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MR. SURRETTE
VAN NUYS HIGH SCHOOL
CHAPTER 7: ROTATION (PART 2)
TORQUE CLASS NOTES
TORQUE
Torque measures the force that causes a body to rotate about a specified axis.
TORQUE EQUATIONS
The torque equations develop from Newton’s second law:
 = I
F x r
(“F x r” means the force component perpendicular to the radius arm.
TORQUE AND NEWTON’S 2ND LAW
Notice the following comparisons between
 = I and F = ma:

 (torque)
~ F x r(force x radius)
I (moment of inertia)
~ m (mass)
 (angular acceleration)
~ a (acceleration)
Examples 1 - 4. A pulley that rotates about a fixed axis O has string wrapped around it and attached to
a mass m that hangs vertically as shown. T is the tension in the string.
Examples 1 - 4. Diagram
1. According to the diagram, what is the net torque on the pulley about O?
1A.
(1)  = F x r
(2)  = TR
2. Choose the rotational equation that applies:
(A) TR = I
(B) T = I
(C) T/R = I
(D) TR = I
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PHYSICS
MR. SURRETTE
VAN NUYS HIGH SCHOOL
2A.
(1)  = F x r
(2)  = I
(3) F x r = I
(4) TR = I(from 1.)
3. According to the diagram, what is the net downward force (Fy) on mass m?
3A.
(1) FNET = ma
(2) T – mg = - ma
(3) -1 (T – mg) = -1 (- ma)
(4) mg – T = ma
Fy: mg – T
4. Express Fy in terms of torque.
4A.
(1) Fy: mg – T (from 3.)
(2)  = TR (from 1.)
(3) T =  / R
(4) Fy: mg – ( / R)
MOMENT OF INERTIA
The moment of inertia of a linear body depends on the distribution of mass relative to the axis of
rotation:
I = mr2)
Example 5. A majorette takes two batons and fastens them together in the middle at right angles to
make an “x” shape. Each baton is 0.8 m long and each ball on the end is 0.30 kg. (Ignore the mass of
the rods.) What is the moment of inertia if the arrangement is spun around an axis formed by one of the
batons?
5A.
(1) I = mr2)
Since there are two batons:
(2) I = 2 [(0.30 kg)(0.4 m)2]
(3) I = 0.096 kg.m2
MOMENT OF INERTIA
The moments of inertia for non-linear bodies are determined by experimental data. Some common
moments of inertia are:
CYLINDER:
I = ½ mr2)
HOOP:
I = mr2)
SPHERE:
I = 2/5 mr2)
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PHYSICS
MR. SURRETTE
VAN NUYS HIGH SCHOOL
ANGULAR MOMENTUM
Angular momentum develops from the concept of linear momentum:
L = I
L = mvr
TYPES OF MOMENTUM
Notice the following comparisons between
L = I and p = mv:
L (angular momentum)
~ p (momentum)


I (moment of inertia)
~ m (mass)


 (angular velocity)
~ v (velocity)
Examples 6 - 8. Two particles move in opposite directions about their center of mass, a point halfway
between them. The CM (center of mass) is 4 m from each particle.
Examples 6 - 8. Diagram
6. Compute the angular momentum of one 2 kg particle about the axis through the CM in kg.m2/s
units. Take counterclockwise as positive.
6A.
(1) L = I
(2) Find the value of I: I = mr2
(3) I = (2 kg)(4 m)2
(4) I = 32 kg.m2
(5) Find the value of : v = R
(6)  = v / R
(7)  = (- 15 m/s) / (4 m)
(8)  = - 3.75 rad /s
(9) L = (32 kg.m2)(- 3.75 rad /s)
(10) L = - 120 kg.m2/s
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PHYSICS
MR. SURRETTE
VAN NUYS HIGH SCHOOL
7. Compute the angular momentum of the other 2 kg particle about the axis through the CM in kg.m2/s
units.
7A.
L = - 120 kg.m2/s (same steps as 6.)
8. Compute the total angular momentum of both particles about the axis through the CM in kg.m2/s
units.
8A.
(1) LT = LM1 + LM2
(2) LT = - 120 kg.m2/s – 120 kg.m2/s
(3) LT = - 240 kg.m2/s
Examples 9 - 12. A girl of mass 60 kg runs to the left at 5 m/s and jumps on the rim of a merry-goround that is rotating clockwise at 4 rad/s. The radius R is 3 m and the moment of inertia about the
center is 400 kg.m2.
Examples 9 - 12. Diagram
9. Let counterclockwise momentum be positive. What is the girl’s angular momentum about O in units
of kg.m2/s?
9A.
(1) L = mvR
(2) L = (60 kg)(5 m/s) (3 m)
(3) L = - 900 kg.m2/s (the girl adds clockwise momentum)
10. What is the merry-go-round’s angular momentum about O before the girl jumps on it (in units of
kg.m2/s)?
10A.
(1) L = I
(2) L = (400 kg.m2)(- 4 rad/s)
(3) L = - 1600 kg.m2/s
11. What is the total angular momentum of the system of girl plus merry-go-round in units of kg.m2/s?
11A.
(1) LT = LM + LG
(2) L = - 1600 kg.m2/s – 900 kg.m2/s
(3) L = - 2500 kg.m2/s
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PHYSICS
MR. SURRETTE
VAN NUYS HIGH SCHOOL
12. Compute the moment of inertia of the whole system with the girl on the rim in kg.m2 units.
12A.
(1) IT = IM + IG
(2) IG = mR2)
(3) IG = (60 kg)(3 m)2
(4) IG = 540 kg.m2
(5) IT = (400 kg.m2) + (540 kg.m2)
(6) IT = 940 kg.m2
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PHYSICS
MR. SURRETTE
VAN NUYS HIGH SCHOOL
CHAPTER 7: ROTATION (PART 2)
ROTATIONAL DYNAMICS CLASS NOTES
CONSERVATION OF ANGULAR MOMENTUM
When the net external torque acting on a system is zero, the angular momentum of the system is
conserved:
LI = LF
III = IFF
CONSERVATION OF ANGULAR MOMENTUM
The conservation of angular momentum can also be expressed in terms of one object:
I11 = I1’1’
Example 1. A phonograph turntable has a moment of inertia of 2.50 x 10-2 kg.m2 and spins freely on a
frictionless bearing at 33.3 rev/min. A 0.25 kg ball of putty is dropped vertically on the turntable and
sticks at a point 0.20 m from the center. What is the new rate of rotation (rev/min) of the 0.20 m
system?
1A.
(1) I11 = I1’1’
(2) 1 = (33.3 rev / min)(1 min / 60 sec)(2 rad / 1 rev)
(3) 1 = 3.49 rad/s
(4) I1’ = ITT + IP
(5) I1’ = ITT + mr2)
(6) 1’ = (I1)(1) / (I1’)
(7) 1’ = [(2.50 x 10-2 kg.m2)(3.49 rad/s)] / [(2.50 x 10-2 kg.m2) + (0.25 kg)(0.20 m)2]
(8) 1’ = 2.49 rad/s
(9) 1’ = (2.49 rad/s) (60 s / 1 minute)(1 rev / 2)
(10) 1’ = 23.8 rev/min
ROTATIONAL KINETIC ENERGY
The concept of energy can be used to analyze rotational motion:
KR = 1/2 I2
[Compare to translational (linear) kinetic energy: K = ½ mv2]
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PHYSICS
MR. SURRETTE
VAN NUYS HIGH SCHOOL
Example 2. A solid cylinder with mass, M, and radius, R, rolls along a level surface without slipping
with a linear velocity, v. What is the ratio of its rotational kinetic energy to its linear kinetic energy?
(For a solid cylinder, I = ½ mr2).)
2A.
(1) KR = ½ I2
(2) KR = ½ (1/2 MR2)(v/R)2
(3) KR = 1/4 Mv2 (rotational kinetic)
(4) K = ½ Mv2 (translational kinetic)
(5) KR ~ K = 1/4 ~ 1/2
(6) 0.25 ~ 0.5
(7) KR ~ K = 1 : 2
ROTATIONAL POWER
The power concept is also useful in simplifying the analysis of rotational motion:
P = KR / t
Example 3. A bus is designed to draw its power from a rotating flywheel that is brought up to its
maximum speed (3000 rpm) by an electric motor. The flywheel is a solid cylinder of mass 1000 kg and
diameter 1 m (ICYLINDER = ½ mr2)). If the bus requires an average power of 10 kilowatts, how long
will the flywheel rotate?
3A.
(1) KR = ½ I2
(2) KR = ½ (1/2 mr2)2
(3) KR = ¼ mr22
(4) Solve for :
(3000 rev/min)(2 rad / rev)(1 min / 60 s)
(5)  = 314 rad / s
(6) KR = ¼ (1000 kg)(0.5 m)2(314 rad/s)2
(7) KR = 6.162 x 106 J
(8) P = KR / t
(9) tP = KR
(10) t = KR / P
(11) t = 6.162 x 106 J / 1.0 x 104 watts
(12) t = 616 s
CONSERVATION OF MECHANICAL ENERGY
The conservation of mechanical energy can be applied to rotational systems:
E = K + KR
STATIC EQUILIBRIUM (re-visited)
In the Dynamics chapter we learned that non-moving objects are in static equilibrium. In other words,
the sum of the forces acting on a stationary object must equal 0: F = 0.
STATIC EQUILIBRIUM (re-visited)
It turns out that this definition is incomplete. To ensure that objects are in static equilibrium, they must
not rotate either. The sum of the torques acting on a stationary object must also equal 0:  = 0.
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PHYSICS
MR. SURRETTE
VAN NUYS HIGH SCHOOL
STATIC EQUILIBRIUM (re-visited)
In order to avoid spin, clockwise rotation about a pivot must equal the counter-clockwise torque around
the same pivot.
TORQUE EQUATIONS
In general,  = 0 can be expressed:
cw = ccw
(clockwise torque [N.m]= counter-clockwise torque [N.m])
TORQUE EQUATIONS
Specifically,  = 0 can be expressed:
FCW x r = FCCW x r
(force in the clockwise direction [N] x distance from pivot [m] =
force in the counter-clockwise direction [N] x distance from pivot [m])
FREE-BODY DIAGRAMS
Free-body diagrams can be used to analyze both conditions of static equilibrium.
Examples 4 - 5. A uniform 10 m long plank weighing 150 N is supported by two blocks, A and B, as
shown:
4. What is the upward force supplied by block A?
4A. Solve in two steps.
Step 1:
(1) F = 0
(2) FUP = FDOWN
(3) FBLOCK A + FBLOCK B = 150 N
Step 2:
(4)  = 0
(5) CW = CCW
Choose block B as a pivot:
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PHYSICS
MR. SURRETTE
VAN NUYS HIGH SCHOOL
4A. (continued…)
(6) FCW x r = FCCW x r
(7) (FBLOCK A)(8 m) = (FBOARD)(5 m)
(8) (FBLOCK A)(8 m) = (150 N)(5 m)
(9) FBLOCK A = (150 N)(5 m) / (8 m)
(10) FBLOCK A = 94 N
5. A 10 N can of paint is placed on the plank directly over block B. What is the upward force supplied
by block B?
5A.
Before the paint can was added, the upward force supplied by block B was previously FBLOCK B = 150 N
– 94 N = 56 N. Adding the can over block B supplies no additional torque if we still use B as our pivot.
Therefore, now FBLOCK B = 56 N + 10 N = 66 N.
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PHYSICS
MR. SURRETTE
VAN NUYS HIGH SCHOOL
CHAPTER 7: ROTATION (PART 2)
GRAVITY CLASS NOTES
GRAVITY
There are three major features to Newton’s Universal Law of Gravitation.
1. Gravity is an action-at-a-distance force that always exists between two particles regardless of the
medium that separates them.
2. The force varies as the inverse square of the distance between the particles.
3. The force is proportional to the product of their masses.
UNIVERSAL LAW OF GRAVITATION
FG = - (Gm1m2) /r2
(G = constant = 6.67 x 10-11 m3/kg.s2)
(The negative sign indicates that gravity is always an attractive force)
Example 1. Two lead balls of mass 5 kg and 10 kg are 1.41 m apart. Calculate the magnitude of the
gravitational force they exert on one another.
1A.
(1) FG = Gm1m2/r2
(2) (6.67 x 10-11 m3/kg.s2)(5 kg)(10 kg) / (1.41 m)2
(3) FG = 1.67 x 10-9 N
GRAVITY AS CENTRIPETAL FORCE
Example 2. An artificial satellite of mass m travels at a constant speed in a circular orbit of radius r
around the Earth mass M. What is the speed of the satellite?
2A.
(1) FG = Gm1m2/r2
(2) FG = GmM/r2
(3) Fc = mv2/r
(4) Set Fc to FG: Gm1m2/r2 = mv2/r
(5) (r)GmM/r2 = (r)mv2/r
(6) GmM/r = mv2
(7) GM/r = v2
(8) v2 = GM/r
(9) v = (GM/r)1/2
GRAVITATIONAL POTENTIAL ENERGY (re-visited)
Earlier we learned that gravitational potential energy Ug can be calculated: Ug = mgh. This is true for
objects near or at the surface of the earth. This equation does not hold true for satellites high above the
surface of the earth.
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PHYSICS
MR. SURRETTE
VAN NUYS HIGH SCHOOL
GRAVITATIONAL POTENTIAL ENERGY
Satellites that orbit the earth have both kinetic and gravitational potential energy. Their kinetic energy K
is calculated: K = ½ mv2. Their gravitational potential energy is calculated using a variation of
Newton’s law of gravitation.
GRAVITATIONAL POTENTIAL ENERGY
UG = - (Gm1m2) / r
(r = height above the earth)
Example 3. A satellite of mass m is in a circular orbit of radius R around the earth (mass M). What is
its total mechanical energy?
3A.
(1) FC = mv2 / r
(2) FG = - GmM / r2
(3) FC = FG
(4) mv2 / r = - GmM /r2
(5) r [ mv2 / r = - GmM / r2]
(6) mv2 = - GmM / r
(7) K = ½ mv2
(8) ½ [mv2 = - GmM / r]
(9) ½ mv2 = - GmM / 2r
(10) K = | - GmM / 2r | (kinetic energy is scalar and positive)
(11) K = GmM / 2r
(12) E = K + U
(13) E = GmM /2r – GmM / r
(14) E = - GmM / 2r
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PHYSICS