Motion in a Plane

Chapter 8

Centripetal Acceleration

• Centripetal Acceleration – acceleration that points towards the center of a circle.– Also called Radial Acceleration (aR)

vBall rolling in a straight line (inertia)

v Same ball, hooked to a string

aR

vaR aR = v2

r

If you are on a carousel at constant speed, are you experiencing acceleration?

If you twirl a yo-yo and let go of the string, what way will it fly?

Period and Frequency

Period (T)– Time required for one complete (360o) revolution– Measured in seconds

Frequency– Number of revolutions per second– Measured in rev/s or Hertz (Hz)

T = 1 f

Formulas

v = 2r v = r T

aR = v2 a = r

r

A 150-kg ball is twirled at the end of a 0.600 m string. It makes 2.00 revolutions per second. Find the period, velocity, and acceleration.

(0.500 s, 7.54 m/s, 94.8 m/s2)

The moon has a radius with the earth of about 384,000 km and a period of 27.3 days.

A.Calculate the acceleration of the moon toward the earth. (2.72 X 10-3 m/s2)

B.Calculate the previous answer in “g’s” (2.78 X 10-4 g)

Centripetal Force – the “center seeking” force that pulls an object in a circular path.– Yo-yo– Planets– Merry-go-round– Car rounding a curve

Centrifugal Force

A word about Centrifugal Force

• Doesn’t really exist.• “apparent outward

force”• Water in swinging cup

example Centripetal Force of string

Direction water wants to go

Centripetal Motion

F = maR = mv2

rA 0.150 kg yo-yo is attached to a 0.600 m string

and twirled at 2 revolutions per minute.a. Calculate the velocity in m/s ()b.Calculate the centripetal force in the string

(14.2 N)

Thor’s Hammer (mjolnir) has a mass of 10 kg and the handle and loop have a length of 50 cm. If he can swing the hammer at a speed of 3 m/s, what force is exerted on Thor’s hands?

(Ans: 180 N)

Can Thor swing his hammer so that it is perfectly parallel to the ground?

FR

What angle will the hammer take with the horizontal?

F R

mg

Let’s resolve the FR vector into it’s components:

FRx = FRsin

FRy = FRcos

Fy = 0 (the hammer is not rising or falling)

Fy = 0 = FRcos – mg

FRcos = mg

cos = mg/FR

= 57o

How about if he swings faster?

A father places a 20.0 kg child on a 5.00 kg wagon and twirls her in a circle with a 2.00 m rope of tension 100 N. How many rpms does the wagon make ()? (14 rpm)

A 0.150 kg ball is swung on a 1.10-m string in a vertical circle. What minimum speed must it have at the top of the circle to keep moving in a circle?

At the top of the circle, both the weight and the tension in the string contribute to the centripetal force

F = FT + mg

mg FT

F = FT + mg

FR = FT + mgmv2 = FT + mg r(tricky part: assume FT = 0, just as the cord

goes slack, but before the ball falls)mv2 = mg rv2 = grv = 3.28 m/s

Note: this equation is also the minumum velocity for orbit of a satellite

v = \/rg

What is the tension in the cord at the bottom of the arc if the ball moves at twice the minimum speed? (v = 6.56 m/s)

mg

FT

At the bottom of the circle, the weight opposes the centripetal force.

F = FT – mgmv2 = FT - mg rFT = mv2 + mg rFT = 7.34 N

Car Rounding a Turn

• Friction provides the centripetal force• Use the coefficient of static friction (s). The

wheels are turning, not sliding, across the surface

• Wheel lock = kinetic friction takes over. k is always less than s, so the car is much more likely to skid.

A 1000-kg car rounds a curve (r=50 m) at a speed of 14 m/s. Will the car skid if the road is dry and s=0.60?

Ffr = FR

mg

FNLet’s first solve for the Normal Force

FN = mg = (1000 kg)(9.8 m/s2)

FN = 9800 N

Fx = Ffr

FR = Ffr

mv2 = sFN

r(1000 kg)(14m/s)2 = (0.60)(9800 N) (50 m)3920 N < 5800 N

The car will make it. 3920 N are required, and the frcition provides 5800 N.

Will the car make it if it is icy and the s = 0.25

Fx = Ffr

FR = Ffr

mv2 = sFN

r(1000 kg)(14m/s)2 = (0.25)(9800 N) (50 m)3920 N > 2450 N

The car will not make it. 3920 N are required, and the friction only provides 2450 N.

What is the maximum speed a 1500 kg car can take a flat curve with a radius of 50 m (s = 0.80)

BANKED CURVES• Banked to reduce the reliance on friction• Part of the Normal Force now contributes

to the centripetal force

FR = Ffr + FNsin

(ideally, we bank the road so that no friction is required: Ffr = 0)

Banked Curves: Example 1A 1000-kg car rounds a 50 m radius turn at 14

m/s. What angle should the road be banked so that no friction is required?

mg

FN

FN = mgcos

Now we will simply work with the Normal Force to find the component that points to the center of the circle

mg

FN

First consider the y forces.

Fy = FNcos - mg

Since the car does not move up or down:

Fy = 0

0 = FNcos – mg

FNcos = mg

FN = mg/cos

FNcos

FNsin

mv2 = FNsin rmv2 = mgsin r cosv2 = gtan rv2 = gtanrv2 = tangr

tan = (14 m/s)2 = 0.40 (50 m)(9.8m/s2)

= 22o

Fred Flintstone places a 1.00 kg rock in a 1.00 m long sling. The vine breaks at a tension of 200 N.

a. Calculate the angle below the horizontal plane that the rock will take. (2.81o)

b.Calculate the maximum linear velocity the rock can twirl. (14.1 m/s)

c. Calculate the angular velocity in rpm’s. (135 rpm)

Circular Orbits• Orbits are freefall (not true weightlessness)• Orbital velocity must match the weight

mg = mv2

r g = v2 v = √ gr r

A satellite wishes to orbits at a height of 200 miles above the earth’s surface.

a. Calculate the height above the center of the earth if Rearth = 6.37 X 106 m. (6.69 X 106 m)

b.Calculate the orbital velocity. (8098 m/s)c. Calculate the period in minutes. (86.5 min)

Review of Angular KinematicsA motor spins a 2.0 kg block on an 80.0 cm arm at

200 rpm. The coefficient of kinetic friction is 0.60.a. Draw a free body diagram of the block.b. Calculate the tangential acceleration of the block

(due to friction). (-5.88 m/s2)c. Calculate the angular acceleration. (-7.35 rad/s2)d. Calculate the time until the block comes to a rest.

(2.8 s)e. Calculate the number of revolutions. (4.7 rev)