GRABSUM School Inc.

SENIOR HIGH SCHOOL DEPARTMENT

Name: Grade level and section: 12 - STEM

Unit: 2 Subject: General Physics 1 Module: 8-9

ROTATIONAL EQUILIBRIUM AND ROTATIONAL DYNAMICS

Learning Competencies

At the end of the lesson, you will be able to:

 Calculate the moment of inertia about a given axis of single-object and multiple-object systems
 Calculate magnitude and direction of torque using the definition of torque as a cross product
 Describe rotational quantities using vectors
 Determine whether a system is in static equilibrium or not
 Apply the rotational kinematic relations for systems with constant angular accelerations
 Determine angular momentum of different systems
 Apply the torque-angular momentum relation

Overview

This module was designed and written with you in mind. The scope of this module permits it to be
used in many different learning situations. The language used recognizes the diverse vocabulary level of
students. The lessons are arranged to follow the standard sequence of the course.

Introduction

Have you ever wondered why tornadoes are so devastating? Is it the speed of the cyclone that
engulfs the surroundings or there is something else to it! Well, a tornado is a mixture of force, power,
and energy. These govern the rotational motion of a tornado, resulting in destructions.
We come across many objects that follow rotational movements. No matter whether fixed
or moving, these objects follow special dynamism which lets them perform their specific activity. Whether it is a
ceiling fan or a potter’s wheel, these rotating objects are a system of particles that consider the motion as a
whole. In the introduction to rotational dynamics of a system, we shall emphasize on the center of mass of that
particle and use the same in understanding motion as a whole.
Before going deeper into the subject, we should first understand the term “Extended body”. When
we refer to an object as an extended body we intend to signify it as a system of particles. Rigid bodies are
those bodies with definite shape and size. In rigid bodies, the distance between the constituting pairs of particles
does not change.

Moment of Inertia

Moment of inertia, also known as rotational inertia, is the rotational analog for mass and is
represented by a capital letter I. it is defined as the property of a rotating body to resist change in its state
of rotation. The larger the moment of inertia, the greater the resistance it offers to angular acceleration. The
SI unit for moment of inertia is kg•m2.
The moment of inertia I of a particle about an axis is obtained by multiplying the mass m by the
square of its distance r from the axis.

For a system made up of several particles, the moment of inertia of the system I, is the sum of the
individual moments of inertia.
+

Radius of gyration (k) is the distance from an axis of rotation where the mass of a body may be
assumed to be concentrated without altering the moment of inertia of the body about that axis. Radius of
gyration is analogous to the center of mass.

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 The moments of inertia of composite bodies of simple geometric shapes about a specified axis
are given below. The specified axes, except for the rod, are generally taken along the axes of symmetry
that is, running through the center of mass resulting in symmetrical mass distribution.

EXAMPLE PROBLEM 1:

A 100 g ball is connected to one end of a cord with a length of 30

cm. What is the moment of inertia of ball about the axis of rotation AB? Ignore
cord’s mass.

SOLUTION:

EXAMPLE PROBLEM 2:

A familiar sight during a town fiesta is a majorette twirling a

baton in her hand and leading a group of musicians. Suppose the baton
consists of a metal ball of mass of 0.25 kg and diameter of 0.12 m at each
end of a slim rod as shown here. The rod is 0.50 m long. Find the moment
of inertia and radius of gyration of the baton about an axis perpendicular
to the rod and intersecting it at (a) the middle and (b) 0.4 m from one
end of the rod. Assume that the rod has negligible mass.

SOLUTION:
We have a 0.5 m long baton with the following attached
masses and diameters: To calculate the moment of
inertia and the radius of gyration about the axis indicated, we shall use and .

a. Axis at the middle or at axis a

+

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 b. Axis 0.4 m from one end or at axis b
+

EXAMPLE PROBLEM 3:
What is the moment of inertia of a 2 kg long uniform rod with a length of 2 m? the axis of rotation
is located at one end of the rod.

SOLUTION:
The formula of the moment of inertia when the axis of rotation is located at one end of the rod:

CHECKPOINT 1!
A 100-g ball, m1, and a 200-g ball, m2, are connected by a rod with a length of 60 cm. The mass of
the rod is ignored. The axis of rotation is located at the center of the rod. What is the (a) moment of
inertia of the balls about the axis of rotation (b) and its radius of gyration?

CHECKPOINT 2!

Find the moment of inertia of a 20 kg solid sphere with a length of 0.1 m. The axis of rotation is located
at the center of the sphere.

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 Torque
To understand the concept of torque, we begin by investigating the rotation of a door hinged at
one edge. Try pulling the door with a constant force applied at different points (a) at varying angles (b).
This simple activity shows that it is easiest to rotate the door when the force is applied farthest from the
hinge. That is the reason why doorknobs are placed at the other edge of the door, farthest from the hinge.
Furthermore, rotation is greatest when force s applied perpendicularly to the door.

The effectiveness of a force in rotating a body on which it acts is called torque or moment of
force. Torque can be determined by multiplying the force applied F by the perpendicular displacement of
its line of action from the pivot point. This perpendicular displacement is called moment arm or leave arm,
represented by r.

To find the lever arm, the line of action of force should be extended by drawing a line from the
axis of rotation perpendicular to the line of action of force. A right triangle is then obtained. The lever arm is
found to be ( )

Where is the angle between the line of action of force and the distance from axis rotation.
Maximum torque can be obtained if .
An equivalent way of finding the torque associated with a given force F is to resolve the force
into components parallel and perpendicular to the axis of rotation. The parallel component exerts no
torque because its lever arm is zero. The perpendicular component of the force produces torque.
Thus, we can generalize that the magnitude of the torque produced by a force F at a distance r
from the axis of rotation is given by

This formula is consistent with the definition of torque as a cross product of force and
displacement.
The SI unit of torque is mete-newton (m•N). Torque is a vector quantity. Torque may also be
positive or negative, depending on the sense of rotation. By conversion, torque is positive if it tends to
produce a counterclockwise rotation, and it is negative if it tends to produce a clockwise rotation. The
greater torque applied to an object, the greater its tendency to rotate.

EXAMPLE PROBLEM 4:
A force of 5.0 N is applied at the end of a lever that has a length of 2.0 m. If
the force is applied directly perpendicular to the lever, as shown in the above diagram,
what is the magnitude of the torque acting the lever?

SOLUTION:
This is a simple matter of plugging the values into the equation:

EXAMPLE PROBLEM 5:
If the same force as in example problem 4 is applied at an angle of 30⁰ at
the end of the 2.0 m lever, what will be the magnitude of the torque?

SOLUTION:

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 CHECKPOINT 3!

A 3.0kg mass is place 2.00m to the right of the pivot point of a see-
saw. What is the magnitude and the sign of the torque applied?

CHECKPOINT 4!

What force is necessary to generate a 20 N•m torque at an angle of 50⁰ from along a 3.0 m rod?

Angular Momentum
By analogy to the definition of linear momentum (as the product of mass and velocity of a
body), angular momentum (L) of a body rotating about a fixed axis is defined as the product of its moment
of inertia (I) about this axis and its angular velocity (ω).

The SI unit of angular momentum is kg•m2/s. Note that the unit radian is omitted. In the absence
of a net external torque, the total angular momentum of a system is conserved. The initial angular
momentum (LO) of the system is equal to its final angular momentum (L).

EXAMPLE PROBLEM 6:
An object with the moment of inertia of 2 kg•m2 rotates at 1 rad/s. What is the anguluar
momentum of the object?

SOLUTION:

EXAMPLE PROBLEM 7:
A 2-kg solid sphere with a radius of 0.2m rotates at 4 rad/s. What is the angular momentum of the ball?

SOLUTION:

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 CHECKPOINT 5!

A 2-kg cylinder pulley about the solid axis with a radius of 0.1 m rotates at a constant angular
speed of 2 rad/s. What is the angular momentum of the pulley?

Rational Work and Kinetic Energy

Work done in rotation is the product of torque and angular displacement. Power is work done
divided by time.

The kinetic energy of a rotating rigid body (KR) is

For a rolling body, the total kinetic energy (KT) is the sum of its translational kinetic energy (K) and
rotational kinetic energy (KR).

EXAMPLE PROBLEM 8:
When you open the lid of spin dryer, a retarding torque of 12.5 N•m stops the dryer after 8 rev.
(a) How much work is done by the torque to stop the dryer? (b) If the dryer stops at 4.0 s, how much power
is done by the stopping device that provided the torque?

SOLUTION:
We are given the following T= 12.5 N•m, = 8 rev, and t=4.0s. First, we need to convert 8 rev to radians.

The work done is computed. Power is obtained by dividing work by time.

a.
b. =

CHECKPOINT 6!

A motor exerts torque on a carousel for it to attain a speed of 2.5 rev/s starting from rest in 3.5 s.
Find the work done by the motor if the carousel has a radius of 27 m and a mass of 1.75 x105 kg.
Consider the carousel to be a thin walled cylinder about its center.

References:
Baltazar & Tolentino. General Physics 1 Senior High School. 2017. Phoenix Publishing House Inc. 927 Quezon Avenue, Quezon City

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