding. That is,

2 22 s

s max max(0 80)(9 8 m/s )( ) 7 23 rad/s

0 15 mtv gmg m m rr r

µµ ω ω . .= = ⇒ = = = .

.

rad 1 rev 60 s7 23 69 rpms 2 rad 1 minπ

= . × × =

So, the coin will stay still on the turntable. Assess: A rotational speed of approximately 1 rev per second for the coin to stay stationary seems reasonable.

8.35. Model: Use the particle model for the car, which is in uniform circular motion. Visualize:

Solve: Newton’s second law is 2

sin 20r rmvF T ma

r∑ = ° = = Gcos20 0zF T F∑ = ° − = N

These equations can be written as 2

sin 20 mvTr

° = cos20T mg° =

Dividing these two equations gives 2 2tan 20 / tan 20 (4 55 m)(9 8 m/s ) tan 20 4 03 m/s 4 m/sv rg v rg° = ⇒ = ° = . . ° = . ≈

8.36. Use the particle model for the ball, which is undergoing uniform circular motion. Visualize: We are given L, r, and m, so our answers must be in terms of those variables. L is the hypotenuse of the right triangle. The ball moves in a horizontal circle of radius cos .r L θ= The acceleration and net force point toward the center of the circle, not along the string.



Solve: (a) Apply Newton’s second law in the -direction.z

cos 0coszmgF T mg Tθ

θ∑ = − = ⇒ =

From the right triangle 2 2cos / .L r Lθ = −

2 2cosmg mgLT

L rθ= =

−

(b) Apply Newton’s second law in the -direction.r 2 2 2sin ( sin ) .rF T m r m L T m Lθ ω ω θ ω∑ = = = ⇒ =

Set the two expressions for T equal to each other, cancel m and one L, and solve for .ω

22 2 2 2

mgL gm LL r L r

ω ω= ⇒ =− −

(c) Insert 1 0 m,L = . 0 20 mr = . and 0 50 kg.m = . 2

2 2 2 2

2

2 2 2 2

(0 50 kg)(9 8 m/s )(1 0 m) 5 0 N(1 0 m) (0 20 m)

9 8 m/s 1 rev 60 s3 163 rad/s 30 rpm2 rad 1 min(1 0 m) (0 20 m)

mgLTL r

g

L rω

π

. . .= = = .

− . − .

.= = = . =

− . − .

Assess: Notice that the mass canceled out of the equation for ,ω but not for T, so the 500 g was necessary information.

8.37. Model: Assume the particle model for a sphere in circular motion at constant speed. Visualize:

Solve: (a) Newton’s second law along the r and z axes is: 2

1 2sin30 sin 60 tr

mvF T Tr

∑ = ° + ° = 1 2 Gcos30 cos60 0 NzF T T F∑ = ° + ° − =

Since we want 1 2 ,T T T= = these two equations become 2

(sin30 sin 60 ) tmvTr

° + ° = (cos30 cos60 )T mg° + ° =

8-24 Chapter 8


Since sin30 sin 60 cos30 cos60 ,° + ° = ° + ° 2t

tmvmg v rg

r= ⇒ =

The triangle with sides 1 2, ,L L and 1.0 m is isosceles, so 2 1 0 mL = . and 2 cos30r L= °. Thus 2

2 cos 30 (1 0 m)cos 30 (0 866 m)(9 8 m/s ) 2 9 m/sL g g ° = . ° = . . = . (b) The tension is

2(2 0 kg)(9 8 m/s ) 14 3 N 14 Ncos30 cos60 0 866 0 5

mgT . .= = = . ≈

° + ° . + .

8.38. Model: Consider the passenger to be a particle and use the model of static friction. Visualize:

Solve: The passengers stick to the wall if the static friction force is sufficient to support the gravitational force on them: s G.f F= The minimum angular velocity occurs when static friction reaches its maximum possible value

s max s( )f nµ= . Although clothing has a range of coefficients of friction, it is the clothing with the smallest coefficient

s( 0 60)µ = . that will slip first, so this is the case we need to examine. Assuming that the person is stuck to the wall, Newton’s second law is

2s s0r zF n m r F f w f mgω∑ = = ∑ = − = ⇒ =

The minimum frequency occurs when 2

s s max s s min( )f f n mrµ µ ω= = = Using this expression for sf in the z-equation gives

2s s min

2

mins

9 80 m/s 1 rev 60 s2 56 rad/s 2 56 rad/s 24 rpm0 60(2 5 m) 2 rad 1 min

f mr mg

gr

µ ω

ωµ π

= =

.⇒ = = = . = . × × =

. .

Assess: Note the velocity does not depend on the mass of the individual. Therefore, the minimum mass sign is not necessary.

8.39. Model: Use the particle model for the marble in uniform circular motion. Visualize:



Solve: The marble will roll in a horizontal circle if the static friction force is sufficient to support the gravitational on it: s G.f F= If s max( )mg f> then static friction is not sufficient and the marble will slip down the side as it rolls around the circumference. The r-equation of Newton’s second law is

22 2 rad 1 min(0 010 kg)(0 060 m) 150 rpm 0 148 N

1 rev 60 srF n mr πω ∑ = = = . . × × = .

Thus the maximum possible static friction is s max s( ) (0 80)(0 148 N) 0 118 Nf nµ= = . . = . . The friction force needed to support a 10 g marble is s 0 098 Nf mg= = . . We see that s s max( ) ,f f< therefore friction is sufficient and the marble spins in a horizontal circle. Assess: In reality, rolling friction will cause the marble to gradually slow down until s max( )f mg< . At that point, it will begin to slip down the inside wall.

8.40. Model: Assume uniform circular motion. Visualize: We expect the centripetal acceleration to be very large because ω is large. This will produce a significant force even though the mass difference of 10 mg is so small. A preliminary calculation will convert the mass difference to kg: 510 mg 1.0 10 kg.−= × If the two samples are equally balanced then the shaft doesn’t feel a net force in the horizontal plane. However, the mass difference of 10 mg is what causes the force. We’ll do another preliminary calculation to convert 70,000 rpm into rad/sω = .

rev 2 rad 1 min70,000 rpm 70,000 7330 rad/smin 1 rev 60 s

πω = = =

Solve: The centripetal acceleration is given by Equation 6.9 and the net force by Newton’s second law. 2 5 2

net ( )( ) ( )( ) (1 0 10 kg)(7330 rad/s) (0 12 m) 64 NF m a m rω −= ∆ = ∆ = . × . = Assess: As we expected, the centripetal acceleration is large. The force is not huge (because of the small mass difference) but still enough to worry about. The net force scales with this mass difference, so if the mistake were bigger it could be enough to shear off the shaft.

8.41. Model: Use the particle model for the car and the model of kinetic friction. Visualize:

8-26 Chapter 8


Solve: We will apply Newton’s second law to all three cars. Car A:

k G k( ) ( ) 0 N 0 Nx x x x xF n f F f ma∑ = + + = − + =

k( ) 0 N 0 Ny y y yF n f y n mg∑ = + + = + − =

The y-component equation means n mg= . Since k k ,f nµ= we have k kf mgµ= . From the x-component equation,

2k kkg 9 8 m/sx

f mgam m

µ µ− −= = = − = − .

Car B: Car B is in circular motion with the center of the circle above the car. 2

k G( ) ( ) 0 Nr r r r rmvF n f F n mg ma

r∑ = + + = + − = =

k G k( ) ( ) 0 N 0 Nt t t t tF n f F f ma∑ = + + = − + = + From the r-equation

2 2

k k kmv vn mg f n m g

r rµ µ

= + ⇒ = = +

Substituting back into the t-equation, 2 2

2 2k kk

(25 m/s)9 8 m/s 12 9 m/s200 mt

f m va gm m r

µ µ

= − = − + = − . + = − .

Car C: Car C is in circular motion with the center of the circle below the car. 2

k G( ) ( ) 0 Nr r r r rmvF n f F n mg ma

r∑ = + + = − + + = =

k G k( ) ( ) 0 N 0 Nt t t t tF n f F f ma∑ = + + = − + =

From the r-equation 2( / )n m g v r= − . Substituting this into the t-equation yields

2 2k kk ( / ) 6 7 m/st

f na g v rm m

µ µ− −= = = − − = − .

8.42. Model: Model the ball as a particle that is moving in a vertical circle. Visualize:

Solve: (a) The ball’s gravitational force 2G (0 500 kg)(9 8 m/s ) 4 9 NF mg= = . . = . .

(b) Newton’s second law at the top is 2

1 Gr rvF T F ma mr

∑ = + = =

( )222

14 0 m/s

(0 500 kg) 9 8 m/s1 02 m

vT m gr

. ⇒ = − = . − . .

2 9 N= .



(c) Newton’s second law at the bottom is 2

2 GrmvF T F

r∑ = − =

2 22

2(7 5 m/s)(0 500 kg) 9 8 m/s 32 N

1 02 mvT m gr

.⇒ = + = . . + = .

8.43. Model: Model a passenger as a particle rotating in a vertical circle. Visualize:

Solve: (a) Newton’s second law at the top is 2

T Gr rmvF n F ma

r∑ = + = =

2

Tmvn mg

r⇒ + =

The speed is 2 2 (8 0 m) 11 17 m/s

4 5 srv

Tπ π .

= = = ..

2 22

T(11 17 m/s)(55 kg) 9 8 m/s 319 N

8 0 mvn m gr

.⇒ = − = − . = .

That is, the ring pushes on the passenger with a force of 23 2 10 N. × at the top of the ride. Newton’s second law at the bottom:

2 2 2

B G B

22(11 17 m/s)(55 kg) 9 8 m/s 1397 N

8 0 m

r rmv mv vF n F ma n mg m g

r r r

∑ = − = = ⇒ = + = +

.= + . =

.

Thus the force with which the ring pushes on the rider when she is at the bottom of the ring is 1.4 kN. (b) To just stay on at the top, T 0 Nn = in the r-equation at the top in part (a). Thus,

222

max

2mvmg mr mrr T

πω

= = =

max 28 0 m2 2 5 7 s

9 8 m/srTg

π π .⇒ = = = .

.

8.44. Model: Model the chair and the rider as a particle in uniform circular motion. Visualize:

8-28 Chapter 8


Solve: Newton’s second law along the r-axis is

G( )r r r rF T F ma∑ = + = 2sin 0 NT mrθ ω⇒ + = Since sin ,r L θ= this equation becomes

22 2 rad(150 kg)(9 0 m) 3330 N

4 0 sT mL πω = = . = .

Thus, the 3000 N chain is not strong enough for the ride.

8.45. Model: Model the ball as a particle in motion in a vertical circle. Visualize:

Solve: (a) If the ball moves in a complete circle, then there is a tension force T

when the ball is at the top of the circle. The tension force adds to the gravitational force to cause the centripetal acceleration. The forces are along the r-axis, and the center of the circle is below the ball. Newton’s second law at the top is

2

net G( )rmvF T F T mg

L= + = + =

topLTv Lgm

⇒ = +

The tension T can’t become negative, so 0 NT = gives the minimum speed minv at which the ball moves in a circle. If the speed is less than min ,v then the string will go slack and the ball will fall out of the circle before it reaches the top. Thus,

minmin min

Lgv gv LgL L L

ω= ⇒ = = =



(b) Insert L = 1.0 m. 2

min(9 8 m/s ) 3 13 rad/s 30 rpm

(1 0 m)gr

ω .= = = . =

.

Assess: Notice that the mass doesn’t appear in the answer, so minω is independent of the mass.

8.46. Model: The ball is a particle on a massless rope in circular motion about the point where the rope is attached to the ceiling. Visualize:

Solve: Newton’s second law in the radial direction is 2

G( )rmvF T F T mg

r∑ = − = − =

Solving for the tension in the rope and evaluating,

( )222 5 5 m/s

(10 2 kg) 9 8 m/s 168 N4 5 m

vT m gr

. = + = . . + = .

Assess: The tension in the rope is greater than the gravitational force on the ball in order to keep the ball moving in a circle.

8.47. Model: Model the ball as a particle in uniform circular motion. Rolling friction is ignored. Visualize:

Solve: The track exerts both an upward normal force and an inward normal force. From Newton’s second law, 2

1 (0 030 kg)(9 8 m/s ) 0 294 N, upn mg= = . . = .

22n mrω=

260 rev 2 rad 1 min(0 030 kg)(0 20 m) 0 2369 N, inmin 1 rev 60 s

π = . . × × = .

2 2 2 2net 1 2 (0 294 N) (0 2369 N) 0 38 NF n n= + = . + . = .

8.48. Model: Masses 1m and 2m are considered particles. The string is assumed to be massless. Visualize:

8-30 Chapter 8


Solve: The tension in the string causes the centripetal acceleration of the circular motion. If the hole is smooth, it acts like a pulley. Thus tension forces 1T

and 2T

act as if they were an action/reaction pair. Mass 1m is in circular motion of radius r, so Newton’s second law for 1m is

21

1rm vF T

r∑ = =

Mass 2m is at rest, so the y-equation of Newton’s second law is

2 2 2 20 NyF T m g T m g∑ = − = ⇒ =

Newton’s third law tells us that 1 2.T T= Equating the two expressions for these quantities: 2

1 22

1

m v m rgm g vr m

= ⇒ =

8.49. Model: Model yourself as a particle in circular motion atop a leg of length .L Visualize: Set up the coordinate system so that the -directionr is down, toward the center of the circle.



Solve: (a) Apply Newton’s second law in the downward vertical direction, which is the -direction.r+

2

rmvF mg n

L∑ = − =

Notice that .n mg< Your body tries to “lift off” as it pivots over your foot, decreasing the normal force exerted on you by the ground. The normal force becomes smaller as you walk faster, but n cannot be less than zero. Thus the maximum possible walking speed maxv occurs when 0.n = Setting 0,n =

2max

maxmvmg v gL

L= ⇒ =

(b) Insert 0 70 m.L = . 2

max (9 8 m/s )(0 70 m) 2 6 m/s 5 9 mphv gL= = . . = . = . Assess: The answer of 5.9 mph is faster than the “normal” walking speed of 3 mph, but we expect maxv to be greater than the normal speed. This seems reasonable. The units check out. The maximum walking speed depends on ,L so taller people can walk faster than shorter people.

8.50. Model: Model the ball as a particle swinging in a vertical circle, then as a projectile. Visualize:

Solve: Initially, the ball is moving in a circle. Once the string is cut, it becomes a projectile. The final circular-motion velocity is the initial velocity for the projectile. The free-body diagram for circular motion is shown at the bottom of the circle. Since G,T F> there is a net force toward the center of the circle that causes the centripetal acceleration. The r-equation of Newton’s second law is

2

net G( )rmvF T F T mg

r= − = − =

2bottom

0 60 m( ) [5 0 N (0 10 kg)(9 8 m/s )] 4 91 m/s0 100 kg

rv T mgm

.⇒ = − = . − . . = .

.

As a projectile the ball starts at 0 1 4 my = . with 0 ˆ 4 91 m/sv i= . . The equation for the y-motion is

2 21 11 0 0 0 12 20 m ( )yy y v t g t y gt= = + ∆ − ∆ = −

This is easily solved to find that the ball hits the ground at time

01

2 0 535 sytg

= = .

During this time interval it travels a horizontal distance 1 0 0 1 (4 91 m/s)(0 535 s) 2 63 mxx x v t= + = . . = .

So the ball hits the floor 2.6 m to the right of the point where the string was cut.

8-32 Chapter 8


8.51. Model: Use the particle model for a ball in motion in a vertical circle and then as a projectile. Visualize:

Solve: For the circular motion, Newton’s second law along the r-direction is 2

Gt

rmvF T F

r∑ = + =

Since the string goes slack as the particle makes it over the top, 0 NT = . That is, 2

2G (9 8 m/s )(0 5 m) 2 21 m/st

tmvF mg v gr

r= = ⇒ = = . . = .

The ball begins projectile motion as the string is released. The time it takes for the ball to hit the floor can be found as follows: 2 2 21 1

1 0 0 1 0 1 0 1 12 2( ) ( ) 0 m 2 0 m 0 m ( 9 8 m/s )( 0 s) 0 639 sy yy y v t t a t t t t= + − + − ⇒ = . + + − . − ⇒ = .

The place where the ball hits the ground is 1 0 0 1 0( ) 0 m ( 2 21 m/s)(0 639 s 0 s) 1 41 mxx x v t t= + − = + + . . − = + .

The ball hits the ground 1.4 m to the right of the point beneath the center of the circle.

8.52. Model: Model yourself as a particle in circular motion. Visualize: Apply Newton’s second law for circular motion in the vertical direction. The upward normal force is the scale reading. We are given 588 N.mg = This means 60 kg.m = We seek r. Solve:

2mvF n mgr

∑ = − =

We can easily get the spreadsheet to subtract 588 Nmg = from each of the scale readings on the left side of the

equation. Then if we graph n mg− vs. 2v we should get a straight line whose slope is / .m r



We see from the spreadsheet that the fit is very good and that the slope is / 0 3999 kg/m.m r = . With 60 kg,m =

60 kg0 3999 kg/m 150 m0 3999 kg/m

m rr

= . ⇒ = =.

Assess: This radius of curvature does not seem extreme.

8.53. Model: Model the ball as a particle undergoing circular motion in a vertical circle. Visualize:

Solve: Initially, the ball is moving in circular motion. Once the string breaks, it becomes a projectile. The final circular-motion velocity is the initial velocity for the projectile, which we can find by using the kinematic equation

2 2 2 2 2 21 0 1 0 0 02 ( ) 0 m /s ( ) 2( 9 8 m/s )(4 0 m 0 m) 8 85 m/syv v a y y v v= + − ⇒ = + − . . − ⇒ = .

This is the speed of the ball as the string broke. The tension in the string at that instant can be found by using the r-component of the net force on the ball:

rF∑ =20 yv

T mr

= ⇒

2(8 85 m/s)(0 100 kg) 13 N0 60 m

T .= . =

.

8.54. Model: Model the car as a particle on a circular track. Visualize:

8-34 Chapter 8


Solve: (a) Newton’s second law along the t-axis is

t t tF F ma∑ = = 21000 N (1500 kg) 2/3 m/st ta a⇒ = ⇒ = With this tangential acceleration, the car’s tangential velocity after 10 s will be

21 0 1 0( ) 0 m/s (2/3 m/s )(10 s 0 s) 20/3 m/st t tv v a t t= + − = + − =

The radial acceleration at this instant is 2 2

21 (20/3 m/s) 16 m/s25 m 9

tr

var

= = =

The car’s acceleration at 10 s has magnitude

2 2 2 2 2 2 21 (2/3 m/s ) (16/9 m/s ) 1 90 m/st ra a a= + = + = . 1 1 2/3tan tan 21

16/9t

r

aa

θ − − = = = °

where the angle is measured from the r-axis. (b) The car will begin to slide out of the circle when the static friction reaches its maximum possible value

s max s( )f nµ= . That is, 22

s max s s( ) tr

mvF f n mgr

µ µ∑ = = = = 22 (25 m)(9 8 m/s ) 15 7 m/stv rg⇒ = = . = .

In the above equation, n mg= follows from Newton’s second law along the z-axis. The time when the car begins to slide can now be obtained as follows:

22 0 2 0 2 2( ) 15.7 m/s m/s (2/3 m/s )( 0) 24 st t tv v a t t t t= + − ⇒ = 0 + − ⇒ =

8.55. Model: Model the steel block as a particle and use the model of kinetic friction. Visualize:

Solve: (a) The components of thrust ( )F

along the r-, t-, and z-directions are

sin20 (3.5 N)sin20 1.20 N cos20 (3.5 N)cos20 3.29 N 0 Nr t zF F F F F= ° = ° = = ° = ° = = Newton’s second law is

2net( )r rF T F mrω= + = net k( )t t tF F f ma= − =

net( ) 0 NzF n mg= − = The z-component equation means n mg= . The force of friction is

2k k k (0 60)(0 500 kg)(9 8 m/s ) 2 94 Nf n mgµ µ= = = . . . = .

Substituting into the t-component of Newton’s second law 2(3 29 N) (2 94 N) (0 500 kg) 0 70 m/st ta a. − . = . ⇒ = .



Having found ,ta we can now find the tangential velocity after 10 revolutions 20π= rad as follows:

2 11 1 1

1 0 1

1 2 18 95 s2

6 63 rad/s

t

t

t

a rt tr a

a tr

θθ

ω ω

= ⇒ = = .

= + = .

The block’s angular velocity after 10 rev is 6.6 rad/s. (b) Substituting 1ω into the r-component of Newton’s second law yields:

21 1rT F mrω+ = 2

1 1(1 20 N) (0 500 kg)(2 0 m)(6 63 rad/s) 44 NT T⇒ + . = . . . ⇒ =

8.56. Model: Assume the particle model for a ball in vertical circular motion. Visualize:

Solve: (a) Newton’s second law in the r- and t-directions is 2

net( ) cos tr r

mvF T mg mar

θ= + = = net( ) sint tF mg maθ= − =

Substituting into the r-component, 2

2(20 N) (2 0 kg)(9 8 m/s )cos30 (2 0 kg)(0 80 m)

tv+ . . ° = .

.3 85 m/stv⇒ = .

The tangential velocity is 3.8 m/s. (b) Substituting into the t-component,

2(9 8 m/s )sin30 ta− . ° = 24 9 m/sta⇒ = − . The radial acceleration is

2 22(3 85 m/s) 18 5 m/s

0 80 mt

rvar

.= = = .

.

Thus, the magnitude of the acceleration is 2 2 2 2 2 2 2 2(18 5 m/s ) ( 4 9 m/s ) 19 1 m/s 19 m/sr ta a a= + = . + − . = . ≈

The angle of the acceleration vector from the r-axis is 1 1 4 9tan tan 14 8 15

18 5t

r

aa

.= = = . ° ≈ °

.2 2o

The angle is below the r-axis.

8.57. Solve: (a) You are spinning a lead fishing weight in a horizontal 1.0 m diameter circle on the ice of a pond when the string breaks. You know that the test weight (breaking force) of the line is 60 N and that the lead weight has a mass of 0.30 kg. What was the weight’s angular velocity in rad/s and in rpm?

(b) 2 60 N rev 60 s20 rad/s 191 rpm(0.3 kg)(0.5 m) 2 rad min

ω ωπ

= ⇒ = × × =

8-36 Chapter 8


8.58. Solve: (a) At what speed does a 1500 kg car going over a hill with a radius of 200 m have a weight of 11,760 N? (b) The weight is the normal force.

21500 kg2940 N 19.8 m/s200 m

v v= ⇒ =

8.59. Model: Assume the particle model and apply the constant-acceleration kinematic equations. Visualize:

Solve: (a) Newton’s second law for the projectile is

( )net wind x xxFF F ma a

m−

= − = ⇒ =

where windF is shortened to F. For the y-motion:

211 0 0 1 0 1 02( ) ( )y yy y v t t a t t= + − + − 21

0 1 120 m 0 m ( sin )v t gtθ⇒ = + − 1 0 st⇒ = and 01

2 sinvtg

θ=

Using the above expression for 1t and defining the range as R we get from the x motion: 21

1 0 0 1 0 1 02( ) ( )x xx x v t t a t t= + − + −

211 0 0 1 12x

Fx x R v t tm

⇒ − = = + −

20 0

02 sin 2 sin( cos )

2v F vv

g m gθ θθ

= −

2 220 0

22 2cos sin sinv v Fg mg

θ θ θ= −

We will now maximize R as a function ofθ by setting the derivative equal to 0: 2 2

2 20 02

2 2(cos sin ) 2sin cos 0dR v Fvd g mg

θ θ θ θθ

= − − =

22 2 0

2 20

2cos sin cos2 sin 22

Fv gmg v

θ θ θ θ

⇒ − = =

tan 2 mgF

θ⇒ =

Thus the angle for maximum range is 112 tan ( / ).mg Fθ −=

(b) We have 2(0.50 kg)(9.8 m/s ) 8.167

0.60 NmgF

= = 112 tan (8.167) 41.51θ −⇒ = = °

The maximum range without air resistance is 2 20 02 sin 45 cos45v vR

g g° °′ = =

Therefore, we can write the equation for the range R as 222 sin 41.51 cos41.51 sin 41.51FR R R

mg′ ′= ° ° − ° (0.9926 0.1076) 0.885R R′ ′= − =

0.8850RR

⇒ =′

⇒ 1 0.8850 0.115R RR′ −

= − =′

Thus R is reduced from R′ by 11.5%.



Assess: The condition for maximum range (tan 2 / )mg Fθ = means 2 90θ → ° as 0.F → That is, 45θ = ° when F = 0, as is to be expected.

8.60. Model: Use the particle model for the (cart + child) system which is in uniform circular motion. Visualize:

Solve: Newton’s second law along r and z directions can be written: cos20 sin 20r rF T n ma∑ = ° − ° = sin 20 cos20 0zF T n mg∑ = ° − ° − =

The cart’s centripetal acceleration is 2

2 2rev 1 min 2 rad(2.0cos20 m) 14 4.04 m/smin 60 s 1 revra r πω = = ° × × =

The above force equations can be rewritten as 2

2

0.94 0.342 (25 kg)(4.04 m/s ) 101 N

0.342 0.94 (25 kg)(9.8 m/s ) 245 N

T n

T n

− = =

+ = =

Solving these two equations yields 179 N 180 NT = ≈ for the tension in the rope. Assess: In view of the (child + cart) weight of 245 N, a tension of 179 N is reasonable.

8.61. Model: Use the particle model for the ball, which is in uniform circular motion. Visualize:

8-38 Chapter 8


Solve: From Newton’s second law along r and z directions, 2

cosrmvF n

rθ∑ = = sin 0 sinzF n mg n mgθ θ∑ = − = ⇒ =

Dividing the two force equations gives

2tan grv

θ =

From the geometry of the cone, tan / .r yθ = Thus

2r gr v gyy v

= ⇒ =

8.62. Model: Model the block as a particle and use the model of kinetic friction. Visualize:

Solve: The only radial force is tension, so we can use Newton’s second law to find the angular velocity maxω at which the tube breaks:

2 maxmax

50 N 9.12 rad/s(0.50 kg)(1.2 m)r

TF T m rmr

ω ω∑ = = ⇒ = = =

The compressed air and friction exert tangential forces, and the second law along the tangential direction is k k k

2 2k

4.0 N (0.60)(9.80 m/s ) 2.12 m/s0.50 kg

t t t t t

tt

F F f F n F mg maFa gm

µ µ

µ

∑ = − = − = − =

= − = − =

The time needed to accelerate to 9.12 rad/s is given by

max1 max 1 1 2

(1.2)(9.12 rad/s)0 5.16 s2.12 m/s

t

t

a rt tr a

ωω ω = = + ⇒ = = =

During this interval, the block turns through angle 2

2 21 11 0 0 1 12 2

2.12 m/s 1 rev0 (5.16 s) 23.52 rad 3.7 rev1.2 m 2 rad

tat tr

θ θ θ ωπ

∆ = − = + = + = × =

8.63. Model: Use the particle model for a sphere revolving in a horizontal circle. Visualize:



Solve: Newton’s second law in the r- and z-directions is 2

1 2 1 2 G( ) cos30 cos30 ( ) sin30 sin30 0 Ntr z

mvF T T F T T Fr

∑ = ° + ° = ∑ = ° − ° − =

Using (1.0 m)cos30 0.886 m,r = ° = these equations become 2 2

1 2(0.300 kg)(7.5 m/s) 22.5 N

cos30 (0.866 m)(0.866)tmvT T

r+ = = =

°

2

1 2(0.300 kg)(9.8 m/s ) 5.88 N

sin30 (0.5)mgT T− = = =

°

Solving for 1T and 2T yields 1 14.2 N 14 NT = ≈ and 2 8.3 N.T =

8.64. Model: Use the particle model for the ball. Visualize:

Solve: (a) Newton’s second law along the r- and z-directions is 2

Gcos sin 0 Nr zF n mr F n Fθ ω θ∑ = = ∑ = − = Using GF mg= and dividing these equations yields:

2tan g R yrr

θω

−= =

where you can see from the figure that tan ( )/ .R y rθ = − Thus .gR y

ω =−

(b) ω will be minimum when ( )R y− is maximum or when 0 m.y = Then min / .g Rω = (c) Substituting into the above expression,

29.8 m/s rad 60 s 1 rev9.9 95 rpm0.20 m 0.10 m s 1 min 2 rad

gR y

ωπ

= = = × × =− −

8.65. Model: Use the particle model for the airplane. Visualize:

8-40 Chapter 8


Solve: In level flight, the lift force L

balances the gravitational force. When turning, the plane banks so that the radial component of the lift force can create a centripetal acceleration. Newton’s second law along the r- and z-directions is

2sin cos 0 Nt

r zmvF L F L mg

rθ θ∑ = = ∑ = − =

These can be written: 2

sin cosmv mgrL L

θ θ= =

Dividing the two equations gives: 2

2 2

2

miles 1 hr 1610 m400hour 3600 s 1 miletan 18.5 km

tan 9.8 m/s tan10v vrgr g

θθ

× × = ⇒ = = = °

The diameter of the airplane’s path around the airport is 2 18.5 km 37 km.× =

8.66. Model: Use the particle model for a small volume of water on the surface. Visualize:

Solve: Consider a particle of water of mass m at point C on the surface. Newton’s second law along the r- and z-directions is

22

net

net

( ) cos cos

( ) sin 0 N sin

r

z

mrF n mrnmgF n mgn

ωθ ω θ

θ θ

= = ⇒ =

= − = ⇒ =

Dividing both equations gives 2tan / .g rθ ω= For a parabola 2.z ar= This means

2dz ardr

= = slope of the curve at 1 1tan tan(90 ) tantan 2

Car

φ θ θθ

= = ° − = ⇒ =

Equating the two equations for tan ,θ we get 2

21

2 2g a

ar grω

ω= ⇒ =

Thus the surface is described by the equation 2

2

2z r

gω

=

which is the equation of a parabola.



8.67. Model: Model the block as a particle in circular motion. Visualize: Apply Newton’s second law in the r- and t-directions. We seek ( )v t where v is the tangential speed.

Solve: 2

r r t k tmvF n ma F f ma

r∑ = = = ∑ = − =

Use 2/n mv r= in kkf nµ= . 2

k kk tmvf n ma

rµ µ− = − = − =

Cancel m and use tdvadt

= .

2k dvvr dt

µ− =

Separate variables and integrate.

0

k20

t v

v

dvdtr v

µ− =Ñ Ñ

0

k0

1[ ]v

t

vt

r vµ − − =

k

0

1 1tr v v

µ− = −

Solve for v which is a function of t . 1

k k 0

0 0 0 k

1 1 1 rvt v tv v r v r r v t

µ µµ

−

= + ⇒ = + = +

Assess: The dependencies seem to go in the right direction: as t increases ( )v t decreases; as kµ increases ( )v t decreases. The m canceled out, so the result does not depend on the mass.

dynamics ii: otion in a lane - santa rosa junior...

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