Example of Air Drag: cowboy to go

after the bike-riding, escaping bull…

The drag force (due to air/water) can be

computed from

Where C is called the drag coefficient (‘thick’ air

on a summer night has a bigger value than ‘thin’

air in Mile-High city of Colorado). A is the frontal

surface area of moving object (cowboy) exposed

to the air, ! is density of the medium (air/water),

v is his speed of object (cowboy).

!

fair =1

2CA"v 2

In this problem: Suppose drag coefficient C=1.0, frontal area of

cowboy (contact area with air) is 0.463 m2, the coefficient of

rolling friction of bike is 0.0045. The cowboy has mass of 50 kg, the bike’s mass is 12kg. Density of air = 1.2 kgm-3

.

frontal

Q1: To maintain a speed of 12 m/s on level ground, what must

be the cowboy’s power output to the rear wheel?.

Q1: Suppose the cowboy reduces speed to 6 m/s. What must

his power output to the rear wheel?

Solution: Work must be done to counteract two forces acting on

the cowboy (1) air drag (2) ground rolling friction.

!

fair =1

2CA"v 2

!

froll = µrsn = 0.0045 " (mBike + mcowboy )g

!

=1

2" (1.0) " (0.463) " (1.2) "12

2= 40.0N

!

= 2.74 N

Power needed:

!

P = Ftotalv = fair + f roll( ) " (12) = 512.88 W

!

fair =1

2CA"v 2

!

=1

2" (1.0) " (0.463) " (1.2) " 6

2=10.00 N

Power needed:

!

P = Ftotalv = fair + f roll( ) " (6) = 76.44 W

!

froll = µrsn = 2.74 N (unchanged)

Moral: Air drag is proportional to v2, while P is proportional to v, so there is a cubic dependence of power on velocity as far as drag force is concerned.

Example: The engine of a car with mass m supplies a constant power P0 to the wheels to accelerate the car. Ignore rolling friction and air resistance. The car is initially at rest. !

a)! What is the speed as a function of time?!b) What is the acceleration of the car?. !c) Show that the displacement as a function of time is given by x-x0= (8P0/9m)

1/2 t3/2.!

!"#$%&'$())*+,)-$).//#0#1.%&')/&'$2.'23))*4,)/.')5#)6'7).//#0#1.%&')"$-'8)9:;)<)=3)

>&0"%&'()

!

P =W

"t

!

P0

=0.5mv

2

t

!

v =2P

0t

m

!

a =dv

dt=1

2

2P0t

m

"

# $

%

& '

(1

2 2P0

m

"

# $

%

& ' =

P0

2mt

!

x " x0

= v

0

t

# dt

!

=2P

0t

m0

t

" dt =2P

0

mt

0

t

" dt =2P

0

m

2

3t3 / 2

=8P

0

9mt3 / 2

!"#$

!%#$

!&#$

Example A 2.00 kg block is pushed against a spring with negligible mass

and spring constant k= 400 N/m, compressing it 0.220 m. When the block

is released it moves along a frictionless horizontal surface and then up a

frictionless incline with slope 37.0°. a) What is the speed of the block as

it slides along the horizontal surface after having left the spring? b)

How far along the incline does it travel before sliding down?

Solution:

First, only gravity and spring will do work (both are conservative), !

!

Ugrav+Uel +Uother = K

2"K

1

!

For the segment up the board, Initial K equals final Ugrav (or Uel).

!

Uel

=1

2mv

2

At the horizontal surface, let y=0 --! Ugrav = 0

!

Ugrav+Uel = K

2"K

1

!

v =k

mx

!

kx2

2=1

2mv

2

!

=400

2" 0.22 = 3.11 m /s

(a)

(b)

!

1

2mv

0

2

= mgh

!

1

2mv

0

2

= mgLsin37

!

L =1

2gsin37v0

2

!

= 0.82 m

Example: A system of two buckets connected by a light-weight rope is released from rest with the 12.0 kg bucket 2.00 m above the floor. !a) Use the principle of conservation of energy to find the speed with which this bucket strikes the floor. You can ignore friction and the inertia of the pulley.!b) How much higher does the 4.0 kg bucket rise after the 12.0 kg bucket has hit the floor?!

Solutions:

(a) Work done by Tension force = 0 (since

work tension force does to bucket B is

negated by the work it does on A). Work is done only by Gravity.

A

B

!

U1

= mAgyA ,1=12 " 9.81" 2.00 = 235.44 J

!

U2 = mBgyB,2 = 4 " 9.81" 2 = 78.48J

For the pulley system (consisting of A and B), the Initial

kinetic energy K1= 0

After the masses are in motion and before A lands on ground, the final kinetic energy

K2= 1/2mAvA2+1/2mBvB

2

But the two masses are connected by a string, hence same speed v,

K2= !(mA+mB) v2=0.5x(16)v2=8v2

Conservation of energy implies total energy before must be equal to total

energy after

!

K1

+U1

+Wother

= K2

+U2

!

U1

= K2

+U2

!

235.44 = 8v2+ 78.48

!

v = 157.96/8 = 4.45 m/s

(b) How much higher does the 4.0 kg bucket rise after the 12.0 kg bucket has hit the floor?

'($

)$

Solution: Remember the soccer-ball problem? Same thing…

*++,-"&.$/0$$123$4.3$&-524"54$"&&363,"7-5$

389"7-52$:;4.$

"! "<=)>$'(<?@?A$B2=/>$'<($B2=/$

"! 123$'C<'(CDC"2$

!

0 = 4.452" 2 # 9.81# s

!

s =1.01 m*++,-"&.$C0$

!

1

2mAv0A

2= mAghA ,additional

!

hA ,additional =1

2gv

0A

2=1.01 m

Example: *$,-&E$-F$B"22$B$;2$73G$4-$"$&-,G$-F$635)4.$H$"5G$4.3$-4.3,$35G$-F$4.3$&-,G$;2$.36G$IJ3G@$$K.3$,-&E$;2$);'35$"5$;5;7"6$4"5)357"6$'36-&;4L$4."4$&"9232$;4$4-$

,-4"43$;5$"$'3,7&"6$&;,&63@$$MF$K*$;2$4.3$4352;-5$;5$4.3$&-,G$"4$4.3$4-+>$"5G$KN$;2$4.3$

4352;-5$"4$4.3$%-O-B>$4.35$G343,B;53$KN$P$K*@$$

Solution:

!

UA

+ KA

=UB

+ KB

!

mg(2R) +mvA

2

2= 0 +

mvB2

2

Need tension forces, which requires Newton II,

!

! F = m

! a " = m

v2

! r

To compare tension forces, need velocity relationship between VA and VB,

which can be deduced from energy relationship (choose bottom U=0):

!

(vB2" vA

2)

2= 2gR

!

(vB2" vA

2) = 4gR

Now, Newton II on top and bottom

!

"TA "mg = "mvA

2

R

!

TB "mg = mvB

2

R

positive

!

TA + mg = mvA

2

RSubtract the two equations:

!

TB "TA " 2mg = m(vB

2" vA

2)

R

!

TB "TA = 4mg+ 2mg = 6mg