Rotational Motion vs Linear Motion
Rotational Motion 
Linear Motion 
θ (rad) 
d (m) 
ω (rad/s) 

α (rad/s^{2}) 
a (m/s^{2}) 
τ = Iα 
F = ma 
KE = (1/2)Iω^{2} 
KE = (1/2)mv^{2} 
L = Iω 
p = mv 
A. Rotational Kinematics
Example:
example 1. The diameter of a tire is 64.8 cm. A tack is embedded in the tread of the right rear tire. What is the magnitude and direction of the tack's angular velocity vector if the vehicle is traveling at 10.0 km/h?
Answer to Example 1:
(10 km/h)(1000/3600) = 2.78 m/s
r = .648/2 = .324 m
ω = v/r = (2.78)/(.324) = 8.58 rad/s
The direction is given by the right hand rule. Assuming the vehicle 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:
(See below for answers)
1. A spot of paint on a tire moves in a circular path of radius 0.42 m. Through what angle must the tire rotate for the spot to travel 1.48 m?
2. A propeller (arm length 1.2 m) starts from rest and begins to rotate counterclockwise with a constant angular acceleration of size 2.7 rad/s^{2}.
a. How long does it take for the propeller's angular speed to reach 5.7 rad/s?
b. How many revolutions does it take for the propeller's angular speed to reach 5.7 rad/s?
c. What is the linear speed of the tip of the propeller at 5.7 rad/s?
d. What is the linear acceleration of the tip of the propeller at this point?
e.What is the centripetal acceleration of the tip of the propeller?
f. What is the total acceleration of the tip of the propeller?
3. An airplane propeller is rotating at 1900 rev/min.
a. Compute the propeller's angular velocity in rad/s.
b. How long in seconds does it take for the propeller to turn through 30.0 degrees?
4. A tire on a rim rolls along a level surface from A to B. The rim moves from C to D and is always in contact with line CD. Since both travel the same distance and the same number of rotations, how can the circumference of the rim not be the same as the circumference of the tire?
5. A disk with a 1.0m 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 disk to stop?
6. A wagon with wheels having diameter d = 123.0 cm accelerates from rest to 14.5 km/hr in 13.3 s. What is the angular acceleration of the wheels?
7. How many revolutions does a disc make in in 5.44 seconds accelerating from 0 to 7.598 thousand rpm?
8. A wagon's wheels accelerate at 78.1 rad/s^{2}. If the diameter of wheels is 5.62 inches, what is the wagon's acceleration?
9. A truck accelerates from 48 miles/hour to 74 miles/hour through 0.244 miles. What was the angular acceleration of the 3.82 foot diameter wheels?
10. Linear Acceleration and Angular Acceleration are related by ...
11. A 25 inch tire rotates 138 degrees. What distance does the center of the tire move?
12. A drill bit rotating at 3600 rev/min rotates through 50 revolutions while coming to rest. Find the drill's angular acceleration.
13. A 10 inch diameter disk with a nick on its rim moves from rest to an angular speed of 78 rev/min in 3 sec. What is the tangential acceleration of the nick? What are the tangential acceleration, centripetal acceleration, and total acceleration of the nick one second after the disk starts to accelerate? When the disk is reaches full speed, what is the tangential velocity of the nick?
B. Torque and Equilibrium
Example:
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:
(See below for answers)
1. Two 0.51 kg cartons are at one end of a bin that is 0.69 m long. Where should a 1.8 kg carton be placed so that the center of mass of the three cartons is at the center of the bin?
2.
A 231 N sign is supported by a uniform 137 N beam of length 1.76 m. The wire is connected 1.24 m from the wall.
a. Find the tension in the wire. The wire forms an angle of 38.5º with the beam.
b. Find the horizontal force exerted wall on the beam.
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 .An 8 foot board weighs 13 lbs and rests with one end on the ground and the other on a table that is 33 in. in height. What force does the board exert on the table?
5. A uniform horizontal beam 6.0 m long and weighing 4 kg is supported at points p
and q each 1.0 m from opposite ends of the beam. Masses of 10 kg and 8 kg are placed near p and q respectively, one on each end of
the beam. Calculate the reaction forces at p and q.
6. A 500 N rock rests on an 8.0mlong 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 rock standing?
7.When the proper weights are assigned, this mobile is perfectly balanced. What are the 2 missing weights?
8.
In the diagram to the left, a uniform 10.0 kg beam 3.0 m long is hinged to a wall and supported by a horizontal rope to make a 40º angle with the wall. A 15 kg mass hangs from the end of the beam. What is the tension in the rope?
C. Torque and Momentum
Example:
Example 3. Two spheres, one solid, one a thin shell, each have a mass of 0.89 kg and a radius of 11 cm. They are given the same angular impulse (i.e. the same torque of 0.13 N*m acting for 7.8s).
Explain which sphere is rotating faster and why.
Answer to example 3:
The moment of inertia of a solid sphere is 2/5MR^{2} & that of a spherical shell is 2/3MR^{2}.
(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:
ω_{1} = τ*t/I_{1} = 5*τ*t/(2*M*R^{2}) = 5*0.13*7.8/(2*0.89*0.11^{2}) = 235 rad/s
ω_{2} = τ*t/I_{2} = 3*τ*t/(2*M*R^{2}) = 3*0.13*7.8/(2*0.89*0.11^{2}) = 141 rad/s
Torque and Momentum Problems:
(See below for answers)
1. Four small spheres of mass m = 0.110 kg, are connected by light rods and form a square with sides 0.850 m.
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. Two concentric disks, one with radius R_{1} = 2.80 cm and mass M_{1} = 0.90 kg and the other with radius R_{2} = 5.60 cm and mass M_{2} = 1.80 kg, are welded together and mounted on a frictionless axis.
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.
3. A net torque of 36N*m acts on a wheel rotating about a fixed axis for 6 s. During this 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.
a. What is the moment of inertia of the wheel?
b. What is the magnitude of the frictional torque?
c. How many revolutions does the wheel make?
4. A cylinder has a mass of .85 kg and a radius of 4 cm. Friction about its axis supplies a torque of 1.3 N*m. A light string wrapped around the cylinder is pulled and the cylinder spins with an angular acceleration of 66rad/s/s. What is the tension in the string?
5. A cable passes over a pulley. Because of friction, the force on one side is 120 N and the other side is 100 N. The pulley has a mass of 2.1 kg and a radius of .81m. Determine its angular acceleration.
6. Calculate the torque needed to increase the speed of a revolving cylinder from 35 to 70 rpm in 70s. There is a frictional torque of 17 kNm and the cylinder has a moment of inertia of 1800 kg m^{2}.
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 horizontalcircular 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 uniformly distributed mass M.
a. If the cylinder rotates about its axis of symmetry, calculate its moment of inertia.
b. Show that a solid cylinder has the same result given the proper choice 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 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.
a. What is the fictional torque acting on the pulley?
b. What is the moment of inertia of the pulley?
12. A hoop is rotating about an axis that passes through its center, and so
is a hollow thin sphere. They have identical masses, radii, and
kinetic energies. The angular speed of the spherical shell is 5.9 rad/s.
What is the angular speed of the hoop, in rad/s?
D. Energy and Power
Example:
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?
Answer to Example 4:
m = 1.3 kg
r = 0.08 m
ω = angular speed = 66*2π = 415 rad/min = 6.91 rad/s
I = moment of inertia = (2/5)mr^{2} = (2/5)(1.3)(0.08)^{2} = 0.00333 kg m^{2}
E_{k} = (1/2)Iω^{2} = .0796 J
Energy and Power Problems:
(See below for answers)
1. Approximately 0.2 W of power is supplied to a wheel that has a moment of inertia of 0.8 Kgm^{2}. How long does it take the wheel to accelerate from rest to 24 rev/minute?
2. A monocycle traveling at 4 m/s has a 60 cm. diameter wheel with a mass of 4.0 kg and a gyration radius, k=50 cm. What is the rotational kinetic energy of the wheel?
3. Two concentric disks, one with radius R_{1} = 2.80 cm and mass M_{1} = 0.90 kg and the other with radius R_{2} = 5.60 cm and mass M_{2} = 1.80 kg, are welded together and mounted on a frictionless axis.
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.
4. A solid disk with a diameter of 0.600 m and a mass of 50.0 kg is rotating at 1100 rev/min. A pad pressed against the outside edge with a normal force of 160 N brings the disk to rest in 10 s. Find the coefficient of friction between the pad and the disk. Neglect friction in the axle.
5. A solid uniform sphere is released the top of an inclined plane 0.25 m tall.
a. The sphere rolls down the plane without slipping and there is no energy lost from friction. What is the translational speed of the sphere at the bottom of the incline?
b. If the inclined plane above was frictionless and the sphere slid down the plane instead of rolling, what is the translational speed of the sphere at the bottom of the incline?
6. A 270 N sphere 0.20 m in radius at the top of a 6.0 m ramp inclined at 31° with the horizontal. It begins to roll without slipping. What is the angular speed of the sphere at the bottom of the ramp?
7. If a solid cylinder rolls without slipping, what is the ratio of its rotational kinetic energy to its translational kinetic energy?
8. A 10 kg cylinder rolls without slipping. When its translational speed is 10 m/s, What is its translational kinetic energy, its rotational kinetic energy, and its total kinetic energy?
9. A solid disk does 3.9 kJ of work and as a result decelerates from 500.0 rpm to rest.. What is the mass of the disk if its radius is 1.2m?
10. A vertical pole 2.81 m in height topples, starting from
rest. Find the tangential speed of the free end of the pole, just before the
pole hits the floor.
Answers:
For solutions to all the problems on this page click here.
A5.
ω_{f}^{2} = ω_{i}^{2} + 2αθ
0^{2} = 18 ^{2} + 2α(220)
α = 0.736 rad/s/s
t = ( ω_{f}  ω_{i}) / α = ( 0  18) / (0.736) = 24.4 s
For solutions to all the problems on this page click here.
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:
√(20^{2} + 2.5^{2}  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 inchpounds
For solutions to all the problems on this page click here.
C8.
a = acceleration of the mass
b = angular acceleration of the spool
R = radius of the spool
I = moment of inertial of the spool
m = mass
T = tension along the string
a = bR
b = a / R
m g – T = ma
mg = T + ma
torque on the spool = TR = Ib
Substituting:
TR = Ia / R
T = Ia / R^{2}
Since mg = T + ma
mg = ( m + I / R^{2} )a
a = 9.8*35.50 / (35.50+400) = 0.799 m/s^{2}.
Since d = (1/2)at^{2}
t = √(2d/a) = 2.96 s.
For solutions to all the problems on this page click here.
D6.
We assume the sphere is uniform (solid), since we are not told otherwise.
E = (1/2)mv_{cm}^{2} + (1/2)[(2/5)mR^{2}](v_{cm} / R)^{2} where v_{cm} is the (linear) speed of the center of mass
v_{cm} = √[(10/7)gh]
substituting ω = v_{cm} /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)

ω τ α π δ Δ θ φ μ √ º θ ρ ≤ ≥ λ
