AP Physics 5: Circular and Rotational Motion Name ______

AP Physics 5: Circular and Rotational Motion Name ______

A. Circular Motion

1. constant perimeter (tangential) speed: vt = 2pr/T

a. distance = circumference of the circle: 2pr

b. time = time for one revolution: T (period)

2. constant inward (centripetal) acceleration: ac = v2/r

3. centripetal force, Fc = mac = mv2/r

a. turning on a road problems

v = 2pr/T
ac
·  when the road is horizontal: Fc = Ff = msmg
·  roads are banked in order to reduce the amount of friction (component of the Fg is || to Fc)

b. horizontal loop problem (mass on a string)

Ft-x = Fc = mv2/r
q
Ft Ft-y = Fg = mg
v = 2pr/T
·  Ft = (Fc2 + Fg2)½
·  tanq = Fg/Fc (q is measured from horizontal)

c. vertical loop problem (mass on a string)

·  top: Fnet = Fc = Ft + Fg \ Ft = Fc – Fg
Fg Ft
Fg Ft
·  bottom: Fnet = Fc = Ft – Fg \ Ft = Fc + Fg
·  if on a roller coaster: Fn = Ft

4. Newton's law of universal gravity, Fg = GMm/r2

a. G = 6.67 x 10-11 N•m2/kg2

b. M = mplanet and m = msatellite

c. r is the distance, measured from center to center

d. g = GM/r2

e. Fg = Fc: GMm/r2 = mv2/r \ v = (GM/r)½

v = 2pr/T

m Fg = Fg M

r

Mass (kg) / Radius (m) / r from Earth (m)
Earth / 5.98 x 1024 / 6.38 x 106
Moon / 7.35 x 1022 / 1.74 x 106 / 3.84 x 108
Sun / 1.99 x 1030 / 6.96 x 108 / 1.50 x 1011

B. Newton's Laws—Rotation

1. torque, t = r^Fr (tau—Greek letter for t)

point

of

rotation r 90o Fr

a. r = perpendicular distance from axis of rotation to rotating force Fr

b. when r is not perpendicular to Fr, then t = rFrsinq

c. torque units are m•N (not N•m—work)

2. First Law: Object remains at rest or uniform rotation as long as no net torque (tnet) acts on it

a. measured as the moment of inertia, I = bmr2

b. b corrects for mass distribution (b = 1 for a hoop)

3. Second Law: Fr = bma (acceleration at the rim)

(Frolling = ma + bma = (1 + b)ma)

4. equilibrium (tnet = 0)

a. no acceleration (velocity can be non-zero)

b. center of mass: rcm = S(rimi)/Smi

rcm Fcm = Fg1 + Fg2

m1 m2

r1 m1g r2 m2g

tCM = t1 + t2
rcm(m1 + m2)g = r1m1g + r2m2g
rcm = (r1m1 + r2m2)(m1 + m2)

c. first condition: all forces act through the center

1. solving first condition problems (general)

·  draw free body diagram
·  resolve non-||, non-^ forces into || and ^ components
o  v = 0: || and ^ to horizontal
o  v ¹ 0: || and ^ to velocity
·  ||: F||A + F||B + F||C + . . . = 0
·  ^: F^A + F^B + F^C + . . . = 0
·  solve for unknown

2. solving first condition problems (special case)

·  draw free body diagram
·  if there are only three forces and two of the forces are ^ to each other, then proceed
·  rearrange forces into a tail to tip diagram (vector sum)
·  use trigonometry to solve for unknown sides f
o  q + f = 90o C
o  sinq = cosf = B/C B
o  cosq = sinf = A/C q
o  tanq = B/A, tanf = A/B A

d. second condition: forces act away from the center

1. solving center of mass problems

Fg1 r1 cm r2 Fg2
·  system is NOT rotating \ t1 + t2 = 0
·  t1 = t2 ® r1F1 = r2F2 ® r1m1g = r2m2g \ r1m1 = r2m2

2. solving two supports problems

FL m2 FR
rR m1
r1
r2
·  when support bar has mass: assume all of its mass is in its center of gravity (geometric center)
·  assume point of rotation on left end \ rL = 0 and tL = 0
·  tR – t1 – t2 = 0 \ tR = t1 + t2 ® rRFR = r1m1g + r2m2g
·  solve for FL: FL + FR = m1g + m2g

C. Conservation Laws—Rotation

1. rotational kinetic energy, Kr = ½bmv2 (J)

a. v is the velocity at the rim

b. rolling kinetic energy: Krolling = ½(1 + b)mv2

2. rotational momentum, L = rbmv (kg•m2/s)

a. when tnet = 0, then DL = 0

b. change r and/or b will change v

orbiting planet spinning diver

Kepler's Law

(A1 ® 2 = A3 ® 4)

r1v1 = r2v2 r1b1v1 = r2b2v2

3. mixed linear and rotation motion problems

a. Summary of translational and rotational formulas

Variable / Translational / Rotational / Rolling
force / F = ma / Fr = bma / F = (1 + b)ma
momentum / p = mv / L = rbmv / p + L = (1 + rb)mv
kinetic energy / K = ½mv2 / Kr = ½bmv2 / K = ½(1 + b)mv2

b. conservation of energy problems

ball rolling down a ramp m
b = 2/5
h
How fast is the ball moving when descends h m?
Krolling = Ug
½(1 + b)mv2 = mgh
7/10v2 = gh
v = (10/7gh)½
blocks and pulleys
mk m2
r2
x b
What is the system's speed after m3 descends x m?
Ug-3 – Wf = K1 + K2 + K3
m3gx – mkm1gx = ½(m1 + bm2 + m3)v'2
v' = [2(m3 – mkm1)gx/(m1 + bm2 + m3)]½

c. conservation of momentum problems

jumping on a merry-go-round
m vm
M, rM, bM
How fast is the system going after the boy, m, runs at a stationary merry-go-round at velocity, vm and jumps on at the edge?
(convert boy's linear motion to rotational motion, where r is the rM and b = 1: L = rMbmvm)
Lm + LM = L'
rmbmmvm + rMbMMvM = (rmbmm + rMbMM)v'
mvm = (mm + bMM)v'
v' = mvm/(mm + bMM)

D. Simple Harmonic Motion (SHM)

1. oscillating mass on a spring

a. acceleration is NOT constant \ kinematic formulas are invalid

b. displacement, velocity and acceleration oscillate between +A and –A, where A = amplitude

1. x = +A, when t = 0 (pictured above)

time / t = 0 / t = ¼T / t = ½T / t = ¾T / t = T
displacement / +A / 0 / -A / 0 / +A
velocity / 0 / -vmax / 0 / +vmax / 0
acceleration / -amax / 0 / +amax / 0 / -amax

2. x = 0, when t = 0 (heading downward)

time / t = 0 / t = ¼T / t = ½T / t = ¾T / t = T
displacement / 0 / -A / 0 / +A / 0
velocity / -vmax / 0 / +vmax / 0 / -vmax
acceleration / 0 / +amax / 0 / -amax / 0

c. maximum acceleration, amax = A(k/m)

Steps / Algebra
start with
substitute kA for Fs
solve for a / Fs = ma
kA = ma
amax = A(k/m)

d. maximum velocity, vmax = A(k/m)½

Steps / Algebra
start with
substitute ½kA2 for Us and ½mv2 for K
solve for v / Us = K
½kA2 = ½mv2
vmax = A(k/m)½

e. velocity at x, in terms of vmax: vx = vmax[1 – (x2/A2)]½

Steps / Algebra
start with
substitute for Us and K
solve for vx2
multiply-divide by A2
square root both sides
substitute vmax for A(k/m)½ / Kx + USx = Umax
½mvx2 + ½kx2 = ½kA2
vx2 = (k/m)(A2 – x2)
vx2 = A2(k/m)[(A2/A2) – (x2/A2)]
vx = A(k/m)½[(1 – (x2/A2)]½
vx = vmax[1 – (x2/A2)]½

f. time for one cycle, period, T = 2p(m/k)½

Steps / Algebra
start with
substitute 2pA/T for vmax
simplify
solve for T
substitute (m/k)½ for 1/(k/m)½ / vmax = A(k/m)½
2pA/T = A(k/m)½
2p/T = (k/m)½
T = 2p/(k/m)½
T = 2p(m/k)½

g. formulas at midpoint, 0, and extremes, A

midpoint / extreme
x / 0 / xmax = A
v / vmax = 2pA/T = -A(k/m)½ / 0
a / 0 / amax = vmax2/A = -A(k/m)
U / 0 / Umax = ½kA2
K / Kmax = ½mv2 / 0

2. pendulum

a. period of the simple pendulum, T = 2p(L/g)½

Steps / Algebra
start with
substitute mgsinqrad for F
substitute Lqrad for x
for small angles sinqrad = qrad
solve for k / F = kx
mgsinqrad = kx
mgsinqrad = kLqrad
mg = kL
k = mg/L
start with
substitute mg/L for k
simplify / T = 2p(m/k)½
T = 2p(m/mg/L)½
T = 2p(L/g)½

b. notice that m cancels out of the equation, so the period only depends on the L and g

3. damped harmonic motion

a. amplitude of oscillating spring or swinging pendulum will decrease until it stops—damping

b. damping is due to friction and air resistance

1. forces always oppose direction of velocity

2. damping is enhanced if oscillator is placed in viscous fluid (car shock absorbers)

c. forced damping is accomplished with motors that are programmed to oppose velocity (earthquake protected buildings)

4. resonance

a. object can be set to oscillate by an external force—forced vibration

b. when forced vibration matches natural vibration, then amplitude builds with each vibration—resonance

c. examples

1. child swinging

2. building during an earthquake

3. air inside a musical instrument

A. Circular Motion

Centripetal Force Lab

Measure the period of a whirling mass using two techniques, and then vary the tension and radius to see their effects on the period.

a. Collect the following data.

Control
string length, L / 0.5 m
hanging weight, m1 / 100 g
stopper mass, m2
time (10 orbits), t
Double L / Half m1
m1 / 100 g / m1 / 50 g
string length, L / 1.0 m / string length, L / 0.5 m
time (10 orbits), t / time (10 orbits), t

b. Calculate the following from the data.

Formula / Calculation
Control / Double L / Half m1
T1
Fg1
Fg2
q
Fc
r
v
T2
%D

c. Do the results from this experiment seem reasonable?

Double L
Half m1

Questions 1-16 Briefly explain your answer.

1. When a tetherball is whirling around the pole, the net force is directed

(A) toward the top of the pole

(B) toward the ground

(C) horizontally away from the pole

(D) horizontally toward the pole

2. You are standing in a bus that makes a sharp left turn. Which of the following is true?

(A) you lean to the left because of centripetal force

(B) you lean to the right because of inertia

(C) you lean straight ahead because of the net force is forward

(D) you lean to the right because of centrifugal force

3. You drive your car too fast around a curve and the car starts to skid. What is the correct description of this situation?

(A) car's engine is not strong enough to keep the car from being pushed out

(B) friction between the tires and the road is not strong enough to keep the car in a circle

(C) car is too heavy to make the turn

(D) none of the above

4. A steel ball is whirling around in a circle on the end of a string, when the string breaks. Which path will it follow?

A B C

5. Two stones A and B have the same mass. They are tied to strings and whirled in horizontal circles. The radius of the circular path for stones A is twice the radius of stone B's path. If the period of motion is the same for both stones, what is the tension in cord A compared to cord B

(A) TA = TB (B) TA = 2TB (C)TA = ½TB

6. A rider in a "barrel of fun" finds herself stuck with her back to the wall as the barrel spins about a vertical axis. Which diagram shows the forces acting on her?

(A) (B) (C) (D)

7. You are on a Ferris wheel moving in a vertical circle. Which is true when you are at the top of the wheel?

(A) Fn < Fg (B) Fn = Fg (C) Fn > Fg

8. You driving along a rural road. Which is true when you are at the lowest point along a dip in the road?

(A) Fn < Fg (B) Fn = Fg (C) Fn > Fg

9. You swing a ball on the end of a string in a vertical circle. Which is true of the centripetal force at the top of the circle?

(A) Fc = Ft + Fg (B) Fc = Ft – Fg (C) Fc = Fg – Ft

10. Which is stronger the Earth's pull on the Moon or the Moon's pull on the Earth?

(A) Earth's pull (B) Moon's pull (C) they are equal

11. If the distance between the Earth and Moon were doubled, then the force of gravity would be

(A) equal (B) 2 x (C) ½ x (D) ¼ x

12. You weigh yourself in Denver at 1 mile above sea level. How would your weight compare to here?

(A) less (B) the same (C) more (D)

13. Satellites A and B are of equal mass. A experiences twice the force of gravity compared to B. What is the ratio of radius A compared to radius B?

(A) 1/2 (B) 1/Ö2 (C) 1/4 (D) 2/1

14. Is there a net force acting on an astronaut floating in orbit around the Earth while on a space walk?

(A) yes (B) no

15. If you weighed yourself at the equator, would you weigh more or less than at the poles?

(A) less (B) the same (C) more

16. When the Apollo Missions went to the moon they passed a point where the gravitational attractions from the moon and the earth are equal. What is the ratio of the distances to the Moon and Earth where this happened? (mEarth = 100mMoon)

(A) 1/100 (B) 1/10 (C) 10/1 (D) 100/1

17. A car is traveling east on the north side of a circular track. (r = 50 m) takes 16 s to make one lap.

a. Determine

v
ac
direction ac

b. What direction will the car skid on the icy north side?

18. A rock is whirling in a horizontal circle on the end of a 2.0 m string with a 0.50 s period of revolution. Determine

a. What is the direction of centripetal acceleration when the rock is on the north side of the string?

b. What is the rock's velocity?

c. What is the centripetal acceleration?

d. The string breaks when the rock is on the north side of the string. Which way will the rock fly off?

19. The earth is 1.5 x 1011 m from the sun and makes one complete circular orbit in 1 year.

a. What is the period of orbit in seconds?

b. What is the earth’s orbital velocity?

c. What is the centripetal acceleration of the earth toward the sun?

20. A driver of a 1000-kg sports car attempts a turn whose radius of curvature is 50 m on a road where m = 0.8.

a. What is the fastest that the driver can make the turn?

b. Could the driver make the turn at this speed

(1) with a 2,000-kg SUV? Explain