GP2_LAS_Q2_week 1 Rotational Equilibrium.docx

Republic of the Philippines
Department of Education
REGION IV-A CALABARZON
SCHOOLS DIVISION OF BATANGAS
Address: Provincial Sports Complex,Bolbok,4200 Batangas City
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LEARNING ACTIVITY SHEET IN GENERAL PHYSICS I
ROTATIONAL EQUILIBRIUM AND ROTATIONAL DYNAMICS
Name of Learner:
Grade Level: 12
Strand/Track: STEM/ACADEMIC
Section:
Date:
A. Background Information for Learners
This topic demonstrates your understanding on the concepts of Rotational Motion. Rotational
motion can be most easily approached by considering how it is analogous to linear of translational
motion.
B. Learning Competency with code
At the end of this module, you should be able to:
1. Calculate the moment of inertia about a given axis of single-object and multiple object
systems STEM_GP12REDIIa-1
2. Calculate magnitude and direction of torque using the definition of torque as a cross
product STEM_GP12REDIIa-3
3. Describe rotational quantities using vectors STEM_GP12REDIIa-4
4. Determine whether a system is in static equilibrium or not STEM_GP12REDIIa-5
5. Apply the rotational kinematic relations for systems with constant angular
accelerations STEM_GP12REDIIa-6
6. Solve static equilibrium problems in contexts such as, but not limited to, seesaws,
mobiles, cable-hinge-strut system, leaning ladders, and weighing a heavy suitcase
using a small bathroom scale STEM_GP12REDIIa-8
7. Determine angular momentum of different systems STEM_GP12REDIIa-9
8. Apply the torque-angular momentum relation STEM_GP12REDIIa-1
C. Directions/ Instructions
After going through with this unit, you are expected to:
1. Read and follow each directions carefully.
2. Accomplish each activity for the mastery of competency.
3. Use the Learning Activity Sheets with care.
4. Record your scores for each activity
5. Always aim to get at least 80% of the total number of given items.
6. If you have any questions, contact, or see your teacher through messenger or text.
D. Exercises / Activities
a. What I need to know?
Have you ever watched a Ferris wheel as it turns? How do you feel? Did you ever wonder
how it moves? Will you still ride it if it doesn’t turn? This is why rotational motion is a very important
motion. It is important to know how this motion affects the movement of a certain body.
As you go along this lesson, you will be able to:
 Define kinematic rotational variables such as angular position, angular velocity, and
angular acceleration

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 Derive rotational kinematic equations, and
 Solve for the angular position, angular velocity, and angular acceleration of a rotating
body.
b. What’s new?
List some types of rotating objects and how are they important to society. Write your answer
in tabular form below.
Types of Rotating Objects Importance to Society
c. What I know?
Multiple Choice. Answer the question that follows. Choose the best answer from the given
choices.
1. If no external torque acts on a body, its angular velocity remains conserved.
A) True B) False
2. The easiest way to open a heavy door is by applying the force
A) Near the hinges B) In the middle of the door
C) At the edge of the door far from the hinges D) At the top of the door
3. Does a bridge anchored resting on two pillars have any torque?
A) No, it isn't moving B) Yes, but it is at equilibrium
C) Yes, but it will soon break because of the torque D) No, Bridges can't have torque
4. When an object is experiencing a net torque
A) it is in dynamic equilibrium. B) it is in static equilibrium.
C) it is rotating. D) it is translating.
5. A rusty bolt is hard to get turned. What could be done to help get the bolt turned?
A) use a long-arm lever B) decrease the force
C) apply the force at a 30 degree angle D) use a short-arm lever
d. What’s in?
The first part of this module will discuss the kinematics of rotational motion as described in its
angular position, speed and acceleration. The second part will discuss its dynamics and explain
the forces that causes objects to rotate. In this part, you will understand the Physics of simple
events observed daily such as the motion of opening doors, the concept behind see-saw; the
motion of ice-skaters and many more.
e. What is it?
Concept Discussion
Moment of
Inertia
 A rigid body is a solid composed of a collection of particles. These particles
remain static relative to each other and relative to the axis of rotation.
Regardless of the forces acting on a rigid body, it maintains its original
shape and size.
 Newton's first law of motion as applied to rotating systems states that unless
hindered by an external influence, a rigid body rotating about a fixed axis
will remain rotating at the same rate within the same axis (i.e. objects that
tend to rotate will keep rotating and nonrotating objects will maintain their
state of motion).
 Moment of inertia is defined as the quantity that impedes changes in an

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object's rotational state of motion.
 For a discreet number of particles, the moment of inertia I is:
𝐼 = ∑𝑚𝑖𝑟𝑖
2
Where 𝑚𝑖 is the mass of the ith point particle and 𝑟𝑖 is the distance of the ith
particle from the axis of rotation.
 Whereas a rotating rigid body with a continuous mass distribution gives:
𝐼 = ∫𝑟2𝑑𝑚 = ∫𝑟2𝜌𝑑𝑉
Angular
position,
angular
velocity,
and angular
acceleration
 The angular coordinate 𝜃 describes the rotational position of the body.
When a body rotates, a line can be drawn from the axis of rotation to any
point within the body. With the body as the reference frame, this line is fixed.
Hence, the line rotates with the body.
 The angle 𝜃 that this line makes with respect to the +x-axis is called the
angular position of the body.
 We measure the angle in radians. 1 radian is defined as the angle
subtended at the center of the circle by the arc whose length equals the
radius of the circle.
 The relation between radians and degrees is:
𝜃𝑟𝑎𝑑𝑖𝑎𝑛𝑠 =
𝜋
180𝑜 𝜃𝑑𝑒𝑔𝑟𝑒𝑒𝑠
 The arc length of a circle is equal to the radius of the circle multiplied by the
angle subtended by the arc:
𝑠 = 𝑟𝜃
 The angular displacement of the rotating body equals the change in its
angular position.
 The time rate of change of the angular position describes the rotational
motion of the rigid body.
 Define the average angular velocity of the rigid body as the net
displacement divided by the time elapsed. The subscript z tells us that the
body rotates about the z-axis:
𝜔𝑎𝑣𝑒−𝑧 =
∆𝜃
∆𝑡
 The instantaneous angular velocity or simply angular velocity is the
derivative of the angular position with respect to time:
𝜔𝑧 = lim
∆𝑡→0
∆𝜃
∆𝑡
=
𝑑𝜃
𝑑𝑡
 In some cases, the axis of rotation may not point along the z-direction or it
may change with time. In this, the angular velocity becomes a vector
quantity with the following components:
𝜔
⃗
⃗ = 𝜔𝑥𝑖̂+ 𝜔𝑦𝑗̂ + 𝜔𝑧𝑘
̂
therefore, when the body rotates about the z-axis, then 𝜔
⃗
⃗ has only one
component.
 The right-hand rule is a convenient tool in determining the direction of the
angular velocity. Defining the z-axis as the rotation axis, curl your fingers of
your right hand in the direction of rotation, your thumb points in the direction
of the angular velocity.

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 When a body rotates counterclockwise along the xy plane, by the right-
hand rule, 𝝎𝒛 is positive since it points along the positive z-axis, when a
body rotates clockwise, 𝝎𝒛 is then negative.
 The average angular acceleration is the change in angular velocity per unit
of time:
𝛼𝑎𝑣𝑒−𝑧 =
∆𝜔
∆𝑡
 The instantaneous angular acceleration or simply angular acceleration is
the derivative of the angular velocity with respect to time:
𝛼𝑧 = lim
∆𝑡→0
∆𝜔𝑧
∆𝑡
=
𝑑𝜔𝑧
𝑑𝑡
 The angular acceleration may also be expressed in terms of the second
derivative of the angular position with respect to time:
𝜔𝑧 =
𝑑2𝜃
𝑑𝑡2
 Analogous to translational motion, when a rigid body rotates
counterclockwise about the z-axis, when the angular acceleration is
positive, then the angular velocity increases; and when the angular
acceleration is negative, then the angular velocity decreases.
Rotational
Kinematics
 For any rigid body rotating about a fixed axis, suppose that the angular
acceleration is constant, we can find expressions that relate angular
displacement, angular velocity, and angular acceleration.
 Deriving the equations for rotational kinematics would take a similar
approach from its translational counterpart.
 When the angular acceleration is known and we start our measurement at
𝑡𝑜 = 0, then the angular velocity can be obtained from:
𝜔𝑧 = 𝜔𝑜𝑧 + 𝛼𝑧𝑡
 We can then calculate the angular displacement directly from the angular
acceleration and the time elapsed by:
∆𝜃 = 𝜔𝑜𝑧 +
1
2
𝛼𝑧𝑡2
 With only the angular displacement and angular acceleration known, the
angular velocity is:
𝜔𝑧
2 = 𝜔𝑜𝑧
2 + 2𝛼𝑧∆𝜃
 Lastly, when the angular acceleration is unknown, we find the following
relation:
∆𝜃 =
1
2
(𝜔𝑧 + 𝜔𝑜𝑧)𝑡
Bridging
linear and
rotational
variables of
motion
 When a rigid body rotates about a fixed axis, each particle making up the
body moves in a circular path at a distance r from the axis of rotation. The
tangential speed of each particle is directly proportional to the angular speed
of the rotating body.
 Taking the time derivative of the arc length, we obtain the following relation:

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𝑣 = 𝑟𝑤
At any given angular speed of a rotating rigid body, the tangential speed of
the particles increases with distance measured from the axis of rotation (all
particles would have the same period of revolution, but the particles located
farther from the center would cover a larger path).
 The translation acceleration and angular acceleration are also related by
the factor r:
𝑎 = 𝑟𝛼
 When a central force is present (such as the gravitational pull of the sun),
then the centripetal acceleration of each particle in terms of the angular
speed is:
𝑎𝑐 = 𝜔2𝑟
 With tangential and centripetal components of acceleration present, the
linear acceleration of each particle making up the rigid body equals the
vector sum of the tangential and the centripetal components of acceleration.
Torque and
its relation
with angular
acceleration
 Torque is the rotational counterpart of a force. Just like a net force is to
translational motion, a net torque causes a change in the rotational state of
motion of a rigid body, that is, torque causes angular acceleration.
 Torque depends on three parameters:
a. The magnitude of the force
b. The direction of the applied force
c. The distance from the axis of rotation to the line of action of the force.
 When the axis of rotation lies along the z-axis, torque is positive when it
causes a counterclockwise rotation and is negative when it causes a
clockwise rotation.
 The magnitude of the torque is equal to the lever arm multiplied by the force.
The lever arm is the distance from the rotation axis to the line of action of
the force:
𝜏 = 𝐹𝑙
By dimensional analysis, the unit of torque is N-m. Note however that this
unit does not translate to Joules as torque is different from work and energy.
 Call 𝑟 as the position vector directed from the axis of rotation to the point of
application of the force. Suppose that the direction of the force varies while
the position vector and the magnitude of the force are held fixed.
a) When the applied force is perpendicular to 𝑟, then the body rotates
with maximum torque.
b) When the applied force is parallel to 𝑟, then no rotation occurs.
c) When the applied force makes an angle 𝜑 with 𝑟, then the body
rotates but with lesser torque.
 The torque of the force with respect to the axis of rotation is given by:
𝜏 = 𝑟 × 𝐹
The torque that causes a body to rotate about the z-axis only has a single
component which can be evaluated using the cross product:
𝜏𝑧 = (𝑥𝐹
𝑦 − 𝑦𝐹
𝑥)𝑘
̂
Where x and y, and 𝐹
𝑥 and 𝐹
𝑦 are the x and y components of the position
vector and the force, respectively. In general, the torque is directed
perpendicular to the position vector and to the force.
 An equivalent way of finding torque associated with a given force F is to
resolve the force into components parallel and perpendicular to the axis of

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rotation. The parallel component, represented by Fll, exerts no torque
because its lever arm is zero. The perpendicular component of the force
equal to 𝐹 𝑠𝑖𝑛 𝜃 produces torque.
𝜏 = (𝐹𝑠𝑖𝑛𝜃)𝑑
Thus, we can generalize that the magnitude of the torque produced by a
force at a distance d from the axis of rotation is given by
𝜏 = (𝐹𝑑 𝑠𝑖𝑛𝜃)
where 𝜃 is the angle between the line of action of force and distance from
axis of rotation. This formula is consistent with the definition of torque as
cross product of force and displacement.
The SI unit of torque is Newton-meter. Torque is a vector quantity. Torque
may also be positive or negative, depending on the sense of rotation. By
convention, torque is positive if it tends to produce counterclockwise
rotation. It is negative if it tends to produce clockwise rotation. The greater
torque applied to an object, the greater the tendency of the object to rotate.
 To find the direction of the torque using the right-hand rule, point your
fingers in the direction of the position vector then curl them in the direction
of the force. Your thumb then points in the direction of the torque.
 Newton's Second Law as applied to rotating rigid bodies states that the net
torque on a rigid body equals the momentum of inertia multiplied by the
angular acceleration:
∑𝜏𝑧 = 𝐼𝛼𝑧
Static
Equilibrium
 If the following conditions are to be satisfied, then a rigid body is said to be
in equilibrium:
∑𝐹 = 0,∑𝜏 = 0
Moreover, when the body is at rest, it is in static equilibrium.
 The illustrations below show whether a body is in static equilibrium or not.
a. The body is in static equilibrium:
b. Although the net force is zero causing no translation, the body has the
tendency to rotate:

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c. The body will not rotate but has the tendency to accelerate along the
direction of the net force.
Work done
by torque
and
rotational
kinematic
energy
 Under the influence of a centripetal force, we have shown previously that
the translational work done onto the particle is zero.
 Consider a constant force applied tangent to the rim of a disc that has the
tendency to rotate. Setting the z-axis as the rotation axis of the disc, we
define rotational work as the product between this force and the angular
displacement of the disc:
𝑊𝑟𝑜𝑡 = 𝜏𝑧∆𝜃
 When the torque changes with angular position, then the total rotational
work done equals the integral of the torque with respect to an infinitesimal
(very small) displacement.
𝑊𝑟𝑜𝑡 = ∫ 𝜏𝑧(𝜃)𝑑𝜃
𝜃2
𝜃1
 We can also introduce the work-energy relation between rotating rigid
bodies where
1
2
𝐼𝜔2 is called the rotational kinetic energy of the body. This
gives us:
𝑊𝑟𝑜𝑡 =
1
2
𝐼𝜔𝑓
2 −
1
2
𝐼𝜔𝑜
2
The total rotational work done equals the change in the rotational kinetic
energy of the body.
 As an example, suppose that a disc initially rotates counterclockwise. When
the net rotational work done is positive, then the rotational kinetic energy
increases and the disc spins faster; when the net rotational work done is
negative, then the rotational kinetic energy decreases, and the disc slows
down.

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 In the case of rolling without slipping, such as a ball rolling down an inclined
plane, the total kinetic energy of the ball equals the kinetic energy due to
the motion of the center of mass plus the rotation of the particles making up
the ball about the center of mass:
𝐾𝐸𝑡𝑜𝑡𝑎𝑙 = 𝐾𝐸𝑡𝑟𝑎𝑛𝑠 + 𝐾𝐸𝑟𝑜𝑡 =
1
2
𝑀𝑣𝑐𝑚
2 +
1
2
𝐼𝑐𝑚𝜔2
Since gravity pulls the ball downward, the gravitational potential energy is:
𝑃𝐸 = 𝑀𝑔𝑦𝑐𝑚
And that the total mechanical energy of the ball is:
𝐸 =
1
2
𝑀𝑣𝑐𝑚
2 +
1
2
𝐼𝑐𝑚𝜔2 + 𝑀𝑔𝑦𝑐𝑚
Angular
Momentum
 Analogous to objects moving carrying a momentum 𝑝, the rotational
counterpart of momentum is angular momentum denoted by 𝐿
⃗ .
 When a body rotates about a fixed axis of symmetry such as the axis
piercing through the center of a disc rotating in the xy-plane, then the
angular momentum also has a single component and is directed in line with
the angular velocity. Call this component 𝐿𝑧 or the z-component of the
angular momentum.
 In general, for particles revolving about a fixed axis, the angular momentum
of a single particle is:
𝐿
⃗ = 𝑟 × 𝑝
Where 𝑟 is the same position vector defined from the torque-force relation,
and 𝑝 is the linear momentum of the particle tangent to the path taken by
the particle.
 In terms of the moment of inertia, the angular momentum of a rotating rigid
body is:
𝐿
⃗ = 𝐼𝜔
⃗
⃗
In general, the larger the moment of inertia is, the harder it is to change the
state of motion of a rigid body by a net external torque.
 A net external torque causes the angular momentum of the body to change
with time. Therefore, a net external torque is also defined as the time rate
of change of a body's angular momentum:
𝜏 =
𝑑𝐿
⃗
𝑑𝑡
 When a rigid body rotates about any axis except from the axis of symmetry,
then the angular momentum in general does not lie along the axis of
rotation. In this case, although the angular velocity is constant, the angular
momentum changes and traces out a cone. This results to a net torque
being acted onto the body which then maintains the rotation. If the angular
momentum lies along the symmetry axis, then 𝐿
⃗ is in line with 𝜔
⃗
⃗ and no
torque is needed to maintain rotation.
 The law of conservation of angular momentum states that when the net
external torque acting on a rigid body is zero, then the total angular
momentum is conserved. For a non-rigid body with varying moment of
inertia, conservation of angular momentum yields:
𝐼𝑜𝜔𝑜 = 𝐼𝑓𝜔𝑓
And in the case of rotational collision between two rigid bodies A and B,
conservation of angular momentum gives us:

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𝐼𝐴𝜔𝑜𝐴 + 𝐼𝐵𝜔𝑜𝐵 = 𝐼𝐴𝜔𝑓𝐴 + 𝐼𝐵𝜔𝑓𝐵
Examples:
1. A baton is made up of four particles with each end fastened by rods of negligible mass.
Calculate the moment of inertia at an axis where the rods intersect. The distances from
the axis of rotation to particle A (m = 300 g) and particle C (m=350 g) are both 0.5 meters
whereas particles B (m=150g) and D (m 125 g) are separated by a distance of 0.5 meters
with B located 0.24 meters from the axis of rotation.
Solution:
𝑰 = ∑𝒎𝒊𝒓𝒊
𝟐
𝑰 = 𝒎𝑨𝒓𝑨
𝟐
+ 𝒎𝑩𝒓𝑩
𝟐
+ 𝒎𝑪𝒓𝑪
𝟐
+ 𝒎𝑫𝒓𝑫
𝟐
𝑰 = 𝟎. 𝟑 𝒌𝒈(𝟎. 𝟓 𝒎)𝟐
+ 𝟎. 𝟑𝟓 𝒌𝒈(𝟎. 𝟓 𝒎)𝟐
+ 𝟎. 𝟏𝟓 𝒌𝒈(𝟎. 𝟐𝟒 𝒎)𝟐
+ 𝟎. 𝟏𝟐𝟓 𝒌𝒈(𝟎. 𝟐𝟔 𝒎)𝟐
𝑰 = 𝟎. 𝟏𝟖 𝒌𝒈 − 𝒎𝟐
Note: r is measured from the axis of rotation to the
location of the point mass. The rD is 0.26 m since based
on the problem particles B (m=150 g) and D (m 125 g)
are separated by a distance of 0.5 meters with B located
0.24 meters from the axis of rotation. Also, we need to
convert gram to kilogram for the mass before computing
for the moment of inertia.
2. A fruit blender manufactured by a certain company is being tested. The angular position
of the blender is:
𝜽(𝒕) = (𝟓
𝒓𝒂𝒅
𝒔𝟒
)𝒕𝟒 + (𝟏.𝟐
𝒓𝒂𝒅
𝒔𝟑
) 𝒕𝟑 + (𝟎. 𝟓
𝒓𝒂𝒅
𝒔𝟐
)𝒕𝟐 + (𝟐
𝒓𝒂𝒅
𝒔
)𝒕
Suppose that 𝒕𝟏 = 𝟏 𝒔 and 𝒕𝟐 = 𝟑 𝒔.
A. Find the angular position of the blender at both times.
B. Find the angular velocity of the blender at both times
C. Find the angular acceleration of the blender at both times.
D. Suppose that a food particle is stuck at a distance 5 cm from the axis of rotation. Find
the distance that this particle covered during this time interval.
Solution:
A. The angular position at 𝒕𝟏 = 𝟏 𝒔.
𝜽(𝒕 = 𝟏 𝒔) = (𝟓
𝒓𝒂𝒅
𝒔𝟒
) (𝟏 𝒔)𝟒 + (𝟏.𝟐
𝒓𝒂𝒅
𝒔𝟑
)(𝟏 𝒔)𝟑 + (𝟎. 𝟓
𝒓𝒂𝒅
𝒔𝟐
)(𝟏 𝒔)𝟐 + (𝟐
𝒓𝒂𝒅
𝒔
)(𝟏 𝒔)
𝜽(𝒕 = 𝟏 𝒔) = 𝟖.𝟕 𝒓𝒂𝒅𝒊𝒂𝒏𝒔
The angular position at 𝒕𝟐 = 𝟑 𝒔
𝜽(𝒕 = 𝟑 𝒔) = (𝟓
𝒓𝒂𝒅
𝒔𝟒
) (𝟑 𝒔)𝟒 + (𝟏.𝟐
𝒓𝒂𝒅
𝒔𝟑
)(𝟑 𝒔)𝟑 + (𝟎. 𝟓
𝒓𝒂𝒅
𝒔𝟐
)(𝟑 𝒔)𝟐 + (𝟐
𝒓𝒂𝒅
𝒔
)(𝟑 𝒔)
𝜽(𝒕 = 𝟑 𝒔) = 𝟒𝟒𝟑.𝟗 𝒓𝒂𝒅𝒊𝒂𝒏𝒔
B. The angular velocity is the derivative of the angular position with respect to time. Differentiating
the given expression, we get:
𝝎𝒛(𝒕) = (𝟐𝟎
𝒓𝒂𝒅
𝒔𝟒
) 𝒕𝟑 + (𝟑. 𝟔
𝒓𝒂𝒅
𝒔𝟑
)𝒕𝟐 + (𝟏
𝒓𝒂𝒅
𝒔𝟐
)𝒕
The angular velocity at 𝒕𝟏 = 𝟏 𝒔

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𝝎𝒛(𝒕 = 𝟏) = (𝟐𝟎
𝒓𝒂𝒅
𝒔𝟒
)(𝟏)𝟑 + (𝟑.𝟔
𝒓𝒂𝒅
𝒔𝟑
) (𝟏)𝟐 + (𝟏
𝒓𝒂𝒅
𝒔𝟐
) (𝟏)
𝝎𝒛(𝒕 = 𝟏) = 𝟐𝟔.𝟒
𝒓𝒂𝒅
𝒔
The angular velocity at 𝒕𝟐 = 𝟑 𝒔
𝝎𝒛(𝒕 = 𝟑) = (𝟐𝟎
𝒓𝒂𝒅
𝒔𝟒
)(𝟑)𝟑 + (𝟑.𝟔
𝒓𝒂𝒅
𝒔𝟑
) (𝟑)𝟐 + (𝟏
𝒓𝒂𝒅
𝒔𝟐
) (𝟑)
𝝎𝒛(𝒕 = 𝟑) = 𝟓𝟕𝟓.𝟒
𝒓𝒂𝒅
𝒔
C. The angular acceleration is the derivative of the angular velocity with time. Hence,
𝜶𝒛(𝒕) = (𝟔𝟎
𝒓𝒂𝒅
𝒔𝟒
)𝒕𝟐 + (𝟕. 𝟐
𝒓𝒂𝒅
𝒔𝟑
)𝒕 + (𝟏
𝒓𝒂𝒅
𝒔𝟐
)
The angular acceleration at 𝒕𝟏 = 𝟏 𝒔
𝜶𝒛(𝒕 = 𝟏) = (𝟔𝟎
𝒓𝒂𝒅
𝒔𝟒
) (𝟏)𝟐 + (𝟕.𝟐
𝒓𝒂𝒅
𝒔𝟑
)(𝟏) + (𝟏
𝒓𝒂𝒅
𝒔𝟐
)
𝜶𝒛(𝒕 = 𝟏) = (𝟔𝟖.𝟐
𝒓𝒂𝒅
𝒔𝟐
)
The angular acceleration at 𝒕𝟐 = 𝟑 𝒔
𝜶𝒛(𝒕 = 𝟑) = (𝟔𝟎
𝒓𝒂𝒅
𝒔𝟒
) (𝟑)𝟐 + (𝟕.𝟐
𝒓𝒂𝒅
𝒔𝟑
)(𝟑) + (𝟏
𝒓𝒂𝒅
𝒔𝟐
)
𝜶𝒛(𝒕 = 𝟑) = (𝟓𝟔𝟐.𝟔
𝒓𝒂𝒅
𝒔𝟐
)
D. The angular displacement of the blender in the given time interval is
∆𝜽 = 𝜽(𝒕 = 𝟑 𝒔) − 𝜽(𝒕 = 𝟏 𝒔)
∆𝜽 = 𝟒𝟒𝟑.𝟗 𝒓𝒂𝒅𝒊𝒂𝒏𝒔− 𝟖.𝟕 𝒓𝒂𝒅𝒊𝒂𝒏𝒔
∆𝜽 = 𝟒𝟑𝟓.𝟐 𝒓𝒂𝒅𝒊𝒂𝒏𝒔
The distance covered by the food particle during this time interval can be obtained by:
𝒔 = 𝒓𝜽
𝒔 = (𝟎.𝟎𝟓 𝒎)(𝟒𝟑𝟓.𝟐 𝒓𝒂𝒅𝒊𝒂𝒏𝒔)
𝒔 = 𝟐𝟏.𝟕𝟔 𝒎
3. Whenever a cyclist applies force to the pedal of his bicycle, he is imparting torque to the
bicycle. Suppose a cyclist applies a downward force of 60 N to the pedal. Find the
torque produced in each position shown.
Solution:
a. For position a, the lever arm is already given as 0.25 m. The force will produce a
clockwise rotation with respect to O and therefore the torque is negative.
𝝉 = −(𝟔𝟎𝑵)(𝟎.𝟐𝟓 𝒎)

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𝝉 = −𝟏𝟓 𝑵𝒎
b. The torque is again negative in as much as the rotation is clockwise.
𝝉 = −(𝟔𝟎𝑵)(𝟎.𝟐𝟓 𝒎𝐬𝐢𝐧𝟓𝟑𝒐)
𝝉 = −(𝟔𝟎𝑵)(𝟎.𝟐𝟔 𝒎)
𝝉 = −𝟏𝟐 𝑵𝒎
c. The torque this time is positive since a counterclockwise rotation with respect to O is
produced.
𝝉 = (𝟔𝟎𝑵)(𝟎.𝟐𝟓 𝒎𝐬𝐢𝐧 𝟑𝟕𝒐)
𝝉 = −(𝟔𝟎𝑵)(𝟎.𝟏𝟓 𝒎)
𝝉 = 𝟗.𝟎 𝑵𝒎
4. A force is being applied on a machine given by the vector 𝑭
⃗
⃗ = 𝟏𝟎.𝟓𝑵 𝒊̂ − 𝟔𝑵 𝒋̂. The
position vector from the axis of rotation to the point where the force is applied is 𝒓
⃗ =
−𝟎.𝟑 𝒎 𝒊̂ + 𝟎.𝟓 𝒎 𝒋̂.
A. Sketch the force vector, the position vector, and the origin.
B. Determine the direction of the torque using the right-hand rule.
C. What is the torque vector produced by this force? Show that the direction of the
torque is the same with (b).
Solution:
A. With the origin set as the reference point of both vectors, the force vector lies along
the fourth quadrant whereas the position vector lies along the second quadrant.
B. Using the right-hand rule, point your fingers initially in the direction of the position
vector, then curl your fingers in the direction of the force. Your outstretched thumb
points in the direction of the torque. Hence, the torque is directed at the negative z-
direction.
C. Since there are no z components of the force and of the position vector, then the x
and y components of the torque vanish leaving us with:
𝝉𝒛 = (𝒙𝑭𝒚 − 𝒚𝑭𝒙)𝒌
̂
𝝉𝒛 = [(−𝟎.𝟑)(−𝟔) − (𝟎.𝟓)(𝟏𝟎.𝟓)]𝒌
̂
𝝉𝒛 = 𝟑. 𝟒𝟓 𝑵𝒎 (−𝒌
̂)
The torque is indeed directed along the negative z-direction which agrees with the right-hand
rule.
5. A block of mass 10 kg rests on the left end of a lever made up of a homogenous rigid
plank of length 2.5 m. The fulcrum is placed at the center of mass of the rod.
a. Calculate the torque produced by the weight of the block at an axis passing through
the pivot point.
b. Where should a 20-kg mass be placed in order to balance the system?
c. Suppose that the mass of the plank is 7 kg, calculate the normal force exerted by
the fulcrum onto the plank.
d. Choose the left end of the plank to be the pivot point. Where should the same 20-kg
mass be placed to balance the plank?
Solution:
a. The center of mass of the plank is located exactly at its geometric center. Setting the
geometric center to be the origin, the center of mass is at the point 𝒙𝒄𝒎 = 𝟎 𝒄𝒎. Since the 10-
kg block would cause a clockwise rotation, the torque due to its weight is:
𝝉𝒛𝟏 = 𝟏. 𝟐𝟓 𝒎 (𝟏𝟎 𝒌𝒈)(𝟗.𝟖
𝒎
𝒔𝟐)
𝝉𝒛𝟏 = 𝟏𝟐𝟐.𝟓 𝑵𝒎

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b. In order to balance the system, we set the net torque to zero. We have four forces acting on
the plank - the weight of block 1, the weight of block 2, the normal force exerted by the
fulcrum, and the weight of the plank, then:
𝝉𝒛𝟏 + 𝝉𝒛𝟐 + 𝝉𝒛𝑵 + 𝝉𝒛𝒑𝒍𝒂𝒏𝒌 = 0
The last two terms are equal to zero since the forces are located exactly at the chosen axis.
122.5 𝑁𝑚 + (−𝒙 (20 𝒌𝒈)(𝟗.𝟖
𝒎
𝒔𝟐)+ 0 + 0 = 0
𝒙 =
−122.5 𝑁𝑚
−(20 𝒌𝒈)(𝟗.𝟖
𝒎
𝒔𝟐)
𝒙 = 0.625 𝑚 to the right of the geometric center.
c. Since the forces are balanced, we use Newton's first law of motion to calculate the normal
force n.
∑𝑭𝒚 = 𝟎
𝑭𝟏 + 𝐹𝟐 + 𝑭𝒏 + 𝑭𝒑𝒍𝒂𝒏𝒌 = 0
(𝟏𝟎𝒌𝒈)(−𝟗.𝟖
𝒎
𝒔𝟐)+ (𝟐𝟎 𝒌𝒈)(−𝟗.𝟖
𝒎
𝒔𝟐)+ (𝟕 𝒌𝒈)(−𝟗.𝟖
𝒎
𝒔𝟐) + 𝒏 = 𝟎
−𝟗𝟖 𝑵 − 𝟏𝟗𝟔 𝑵 − 𝟔𝟖.𝟔 𝑵 + 𝒏 = 𝟎
𝒏 = 𝟑𝟔𝟑.𝟔 𝑵
d. Choosing the left end of the plank as the chosen axis, the normal force and the weight of the
plank now produces nonzero torques. In turn, the weight of block 1 produces zero torque.
𝝉𝒛𝟏 + 𝝉𝒛𝟐 + 𝝉𝒛𝑵 + 𝝉𝒛𝒑𝒍𝒂𝒏𝒌 = 0
−𝟏𝟗𝟔 𝑵(𝒙 + 𝟏. 𝟐𝟓 𝒎)− 𝟔𝟖.𝟔 𝑵(𝟏.𝟐𝟓 𝒎) + 𝟑𝟔𝟑.𝟔𝑵 (𝟏.𝟐𝟓 𝒎) = 𝟎
𝒙 =
𝟔𝟖.𝟔 𝑵(𝟏.𝟐𝟓 𝒎) − 𝟑𝟔𝟑.𝟔𝑵 (𝟏. 𝟐𝟓 𝒎)
−𝟏𝟗𝟔 𝑵
− 𝟏. 𝟐𝟓 𝒎
𝒙 = 0.63 𝑚 which agrees with our answer in (b).
6. A 35-kg child is sitting along the rim of a merry-go-round that is rotating at 0.35
revolutions per second about its symmetry axis. The mass and the radius of the merry-
go-round are 85 kg and 8 m, respectively. Assume that you can treat the child as a
point particle and you can model the merry-go-round as a disc.
a. Calculate the moment of inertia of the system about its axis of symmetry.
b. Calculate the total angular momentum of the system.
Solution:
a. The moment of inertia of the child is
𝐼 = 𝑚𝑅2
𝐼 = (35 𝑘𝑔)(8 𝑚)2
𝐼 = 2240 𝑘𝑔 − 𝑚2
The moment of inertia of the merry-go-round is
𝐼 =
1
2
𝑀𝑅2
𝐼 =
1
2
(85 𝑘𝑔)(8 𝑚)2
𝐼 = 2720 𝑘𝑔 − 𝑚2
The moment of inertia of the whole system is
𝐼 = 𝐼𝑐ℎ𝑖𝑙𝑑 + 𝐼𝑚𝑒𝑟𝑟𝑦−𝑔𝑜−𝑟𝑜𝑢𝑛𝑑
𝐼 = 2240 𝑘𝑔 − 𝑚2 + 2720 𝑘𝑔 − 𝑚2
𝐼 = 4960 𝑘𝑔 − 𝑚2
b. The total angular momentum is

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𝐿𝑧 = 𝐼𝜔𝑧
𝐿𝑧 = (4960 𝑘𝑔 − 𝑚2)(0.35
𝑟𝑒𝑣
𝑠
)(
2𝜋
1 𝑟𝑒𝑣
)
𝐿𝑧 = 10907.6 𝑘𝑔
𝑚2
𝑠
7. Under certain conditions, a star can collapse into an extremely dense object made
mostly of neutrons and is called a neutron star. The density of a neutron star is roughly
1014 times as great as that of ordinary solid matter. Suppose we represent the star as a
uniform solid, rigid sphere, both before and after collapse. The star's initial radius was
106 km; its final radius is 38 km. If the original star rotated once every 100 days, find
the angular speed of the neutron star.
Solution:
The moment of inertia of a solid sphere is
2
5
𝑀𝑅2. From the Law of Conservation of Angular
Momentum, when the net external torque equals zero, then:
𝐿𝑜 = 𝐿𝑓
𝐼𝑜𝜔𝑜 = 𝐼𝑓𝜔𝑓
Solving for the final angular velocity we get:
𝜔𝑓 = 𝜔𝑜 (
𝐼𝑜
𝐼𝑓
)
2
5
𝑀𝑅𝑜
2
2
5
𝑀𝑅𝑓
2
)
𝑅𝑜
2
𝑅𝑓
2)
𝜔𝑓 =
2𝜋
(100 𝑑𝑎𝑦𝑠) (86400
𝑠
𝑑𝑎𝑦𝑠
)
(
106 𝑘𝑚
38 𝑘𝑚
)
2
𝜔𝑓 = 503.6
𝑟𝑎𝑑
𝑠
f. What is more?
Activity 1. Direction. Solve the following problems in a separate paper. Show your solutions
systematically and clearly.
The angular position of a hoop is described by the function below:
𝜽(𝒕) = (𝟏𝟎
𝒓𝒂𝒅
𝒔𝟓
)𝒕𝟓 + (𝟎. 𝟒
𝒓𝒂𝒅
𝒔𝟒
)𝒕𝟒 + (𝟑
𝒓𝒂𝒅
𝒔𝟑
)𝒕𝟑 + (−𝟓
𝒓𝒂𝒅
𝒔𝟐
)𝒕𝟐 − 𝟏𝟎
𝒓𝒂𝒅
𝒔
𝒕
Suppose that 𝒕𝟏 = 𝟓 𝒔 and 𝒕𝟐 = 𝟏𝟎 𝒔.
1. Find the angular position of the hoop at both times.
2. Find the angular velocity of the hoop at both times.
3. Find the angular acceleration of the hoop at both times.
4. The radius of the hoop is 0.5 m. Find the distance covered by a particle along the rim of
the hoop during this time interval.
g. What I can do?
Activity 2. Calculate the torque vector of the following force (N) and position (m) vectors:
1. 𝑭
⃗
⃗ = 𝟓𝑵 𝒊̂ + 𝟏𝟎𝑵 𝒋̂ ; 𝒓
⃗ = 𝟓𝒎 𝒊̂ + 𝟏𝟎𝒎 𝒋̂

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2. 𝑭
⃗
⃗ = −𝟏𝟓𝑵 𝒊̂ + 𝟑𝑵 𝒋̂ ; 𝒓
⃗ = 𝟎.𝟔𝒎 𝒊̂ − 𝟏. 𝟓𝒎 𝒋̂
3. 𝑭
⃗
⃗ = 𝟏𝟏𝑵 𝒊̂ − 𝟒𝑵 𝒋̂ + 𝟖𝑵 𝒌
̂; 𝒓
⃗ = 𝟑𝒎 𝒊̂ + 𝟎.𝟗𝒎 𝒋̂+ 𝟐𝒎 𝒌
̂
4. 𝑭
⃗
⃗ = 𝟑.𝟗𝑵 𝒊̂ + 𝟏𝟎𝑵 𝒌
̂; 𝒓
⃗ = −𝟎.𝟓𝒎 𝒊̂ + 𝟎.𝟏𝒎 𝒋̂+ 𝟏𝒎 𝒌
̂
h. What other enrichment activities can I engage in?
Activity 3. An engine flywheel and a clutch plate are both connected to a transmission shaft.
Let the moment of inertia of the flywheel be I1 and its angular velocity be 𝝎𝟏, and let the moment
of inertia of the clutch plate be I1 and its angular velocity to be 𝝎𝟏. The two discs have then
been combined by forces by which are applied at their axes of rotation so as not to cause any
torque. The discs then reached a common final angular velocity after rotational collision. Find
an expression for the final angular velocity.
i. What I have learned?
Activity 4. Sum it up
List down key 5 key points that you have learned from the topic.
j. What I can show?
Multiple Choice. Answer the question that follows. Choose the best answer from the given
choices.
1. If no external torque acts on a body, its angular velocity remains conserved.
A) True B) False
2. The easiest way to open a heavy door is by applying the force
A) Near the hinges B) In the middle of the door
C) At the edge of the door far from the hinges D) At the top of the door
3. Does a bridge anchored resting on two pillars have any torque?
A) No, it isn't moving B) Yes, but it is at equilibrium
C) Yes, but it will soon break because of the torque D) No, Bridges can't have torque
4. When an object is experiencing a net torque
A) it is in dynamic equilibrium. B) it is in static equilibrium.
C) it is rotating. D) it is translating.
5. A rusty bolt is hard to get turned. What could be done to help get the bolt turned?
A) use a long-arm lever B) decrease the force
C) apply the force at a 30 degree angle D) use a short-arm lever
E. Rubric for scoring
The following are the scores to be given in each problem
● Given quantity/ies: 1 point
● Unknown quantity/ies: 1 point
● Working equation: 1 point
● Solution: 2 points
● Final Answer: 2 points
F. Reflection
Learners will write on their notebooks or journals their insights about the lesson.
I understand that .
I realized that .
G. References for learners
General Physics I Reader (Department of the Philipines-Bureau of Learning Resources)
GENERAL PHYSICS 1 (Alternative Delivery Mode) Quarter 2 - Module 1:
ROTATIONAL EQUILIBRIUM AND ROTATIONAL DYNAMICS- Published by the
Department of Education – Division of Cagayan de Oro

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