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Chapter 9
ROTATIONAL DYNAMICS
Definition of Torque

Torque is the quantity that measures the ability of a force to rotate
an object around some axis (fulcrum).
◦ SI Units – Nm
 Imagine a cat trying to leave a house by pushing perpendicularly on a cat-flap door, the
door is free to rotate around a line that passes through the hinge. This is the door’s axis
of rotation.
 If a cat pushed on the door with a force at a point closer to the hinge, the door would be
difficult to rotate.
Definition of Torque
Lever arm is the perpendicular distance
from the axis of rotation to a line drawn
along the direction of the force.
 Torque depends on Force and the length
of the lever arm.

◦ Torque = Force x lever arm
◦  = Fr
◦ Only the perpendicular component of the
force produces a torque.
Definition of Torque
The lever arm is the distance between the point where the
force is applied and the axis of rotation (fulcrum).
 Forces do not have to act perpendicular to the object. If they
are at an angle, more force will be required to produce the
desired outcome

◦ Torque = Fr(sinθ)
Torque is a vector quantity
 Torque resulting from a force is positive if the rotation is
counterclockwise and negative if the rotation is clockwise.

9.1 The Action of Forces and Torques on Rigid Objects
DEFINITION OF TORQUE
Magnitude of Torque = (Magnitude of the force) x (Lever arm)
  Fr
Direction: The torque is positive when the force tends to produce a
counterclockwise rotation about the axis.
SI Unit of Torque: newton x meter (N·m)
Balanced Torques

Given two children on a seesaw, the torque of each child
can be calculated
◦ Force of child x distance from pivot

Each child produces their own torque which may result
in a state of equilibrium.
◦ Torque1 = Torque 2
◦ 𝐹1 × 𝑟1 = 𝐹2 × 𝑟2
Example problem 1
•
A basketball is being pushed by two players during tip-off. One player exerts a downward force
of 11 N at a distance of 7.0 cm from the axis of rotation. The second player applies an upward
force of 15 N at a perpendicular distance of 14 cm from the axis of rotation. Find the net
torque acting on the ball.

G: F1 = -15 N, r1 = 0.14 m, F2 = -11 N, r2 = 0.07m

U: ∑ = ?

E:  = Fr
◦ ∑ =1 + 2 = F1r1 + F2r2

S:

S:
= (-15N x 0.14m) + (-11N x 0.07m)
net = -2.87 N.m
SECTION 2
ROTATION AND
INERTIA
Center of Mass

The point at which all the mass of a body can be
considered to be concentrated during translational motion
is defined as center of mass.
◦ Regularly shaped objects such as a spheres or cubes, will have
their center of mass located at the geometric center of the
object.
◦ For more complicated objects, their centers of gravity is more
difficult to locate and is beyond the scope of our book.

The position at which the gravitational force acts on an
object is called the center of gravity of the object
Moment of Inertia (MOI)
Objects can rotate in many ways, some ways more freely
than others.
 Moment of inertia is the tendency of a body to rotate
freely about a fixed axis to resist a change in rotational
motion.

◦ Units: kg.m2
◦ MOI is similar to mass because they are both forms of inertia
Moment of Inertia (MOI) - I
Mass is an intrinsic property – depends on the type and
quantity of matter
 MOI depends on the objects mass as well as the
distribution of the mass around the axis if rotation
(shape)
 The farther the mass is, on average from the axis of
rotation, the greater the object’s MOI and the more
difficult it is to rotate the object.

Calculating the MOI

Pg. 285: Table 8-1 has equations/formulas
for a few common shapes.
Calculating the MOI
M – mass in kilograms
 R – radius in meters
 l – length in meters


Notice that the MOI for a solid sphere is smaller than
the MOI for a thin spherical shell or a hoop

Net torque = MOI x Angular acceleration
 net  I

Remember that if the net torque is zero, the
object can still be rotating at a constant speed.
Newton’s Second Law For Rotation
Synonymous with Newton’s second law, there is a
relationship between the net torque on an object
and the angular acceleration given to the object.
 F = ma
 We know in a rotational system, torque is a function
of force, also:

◦ MOI replaces the mass and
◦ Acceleration is replaced with angular acceleration.
Example Problem #1

A student tosses a dart using only the rotation of her forearm
to accelerate the dart. The forearm rotates in a vertical plane
about an axis at the elbow joint. The forearm and dart have a
combined moment of inertia of 0.075 kg•m2 about the axis,
and the length of the forearm is 0.26 m. If the dart has a
tangential acceleration of 45 m/s2 just before it is released,
what is the net torque on the arm and dart?
Example Problem #1
(l =lever arm = radius of rotation)

 net  I
Example Problem #2
(l =lever arm = radius of rotation)

Simon decides to ride the ‘Gravitron’, it has a radius of 3 m,
if his mass is 79.4 kg. What is the net torque produced
when his angular acceleration is 7 rad/sec2?
G: l = 3m, α = 7 rad/s2, m = 79.4 kg
 U:  = ?
 E:  net  I
 S: Point mass moving about an axis, its MOI =MR2

◦  = (79.4kg x (3m)2) x 7rad/sec^2
◦ = 5002 N.m
Rotational Equilibrium

For an object to be in complete equilibrium, it requires
zero net force and zero net torque.
◦ 1st condition of equilibrium:
 Achieve Zero net force, all the forces should add up to zero.
◦ 2nd condition of equilibrium:

 To achieve Zero net torque, it is first necessary to choose an axis of
rotation around which to calculate the torque.
The 1st Condition leads to two equations: One for the x direction and
one for the y.
◦ Implement
 Solve the system of equations.
Conditions of Equilibrium
◦ 1. The net external force must
be zero.
F
EXT
0
◦ 2. The net external torque must
be zero.
τ
EXT
0
Rotational Equilibrium
 It does not matter which axis is chosen,
however, this fact is convenient when solving
problems because an unknown force acting
along this axis will not produce any torque.
 Beginning a diagram by choosing an axis where a force acts
can eliminate an unknown in the problem.
Example Problem # 1

A uniform 5.00 m long horizontal beam that weighs 315 N
is attached to a wall by a pin connection that allows the
beam to rotate. Its far end is supported by a cable that
makes an angle of 53° with the horizontal, and a 545 N
person is standing 1.50 m from the pin. Find the force in
the cable, FT, and the force exerted on the beam by the
wall, R, if the beam is in equilibrium.
Example Problem # 1
 G:
𝐿 = 5.00𝑚, 𝐹𝑔𝑏 = 315𝑁, 𝑓𝑔𝑝 = 545𝑁, 𝑑 = 1.5𝑚, 𝜃 = 53𝑜
 U: 𝐹𝑇 =? 𝑎𝑛𝑑 𝑅 =?
 Lets work on this together on the board
Example problem # 2

In 1960, a polar bear with a mass of 9.00 ×
102 kg was discovered in Alaska. Suppose this
bear crosses a 12.0 m long horizontal bridge
that spans a gully. The bridge consists of a
wide board that has a uniform mass of 2.50
× 102 kg and whose ends are loosely set on
either side of the gully. When the bear is
two-thirds of the way across the bridge,
what is the normal force acting on the board
at the end farthest from the bear?
Example problem # 2
 G: l = 12m, mb = 250kg, mp = 900kg,
g = 9.81 m/s2
 U: Fn1 = ? & Fn2 = ?
 Lets work on this together on the
board
Angular Momentum (L)
Rotating objects have angular momentum
 This is because a rotating object has inertia, and
therefore also possesses momentum associated with its
rotation.
 This momentum is called angular momentum

◦ Angular momentum = MOI x angular speed
◦ L = Iω
◦ Units: kg.m2/s
Law of Conservation of Angular
Momentum

When the net external torque acting on an object or
objects is zero, the angular momentum does not change.
◦ Li = Lf

When a skater brings arms and feet closer to his body,
the MOI of his body decreases and therefore his, angular
speed increases to conserve angular momentum
Example Problem #3

A figure skater jumps into the air to perform a twisting
maneuver. When she first jumps her moment of inertia is
86 kg-m2. While she’s in the air she brings her arms in
and decreases her MOI to 77 kg-m2. If her initial angular
speed was 2 rad/sec what is her final angular speed, if
angular momentum is conserved?
Example Problem #3

Example Problem #4

A 65 kg student is spinning on a merry-go-round that has
a mass of 525 kg and a radius of 2 m. She walks from the
edge of the merry-go-round toward the center. If the
angular speed is initially 0.2 rads/s, what is the angular
speed when the student reaches a point 0.50 m from the
center?
Example Problem #4
 G: M = 525kg, Ri = 2m, m =65 kg, Rf = 0.5m,
wi = 0.2 rad/s
U: wf = ?
 E:

Li = Lf
Lm,i + Ls,i = Lm,f + Ls,f

◦ Considering the merry-go-round as a circular disc and the student
as a point mass. Their MOI’s are as follows
 S:
I m  MR
1
2
2
Is,i  mR
2
2
Is,f  mr f
Example Problem #4
Lm ,iS: I mwi 
Lm ,i 
1
2
1
2
MR 2wi
2
(525kg )( 2m ) (0.2 rad / s )
Lm,i  210kg.m2 / s
Ls ,i  I sw i  mR w i
2
Ls,i  (65kg)( 2m)2 (0.2rad / s)
Ls ,i  52kg.m2 / s
Li  Lm,i  Ls ,i
Li  210kg.m2 / s  52kg.m2 / s
Li  262kg.m2 / s
Example Problem #4

S:
L f  Lm, f  Ls , f
L f  I mw f  I s , f w f
L f  MR w f  mrf w f
1
2
Lf 


1
2
2
2

MR 2  mrf w f
2

L f  12 (525kg)( 2m)2  (65kg)(0.5m)2 w f
Lf
 1066.25kg.m w
2
f
Example Problem #4

S:
Li  L f
262kg.m2 / s  1066.25kg.m2 w f
262kg.m2 / s
wf 
1066.25kg.m2

wf = 0.25 rad/s
Section Assignment
Rotational Kinetic Energy (KERot)

This is the energy of an object associated with its angular
speed.

KErot = ½ Iw2

Rotational KE = ½ x MOI x (angular speed)2
Example Problem # 5

Conservation of Mechanical Energy
Remember Mechanical Energy may be conserved.
 ME is the sum of all types of KE and PE.
 ME = KEtrans + KErot + PE

ME = ½ mv2 + ½ Iw2 + mgh
 MEinitial = MEfinal

KEtrans,i  KErot,i  PEi  KEtrans, f  KErot, f  PE f
1
2
1
1
1
mv
2
i  2 Iwi  mghi  2 mv f  2 Iw f  mgh f
2
2
2
Example Problem # 6

A solid ball with a mass of 4.10 kg and a radius of 0.050
m starts from rest at a height of 2.00 m and rolls down a
30.0° slope. What is the translational speed of the ball
when it leaves the incline?
Example Problem # 6

mghi  1 2 mv f  1 2 Iw f  mghf
2
mgh 
1
2

2 mv f
where :w f 
2
vf
r
1
2 Iw f
2
Example Problem # 6

S:
mgh  1 2 mv f  1 2
2

2
5
mr
2
mgh  2 mv f  5 mv f
2
1
mgh 
7
10
mv f
1
 
vf 2
r
2
2
v f  10 7 gh
2
vf 
 vf
10
= 5.29 m/s
7
(9.81m / s ^ 2)( 2m)