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Chapter 9 – Rotational Dynamics
9.1 – The Action of Forces and Torques on Rigid Objects
Definition of Torque:
Torque=(Magnitude of the force)×(Lever arm)
𝜏 = 𝐹ℓ
Direction: The torque is positive when the force tends to produce a counterclockwise rotation about the axis, and
negative when the force tends to produce a clockwise rotation.
SI Unit of Torque: newton∙meter (N∙m)
The line of action is an extended line drawn collinear with the force. The lever arm is the distance ℓ between the
line of action and the axis of rotation, measured on a line that is perpendicular to both.
It is easier to open a door with a force of a given magnitude by pushing it at the outer edge that it is pushing it
closer to the axis of rotation. Pushing it nearly into the hinge makes it very difficult to open the door.
The following diagram shows different variations of the line of action versus the axis of rotation and how the lever
arm corresponds.
In our bodies, muscles and tendons produce torques about various joints.
Homework for Section 9.1: 1-6
9.2 – Rigid Objects in Equilibrium
If a rigid body is in equilibrium, neither its linear motion nor its rotational motion changes.
Equilibrium of a Rigid Body:
A rigid body is in equilibrium if it has zero translational acceleration and zero angular acceleration. In equilibrium,
the sum of the externally applied forces is zero, and the sum of the externally applied torques is zero:
Σ𝐹𝑥 = 0 and Σ𝐹𝑦 = 0
Σ𝜏 = 0
If the object is in equilibrium the choice of the axis of rotation is completely arbitrary, it is already in equilibrium
with respect to any axis whatsoever. Meaning the sum of the external torques is zero, no matter where the axis of
rotation is placed. You typically will choose the location so that the lines of action of one or more of the unknown
forces pass through the axis. This will simplify the torque equation making the math easier.
Remember the direction of your force indicates whether it is positive or negative with respect to your problem.
9.3 – Center of Gravity
Definition of Center of Gravity:
The center of gravity of a rigid body is the point at which its weight can be considered to act when the torque due
to the weight is being calculated. This is abbreviated “cg”.
When an object has a symmetric shape and its weight is distributed uniformly, the center of gravity lies at its
geometric center. This does not mean that the center of gravity has to be on or inside the object for an object like
a cd the center of mass is in the hole in the center of the disk.
Center of Gravity:
𝑊1 𝑥1 + 𝑊2 𝑥2 + ⋯
𝑊1 + 𝑊2 + ⋯
This is very similar to the center of mass equations used previously. For ordinary-sized objects (where the
acceleration due to gravity is uniform), like cars and boats the center of gravity coincides with the center of mass.
𝑥𝑐𝑔 =
Practice problems 12, 13, 16
Homework for Sections 9.2 & 9.3: 11, 14, 15, 17 - 19
9.4 – Newton’s Second Law for Rotational Motion about a Fixed Axis
The equation 𝜏 = (𝑚𝑟 2 )𝛼 is the form of Newton’s second law in terms of rotational motion. It indicates that the
net external torque  is directly proportional to the angular acceleration α. The constant of proportionality
is 𝐼 = 𝑚𝑟 2 , which is called the moment of inertia of the particle. The SI unit of the moment of inertia is kg∙m2.
If all objects were single particles it would be convenient to use 𝐹𝑇 = 𝑚𝑎𝑇 , the advantage to using 𝜏 = 𝐼𝛼 is that
it can be applied to any rigid body rotating about a fixed axis, and not just a particle.
The moment of inertia of a body is given as
𝐼 = Σ𝑚𝑟 2
r is the perpendicular radial distance of each particle from the axis of rotation.
ROTATIONAL ANALOG OF NEWTON’S SECOND LAW FOR A RIGID BODY ROTATING ABOUT A FIXED AXIS
Angular
Moment of
Net external torque = ( inertia ) × (acceleration)
Σ𝜏 = 𝐼𝛼
Requirement: α must be expressed in rad/s2.
When calculating the sum of torques it is necessary to include only the external torques, those applied by agents
outside the body. The torques that are produced by internal forces do not need to be considered because they
always combine to produce a zero net torque.
The farther a particle is from the axis, the greater is its contribution to the moment of inertia. The moment of
inertia depends on the location and orientation of the axis relative to the particles that make up the object.
Rotational motion and translational motion sometimes occur together.
You will need the table of Moments of Inertia for 169 (Table 9.1).
Sections 9.4: Practice problems 28, 32
Homework for 29-31, 33-35
9.5 – Rotational Work and Energy
Translational work is found by using W=Fs (where F is a force and s is the distance). Remember that θ=s/r
therefore, s=rθ. Therefore W=Frθ, but Fr is the torque  applied to the wheel by the tension so the rotational
work can be written as follows:
DEFINITION OF ROTATIONAL WORK
The rotational work WR done by a constant torque  in turning an object through an angle θ is
𝑊𝑅 = 𝜏𝜃
Requirement: θ must be expressed in radians.
SI Unit of Rotational Work: joule (J)
The kinetic energy of the entire rotating body, then, is the sum of the kinetic energies of the particles:
1
1
𝑅𝑜𝑡𝑎𝑡𝑖𝑜𝑛𝑎𝑙 𝐾𝐸 = ∑ (2 𝑚𝑟 2 𝜔2 ) = 2 (∑ mr 2 )ω2
DEFINITION OF ROTATIONAL KINETIC ENERGY
The rotational kinetic energy KER of a rigid object rotating with an angular speed ω about a fixed axis and having a
moment of inertia I is
1
𝐾𝐸𝑅 = 𝐼𝜔2
2
Requirement: ω must be expressed in rad/s.
SI Unit of Rotational Kinetic Energy: joule (J)
Kinetic energy is one part of the object’s total mechanical energy. The total mechanical energy is the sum of the
kinetic and potential energies and obeys the principle of conservation of mechanical energy. We need to
remember that translational and rotational motion can occur at the same time. The total mechanical energy of an
object rolling with both translational and rotational kinetic energies is as follows:
𝐸
=
1
𝑚𝑣 2
2
+
1
𝐼𝜔2
2
+
𝑚𝑔ℎ
Total
mechanical
energy
Translational
kinetic energy
Rotational
kinetic energy
Gravitational
potential energy
Where m is the mass of the object, v is the translational speed of its center of mass, I is its moment of inertia
about an axis through the center of mass, ω is its angular speed, and h is the height of the object’s center of mass
relative to an arbitrary zero level.
Mechanical energy is conserved if Wnc, the net work done by external nonconservative forces and torques, is zero.
If the total mechanical is conserved then the final total mechanical energy equals the initial total mechanical
energy: Eo=Ef
Sections 9.5: Practice problems 44, 47
Homework for 43, 45, 46
9.6 – Angular Momentum
DEFINITION OF ANGULAR MOMENTUM
The angular momentum L of a body rotating about a fixed axis is the product of the body’s moment of inertia I
and its angular velocity ω with respect to that axis:
𝐿 = 𝐼𝜔
Requirements: ω must be expressed in rad/s.
SI Unit of angular Momentum: kg∙m2/s
PRINCIPLE OF CONSERVATION OF ANGULAR MOMENTUM
The total angular momentum of a system remains constant (is conserved) if the net average external torque
acting on the system is zero.
When the total angular momentum is conserved, the final angular momentum Lf equals the initial angular
momentum Lo=Lf
Sections 9.6: Practice problems 53
Homework 52, 54-56