Chapter 9: Torque and Rotation
 9.1 Torque
 9.2 Center of Mass
 9.3 Rotational Inertia
Chapter 9 Objectives






Calculate the torque created by a force.
Solve problems by balancing two torques in rotational
equilibrium.
Define the center of mass of an object.
Describe a technique for finding the center of mass of an
irregularly shaped object.
Calculate the moment of inertia for a mass rotating on the
end of a rod.
Describe the relationship between torque, angular
acceleration, and rotational inertia.
Chapter 9 Vocabulary












angular acceleration
center of gravity
center of mass
center of rotation
lever arm
line of action
moment of inertia
rotational
equilibrium
rotational inertia
torque
translation
Inv 9.1 Torque
Investigation Key Question:
How does force create rotation?
9.1 Torque
 Motion in which an entire
object moves is called
translation.
 Motion in which an object
spins is called rotation.
 An object can both rotate
and translate.
9.1 Torque
 A torque is an action
resulting from a force that
causes objects to rotate.
 Torque is not the same
thing as force.
 Torque is created by
force, but torque also
depends on the point of
application and the
direction of the force.
A door pushed at its handle will
easily turn and open, but a door
pushed near its hinges will not
move as easily.
9.1 Torque
 The point or line about which an
object turns is its center of
rotation.
 A force applied far from the
center of rotation produces a
greater torque than a force
applied close to the center of
rotation.
 Doorknobs are positioned far
from the hinges to provide the
greatest amount of torque for a
given force.
9.1 Torque
 Torque is created when
the line of action of a
force does not pass
through the center of
rotation.
 The line of action is an
imaginary line that
follows the direction of a
force and passes though
its point of application.
9.1 Torque
 To get the maximum
torque, the force should
be applied in a direction
that creates the greatest
lever arm.
 The lever arm is the
perpendicular distance
between the line of action
of the force and the
center of rotation
9.1 The torque created by force
 The torque created by a force is equal to the
lever arm length times the magnitude of the
force.
 Torque is usually represented by the lower case
Greek letter “tau,” t.
 When calculating torque, if the line of action
passes through the center of rotation, the lever
arm is zero, and so is the torque, no matter how
large a force is applied.
9.1 Calculating Torque
Lever arm length (m)
Torque (N.m)
t=rxF
Force (N)
9.1 Calculating torque
 If more than one torque
acts on an object, the
torques are combined to
determine the net torque.
 If the torques make an
object spin in the same
direction, clockwise or
counterclockwise, they are
added together.
9.1 Units of torque
 The units of torque are force times distance, or
newton-meters.
 A torque of 1 N⋅m is created by a force of 1
newton acting with a lever arm of 1 meter.
 Torque is always calculated around a particular
center of rotation.
Calculating a torque
A force of 50 newtons is applied to a
wrench that is 30 centimeters long.
Calculate the torque if the force is applied
perpendicular to the wrench so the lever
arm is 30 cm.
1.
You are asked to find the torque.
2.
You are given the force and lever arm.
3.
Use: t = r F.
4.
4. Solve: t = (-50 N)(0.3 m) = -15 N⋅m
9.1 Rotational Equilibrium
 When an object is in rotational equilibrium, the net
torque applied to it is zero.
 Rotational equilibrium is often used to determine
unknown forces.
 What are the forces (FA, FB) holding the bridge up at
either end?
9.1 Rotational Equilibrium
 For the bridge not to move up
or down, the total upward
force must equal the total
downward force.
 This means FA + FB = 1,250 N.
9.1 Rotational Equilibrium
 For the bridge to be in rotational equilibrium, the total
torque around any point must also be zero.
 If we choose the left end of
the bridge, and set the total
of the remaining torques to
zero, the force on the right
support (FB) is calculated to
be 400 newtons.
 Since the total of both
forces must be 1,250 N, that
means the force FA on the
left must be 850 N.
Solve a rotational equilibrium
problem
A boy and his cat sit on a seesaw. The cat has a mass of 4 kg and sits 2 m
from the center of rotation. If the boy has a mass of 50 kg, where should he
sit so that the see-saw will balance?
1.
2.
3.
4.
5.
You are asked to find the boy’s lever arm.
You are given the two masses and the cat’s lever arm.
Use: t = r F, and F = mg
Solve for the cat: t = (2 m)(4 kg)(9.8 N/kg) = +78.4 N⋅m
Solve for the boy: t = (d )(50 kg)(9.8 N/kg) = -490 d

For rotational equilibrium, the net torque must be zero.

0= 78.4 – 490d d = 0.16 m. The boy must sit 16 cm from center.
9.1 When the force and lever arm are NOT
perpendicular
 When the force and lever arm are not
perpendicular, an extra step is required to
calculate the length of the lever arm.
Calculating a torque from an
angled force
A 20-centimeter wrench is used to loosen a bolt.
The force is applied 0.20 m from the bolt.
It takes 50 newtons to loosen the bolt when the force
is applied perpendicular to the wrench.
How much force would it take if the force was applied
at a 30-degree angle from perpendicular?
1.
2.
3.
4.
You are asked to find the force.
You are given the force and lever arm for one condition.
Use: t = r F.
Solve for the torque required to loosen the bolt
 t = (50 N)(0.2 m) = 10 N⋅m
5.
Calculate the torque with a force applied at 30 degrees:
 10 N⋅m = F × (0.2 m) cos 30o = 0.173 F
 F = 10 N⋅m ÷ 0.173 = 58 N. It takes a larger force at 30o.
Chapter 9: Torque and Rotation
 9.1 Torque
 9.2 Center of Mass
 9.3 Rotational Inertia
Inv 9.2 Center of Mass
Investigation Key Question:
How do objects balance?
9.2 Center of Mass
 There are three different axes about
which an object will naturally spin.
 The point at which the three axes
intersect is called the center of mass.
9.2 Finding the center of mass
 If an object is irregularly shaped, the center of mass
can be found by spinning the object and finding the
intersection of the three spin axes.
 There is not always material at an object’s center of
mass.
9.2 Finding the center
of gravity
 The center of gravity of an
irregularly shaped object can
be found by suspending it
from two or more points.
 For very tall objects, such as
skyscrapers, the acceleration
due to gravity may be
slightly different at points
throughout the object.
9.2 Balance and the center of mass
 Objects balance because the torque caused by
the force of the object’s weight is equal on each
side.
9.2 Balance and the center of mass
 For an object to remain upright, its center of
gravity must be above its area of support.
 The area of support includes the entire region
surrounded by the actual supports.
 An object will topple over if its center of mass
is not above its area of support.
Chapter 9: Torque and Rotation
 9.1 Torque
 9.2 Center of Mass
 9.3 Rotational Inertia
Inv 9.3 Rotational Inertia
Investigation Key Question:
Does mass resist rotation the
way it resists acceleration?
9.3 Rotational Inertia
 Previously inertia was defined
as an object’s resistance to a
change in its motion.
 Rotational inertia is the term
used to describe an object’s
resistance to a change in its
rotational motion.
 An object’s rotational inertia
depends not only on the total
mass, but also on the way
mass is distributed.
9.3 Rotational Inertia
 A rotating mass on a rod can be described with
variables from linear or rotational motion.
9.3 Rotational Inertia
 To equate rotational motion to linear motion,
the force is replaced by the torque about the
center of rotation.
 The linear acceleration is replaced by the
angular acceleration.
9.3 Linear and Angular Acceleration
Angular acceleration
(kg)
Linear
acceleration
(m/sec2)
a=ar
Radius of motion
(m)
9.3 Rotational Inertia
 The product of mass × radius squared (mr2) is
the rotational inertia for a point mass where r is
measured from the axis of rotation.
9.3 Moment of Inertia
 The sum of mr2 for all the particles of mass in
a solid is called the moment of inertia (I).
 A solid object contains mass distributed at
different distances from the center of rotation.
 Because rotational inertia depends on the
square of the radius, the distribution of mass
makes a big difference for solid objects.
9.3 Moment of Inertia
9.3 Rotation and Newton's 2nd Law
 If you apply a torque to a wheel, it will spin in the
direction of the torque.
 The greater the torque, the greater the angular
acceleration.
9.3 Rotation and Newton's 2nd Law
 Force causes linear acceleration, torque causes
angular acceleration.
 Mass is a measure of linear inertia.
 The moment of inertia (I ) is a measure of
rotational inertia.
Bicycle Physics
 A modern bicycle is the
most efficient machine
ever invented for
turning human muscle
power into motion.
 Bicycles work by a
series of
transformations from
forces to torques and
back.