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Physics Help >> Index to Physics Homework Help » Rotational Motion
A. Rotational KinematicsExample:example 1. The diameter of a typical minivan tire is 64.8 cm. A stone is stuck in the tread of the right rear tire. What is the magnitude and direction of the stone's angular velocity vector if the van is traveling at 10.0 km/h? Answer to Example 1: (10 km/h)(1000/3600) = 27.8 m/s
r = .648/2 = .324 m
ω = v/r = (27.8)/(.324) = 85.8 rad/s
The direction is given by the right hand rule. Assuming the car is moving forward on the right hand side of the road, the direction of the angular velocity along the car's axle away from the right curb.
Rotational Kinematics Problems:
1. A spot of paint on a tire moves in a circular path of radius 0.42 m. When the spot has traveled a linear distance of 1.48 m, through what angle has the tire rotated?
2. A discus thrower (arm length 1.2 m) starts from rest and begins to rotate counterclockwise with a constant angular acceleration of size 2.7 rad/s2.
a. How long does it take for the discus thrower's angular speed to reach 5.7 rad/s?
b. How many revolutions does it take for the discus thrower's angular speed to reach 5.7 rad/s?
c. What is the linear speed of the discus at 5.7 rad/s?
d. What is the linear acceleration of the discus at this point?
e.What is the size of the discus's centripetal acceleration?
f. What is the size of the discus's total acceleration?
3. An airplane propeller is rotating at 1900 rev/min. b. How long in seconds does it take for the propeller to turn through 30.0 degrees?
4. A volleyball spike begins with the arm overhead, the shoulder and elbow are flexed and the wrist is hyper extended. The upper arm is .6m and the forearm is .3m. The time of spike is .1 seconds, and the changes in position of each joint is: Shoulder = 30 degrees, Elbow = 70 degrees. What is the linear velocity at the end of the distal endpoint of each segment at the end of the segment's rotation?
5. A roulette wheel with a 1.0-m radius reaches a maximum angular speed of 18 rad/s before it stops 35 revolutions (220 rad) after attaining the maximum speed. how long did it take the wheel to stop?
6. A bicycle has wheels of diameter d = 123.0 cm. The bicyclist accelerates from rest with constant acceleration to 14.5 km/hr in 13.3 s. What is the angular acceleration of the wheels?
7. A computer hard-drive accelerates from 0 to 7.598 thousand rpm in 5.44 seconds. How many revolutions has the drive made in this period of time?
8. An electric golf cart motor applies torque to the drive-axle causing the wheels to accelerate at a nice, gentle 78.1 rad/s2. If the diameter of the cart’s wheels is 5.62 inches, what is the vehicle's linear acceleration rate (in feet/s2)?
9. A cyclist hits a downhill grade at 48 miles/hour. At the bottom of the grade the cyclist is doing 74 miles/hour. If the grade is 0.244 miles long, what was the angular acceleration of the cyclist\'s 3.82 foot diameter wheels during this period of time?
10. Linear Acceleration and Angular Acceleration are related by ...
11. A 25 inch tire mounted on an axle rotates through an angle of 138 degrees. How many feet has the axle moved along the ground?
12. A centrifuge in a medical laboratory rotates at an angular speed of 3600 rev/min. When switched off, it rotates through 50 revolutions before coming to rest. Find the constant angular acceleration of the centrifuge.
13. What is the tangential acceleration of a bug on the rim of a 10 inch diameter disk if the disk moves from rest to an angular speed of 78 rev/min in 3 sec? When the disk is at its final speed, what is the tangential velocity of the bug? One second after the bug starts from rest, what are its tangential acceleration, centripetal acceleration, and total acceleration?
B. Torque and EquilibriumExample:
example 2. The normal force acting on the front legs of a cow on a horizontal surface is Nf= 1960 N. The normal force acting on the rear legs is Nr= 2200 N. The distance between the rear legs and the front legs is 1.52m. Calculate the weight of the cow.
Answer to example 2:
The first condition of equilibrium is that forces must balance to zero. Therefore, the weight of the cow must equal the sum of the two normal forces. The weight of the cow is 1960 + 2200 = 4160 N
Torque and Equilibrium Problems:
1. You are holding a shopping basket at the grocery store with two 0.51 kg cartons of cereal at one end of the basket. The basket is 0.69 m long. Where should you place a half gallon of milk (1.8 kg) so that the center of mass of your groceries is at the center of the basket?
2.
A shop sign weighing 231 N is supported by a uniform 137 N beam of length 1.76 m. The guy wire is connected 1.24 m from the backboard.
3. A wrench applied to a lug nut is giving 50 lbs of force 20 degrees from the horizontal.The diameter of the tire is 28 inches and the diameter of the rim is 5 inches, so the distance from the lug nut to the center of the wheel is 2.5 inches. The length of the wrench is 20 inches. What is the torque on the center of the wheel?
4 .A 2 "x 4 "x 8' board which weighs 13 lbs is positioned so that one end rests on the ground and the other rests on a table that is 33 in. tall. What force does the board exert on the table?
5. A uniform beam 6.0 m long and weighing 4 kg rests on supports p and q placed left and right 1.0 m from each end of the beam. Weights of mass 10 kg and 8 kg are placed near p and q respectively, one on each end of the beam. Calculate the reaction forces p and q.
6. A woman who weighs 500 N stands on an 8.0-m-long board that weighs 100 N. The board is supported at each end. The support force at the right end is 3 times the support force at the left end. How far from the right end is the woman standing?
C. Torque and MomentumExample:
Example 3. Two spheres, one solid, one a thin shell, are given the same angular impulse (i.e. the same torque acting for the same amount of time). After the impulse stops acting the spheres are rotating at different rates.
Explain which sphere is rotating faster and why. It may help to note that the moment of inertia of a solid
sphere is 2/5MR2 & that of a spherical shell is 2/3MR2. For calculating purposes, use M=0.89 kg, R=11 cm,
and the torque of 0.13 N*m acts for 7.8s.
Answer to example 3:
(a) The angular acceleration of an object is given by Newton's second law in rotational form α = τ/I where τ is torque applied, and I is moment of inertia
Then the angular speed (the rate of rotation) after time t is given by ω = α*t = τ*t/I (1)
Thus the less the moment of inertia the more the final rate of rotation
Since the moment of inertia of the solid sphere is less than of a spherical shell we conclude that, when the same torque is acting for the same amount of time, the solid sphere will rotate faster
(b) Calculations: Putting the values of I into equation (1) we get for the solid sphere and the spherical shell respectively: ω1 = τ*t/I1 = 5*τ*t/(2*M*R2) = 5*0.13*7.8/(2*0.89*0.112) = 235 rad/s ω2 = τ*t/I2 = 3*τ*t/(2*M*R2) = 3*0.13*7.8/(2*0.89*0.112) = 141 rad/s
Torque and Momentum Problems:
1. Four small spheres, each of which you can regard as a point of mass m = 0.110 kg, are arranged in a
square d = 0.850 m on a side and connected by light rods.
a. Find the moment of inertia of the system about an axis through the center of the square, perpendicular to
its plane. b. Find the moment of inertia of the system about an axis bisecting two opposite sides of the square. c. Find the moment of inertia of the system about an axis that passes through the centers of the upper left and
lower right spheres.
2.
cm and mass M2 = 1.80 kg, are welded together and mounted on a frictionless axis through their common
center.
a. What is the total moment of inertia of the two disks? b. A light string is wrapped around the edge of the smaller disk, and a m = 1.0 kg block, suspended from the
free end of the string. If the block is released from rest a distance of 2.00 m above the floor, what is its speed
just before it strikes the floor? c. Repeat the calculation of part (b), this time with the string wrapped around the edge of the larger disk. d. In which case is the final speed of the block the greatest?
3. The combination of an applied force and a frictional force produces a constant torque of 36N*m on a wheel rotating about a fixed axis. The applied force acts for 6 s during which time the angular speed of the wheel increases from 0 to 12 rad/s. The applied force is then removed, and the wheel comes to rest in 65 s.
4. A cylindrical fishing reel has a mass of .85 kg and a radius of 4 cm. A friction clutch in the reel exerts a restraining torque of 1.3 N*m if a fish pulls on the line. The fisherman gets a bite, and the reel begins to spin with an angular acceleration of 66rad/s/s. What is the force of the fish on the line?
5. A cable passes over a pulley. Because of friction, the force in the cable is not the same on the opposite sides of the pulley. The force on one side is 120 N and the other side is 100 N. Assuming that the pulley is a uniform disk with a mass of 2.1 kg and a radius of .81m determine its angular acceleration.
6. Calculate the uniform torque to raise the speed of revolution of a winding drum from 35 to 70 rpm in 70s if there is a frictional torque of 17 kNm. The winding drum has a moment of inertia of 1800 kg m2.
7. Determine the magnitude of angular momentum of a solid disk of radius 50 cm. and a mass of 2.4 kg. spinning at 6 rev/s about an axis through its center and perpendicular to its plane.
8. A pendulum consists of a solid sphere of mass m and radius R moving in a horizontal-circular path. During the motion the supporting string of length l maintains a constant angle, θ, with the vertical. What is the magnitude and direction of the angular momentum?
9. A thick cylinder of inner radius, a, and outer radius, b, has a mass M uniformly distributed over its volume. a. Calculate the moment of inertia for the thick cylinder for a rotational axis coincident with its axis of symmetry. b. Show that the expected results for the solid cylinder and the thin walled cylindrical shell are obtained with the appropriate choices of a/b.
10. A block of mass m, in is suspended from a frictionless pulley of mass M, radius R, and moment of inertia I. If the mass is released, a. what acceleration does the mass experience? b. what is the tension in the cord?
11.
a. What is the fictional torque acting on the pulley? b. What is the moment of inertia of the pulley?
D. Energy and PowerExample:
Example 4. A solid ball of mass 1.3 kg and diameter 16 cm is rotating about its diameter at 66 rev/min. What is its kinetic energy?
m = 1.3 kg
r = 0.08 m
Ek = (1/2)Iω2 = 287 J
Energy and Power Problems:
1. The output shaft of a gear system is attached to a motor that has a rated power of .2 W mounted on the shaft is a wheel that has a moment of inertia of .8 Kg meter squared. Estimate the minimum time it takes the motor to accelerate the wheel from rest to 24 rev/minute.
2. What is the rotational kinetic energy of a 60 cm. diameter bicycle wheel of mass 4.0 kg. When the bicycle is traveling at 4 m/s. Assume the wheel has a radius gyration k=50 cm.
3. Two metal disks, one with radius R1 = 2.80 cm and mass M1 = 0.90 kg and the other with radius R2 = 5.60 cm and mass M2 = 1.80 kg, are welded together and mounted on a frictionless axis through their common center. a) What is the total moment of inertia of the two disks? 4. A grindstone in the shape of a solid disk with a diameter of 0.600 m and a mass of 50.0 kg is rotating at 1100 rev/min. You press an ax against the rim with a normal force of 160 N, and the grindstone comes to rest in 10.0 s. Find the coefficient of friction between the ax and the grindstone. Neglect friction in the bearings. 5. A solid uniform sphere is released the top of an inclined plane 0.25 m tall.
Answers:
For solutions to all the problems on this page click here.
A5. ωf2 = ωi2 + 2αθ
02 = 18 2 + 2α(220)
α = -0.736 rad/s/s
t = ( ωf - ωi) / α = ( 0 - 18) / (-0.736) = 24.4 s
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B3.
The wrench is 20 degrees from the horizontal. The radial distance to the the point of application of force from the center of the wheel is given by the cosine law:
√(202 + 2.52 - 2*2.5*20*cos20) = 17.67 inches
The angle between the the radial distance and the wrench is given by the sine law:
(sin20º)/17.67 = (sinA)/2.5 .
A, = 2.77º
So the angle between the radial distance and the wrench, θ, is 92.77º .
The moment of force (torque) on the center of the wheel is
τ = Frsinθ = (50)(17.67)(sin92.8) = 882 inch-pounds
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C8.
r = (l + R)sinθ
For a particle in circular motion, angular momentum, L, is given by
L = mvr
For this situation, L = mv(l + R)sinθ
The direction of the angular momentum is perpendicular to the plane of the orbit according to the right hand rule.
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D6. We assume the sphere is uniform (solid), since we are not told otherwise.
E = (1/2)mvcm2 + (1/2)[(2/5)mR2](vcm / R)2 where vcm is the (linear) speed of the center of mass
vcm = √[(10/7)gh] substituting ω = vcm /R ω = (1/R)√[(10/7)gh]
h = 6*sin31° = 3.09 m; R = 0.20 m; g = 9.81 N/kg
ω = 32.9 rad/s
For solutions to all the problems on this page click here. Related Product: Relativity in Rotating Frames: Relativistic Physics in Rotating Reference Frames (Fundamental Theories of Physics)
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ω τ α π δ Δ θ φ μ √ º θ ρ ≤ ≥ λ
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A 5.00 kg mass is attached to a rope of negligible mass wrapped around a pulley. The pulley has a radius of 3.00 cm. The mass falls 1.61 m in 0.700 s. At this point the rope comes off the pulley. The pulley comes to a stop in 0.234 s.
Since the particle has finite radius, R, and is attached to a string of length l, the radius of its orbit, r, is given by