Rotational Motion MCQ Quiz - Objective Question with Answer for Rotational Motion - Download Free PDF
Last updated on Jul 7, 2025
Latest Rotational Motion MCQ Objective Questions
Rotational Motion Question 1:
Statement 1: A person jumping freely from an airplane follows a parabolic path when observed from the ground, assuming no air resistance.
Statement 2: The rotation of the Earth causes the trajectory of the person to follow a parabolic path, accounting for the Coriolis effect.
Choose the correct option:
Answer (Detailed Solution Below)
Rotational Motion Question 1 Detailed Solution
Calculation:
Statement 1 is true: When a person jumps from an airplane, they are subject to the force of gravity and air resistance (neglecting air resistance in this case). The person follows a parabolic path relative to an observer on the ground, as the horizontal velocity remains constant, and the vertical velocity changes due to gravity.
Statement 2 is also true: The Earth's rotation does have an effect on the trajectory of moving objects. The Coriolis effect causes a deflection of the path of the person jumping from the airplane. However, this effect is very small at the altitude of an airplane and does not significantly alter the path's general parabolic shape.
Thus, the correct answer is (B).
Rotational Motion Question 2:
The moment of inertia of a circular ring of mass M and diameter r about a tangential axis lying in the plane of the ring is :
Answer (Detailed Solution Below)
Rotational Motion Question 2 Detailed Solution
Calculation:
Diameter is given as r.
∴ Radius = r / 2
Using parallel axis theorem,
Moment of inertia about tangent, Itangent = 1/2 m(r/2)2 + m(r/2)2 = (3 / 2) × m × (r / 2)2 = (3 / 8) × m × r2
Rotational Motion Question 3:
A wheel of radius 0.2 m rotates freely about its center when a string that is wrapped over its rim is pulled by force of 10 N as shown in figure. The established torque produces an angular acceleration of 2 rad/s2 . Moment of inertia of the wheel is ________ kg m2 .
(Acceleration due to gravity = 10 m/s2 )
Answer (Detailed Solution Below) 1
Rotational Motion Question 3 Detailed Solution
Calculation:
Given:
Force F = 10 N
Radius R = 0.2 m
Angular acceleration α = 2 rad/s2
Torque τ = F × R = 10 × 0.2 = 2 N·m
Using the relation: τ = Iα
⇒ I = τ / α = 2 / 2 = 1 kg·m2
Hence, the correct option is (1).
Rotational Motion Question 4:
If a solid sphere of mass 5 kg and a disc of mass 4 kg have the same radius. Then the ratio of moment of inertia of the disc about a tangent in its plane to the moment of inertia of the sphere about its tangent will be \(\rm \frac{x}{7}\). The value of x is _______.
Answer (Detailed Solution Below) 5
Rotational Motion Question 4 Detailed Solution
Calculation:
Given the moment of inertia formulas:
I₁ = (2/5) m₁ R² + m₁ R² = m₁ R² (7/5)
So, I₁ = 7m₁ R²
Similarly, for I₂:
I₂ = (m₂ R² / 4) + m₂ R² = (5/4) m₂ R²
So, I₂ = 5m₂ R²
The ratio of I₂ to I₁ is:
I₂ / I₁ = (5m₂ R²) / (7m₁ R²) = 5 / 7
Final Answer: x = 5
Rotational Motion Question 5:
ICM is moment of inertia of a circular disc about an axis (CM) passing through its center and perpendicular to the plane of disc. IAB is it's moment of inertia about an axis AB perpendicular to plane and parallel to axis CM at a distance \(\frac{2}{3}\)R from center. Where R is the radius of the disc. The ratio of IAB and ICM is x ∶ 9. The value of x is ______.
Answer (Detailed Solution Below) 17
Rotational Motion Question 5 Detailed Solution
Calculation:
The moment of inertia of disc about center of mass is Icm = (mR2) / 2
The moment of inertia of disc about AB using parallel axis theorem is
IAB = (mR2) / 2 + m × (2R / 3)2 = (17 / 18) mR2
⇒ IAB / Icm = 17 / 9 ⇒ x = 17
Top Rotational Motion MCQ Objective Questions
The work done to increase the velocity of a 1500 kg car from 36 km/h to 72 km/h is
Answer (Detailed Solution Below)
Rotational Motion Question 6 Detailed Solution
Download Solution PDFThe correct option is: 2
Concept Used:
- Work-Energy Theorem: This theorem states that the net work done by the force on a body is equal to the change in its kinetic energy.
- Mathematically, Work Done (W) = Final Kinetic Energy - Initial Kinetic Energy
- Kinetic Energy (K.E) is given by the formula: K.E = (1/2) × m × v2
- So, W = (1/2) × m × (v2 - u2), where:
- m = mass of the object
- v = final velocity
- u = initial velocity
Calculation:
- Given:
- Mass of the car (m) = 1500 kg
- Initial velocity (u) = 36 km/h = 36 × (1000 ÷ 3600) = 10 m/s
- Final velocity (v) = 72 km/h = 72 × (1000 ÷ 3600) = 20 m/s
- Using the work-energy theorem:
W = (1/2) × m × (v2 - u2)
⇒ W = (1/2) × 1500 × (202 - 102)
⇒ W = (1/2) × 1500 × (400 - 100)
⇒ W = (1/2) × 1500 × 300
⇒ W = 750 × 300
⇒ W = 225000 J
⇒ W = 2.25 × 105 J
Additional Information:
- Work done is a scalar quantity and is expressed in joules (J) in the SI system.
- This type of question is commonly asked in mechanics under the topic of energy and work.
- Always convert velocities from km/h to m/s before substituting in the kinetic energy formula: multiply by (1000 ÷ 3600) or simplify as 5/18.
Diagram/Visual Aid Suggestion:
- A diagram showing a car moving initially at a lower speed and then at a higher speed with vectors labeled u and v.
- Include a bar graph representing the initial and final kinetic energy for visual comparison.
In rotational motion, Power = Torque x ________.
Answer (Detailed Solution Below)
Rotational Motion Question 7 Detailed Solution
Download Solution PDFCONCEPT:
- Rotational motion: When a block is moving about a fixed axis on a circular path then this type of motion is called rotational motion.
Torque (τ):
- It is the twisting force that tends to cause rotation.
- The point where the object rotates is known as the axis of rotation.
- Mathematically it is written as,
τ = rFsin θ
EXPLANATION:
- The power associated with torque is given by the product of torque and angular velocity of the body about an axis of rotation i.e.,
⇒ P = τω
Where τ = torque and ω = angular velocity
EXTRA POINTS:
In rotational dynamics
- Moment of inertia is the analogue of mass
- Angular velocity is analogue of linear velocity
- Angular acceleration is analogue of linear acceleration
Thus, in linear motion mass x velocity = momentum
The analogues of the moment of inertia x angular velocity = angular momentum
|
Linear Motion |
Rotational Motion |
Position |
x |
θ |
Velocity |
v |
ω |
Acceleration |
a |
α |
Motion equations |
x = v̅ t |
θ = ω̅t |
|
v = v0 + at |
ω = ω0 + αt |
|
\(x = {v_0}t + \frac{1}{2}a{t^2}\) |
\(\theta = {ω _0}t + \frac{1}{2}\alpha {t^2}\) |
|
\({v^2} = v_0^2 + 2ax\) |
\({ω ^2} = ω _0^2 + 2\alpha \theta\) |
Mass (linear inertia) |
M |
I |
Newton’s second law |
F = ma |
T = Iα |
Momentum |
p = mv |
L = Iω |
Work |
Fd |
T.θ |
Kinetic energy |
\(\frac{1}{2}m{v^2}\) |
\(\frac{1}{2}I{ω ^2}\) |
Power |
Fv |
Tω |
Moment of inertia, rotational kinetic energy, and angular momentum of a body is I, E, and L respectively then :
Answer (Detailed Solution Below)
Rotational Motion Question 8 Detailed Solution
Download Solution PDFThe correct answer is option 4) i.e. L = √(2EI)
CONCEPT:
- Angular momentum: The angular momentum of a rigid object is defined as the product of the moment of inertia and the angular velocity.
- Angular momentum also obeys the law of conservation of momentum i.e. angular momentum before and after is conserved.
Angular momentum, L = I × ω
Rotational kinetic energy: For a given fixed axis of rotation, the rotational kinetic energy is given by:
\(KE = \frac{1}{2} Iω^2\)
Where I is the moment of inertia, ω is the angular velocity.
CALCULATION:
Angular momentum, L = I × ω ----(1)
Roational kinetic energy = \(E = \frac{1}{2} Iω^2\) ⇒ ω = \(\sqrt{\frac{2E}{I}}\) ----(2)
Substituting (2) in (1) we get
L = I × \(\sqrt{\frac{2E}{I}}\)= √(2EI)
If the moment of inertia of a rotating body is increased then what will be the effect on the angular velocity?
Answer (Detailed Solution Below)
Rotational Motion Question 9 Detailed Solution
Download Solution PDFCONCEPT:
- The angular momentum of a particle rotating about an axis is defined as the moment of the linear momentum of the particle about that axis.
- It is measured as the product of linear momentum and the perpendicular distance of its line of action from the axis of rotation.
- The relation between the angular momentum and moment of inertia is given by
L = Iω
Where I = moment of inertia, L = angular momentum, and ω = angular velocity.
EXPLANATION:
- If there is no external torque acting on system then initial angular momentum (Linitial) of system is equal to final momentum (Lfinal).
- Hence, the angular momentum in a closed system is a conserved.
∴ Iω = constant
⇒ I ∝ 1/ω
i.e. Moment of inertia is inversely proportional to the angular velocity.
- Hence if the moment of inertia of a rotating body is increased then the angular velocity decreases.
The correct relationship between Moment of Inertia, Torque, and Angular acceleration is ________.
Answer (Detailed Solution Below)
Rotational Motion Question 10 Detailed Solution
Download Solution PDFCONCEPT:
- Angular acceleration (α): It is defined as the time rate of change of angular velocity of a particle is called its angular acceleration.
- If Δω is the change in angular velocity time Δt, then average acceleration is
\(\vec α = \frac{{{\rm{\Delta }}\omega }}{{{\rm{\Delta }}t}}\)
Moment of Inertia (I):
- Moment of inertia plays the same role in rotational motion as mass plays in linear motion. It is the property of a body due to which it opposes any change in its state of rest or of uniform rotation.
- Moment of inertia of a particle is
I = mr2
where r = perpendicular distance of the particle from the rotational axis.
Torque (τ):
- It is the twisting force that tends to cause rotation.
- The point where the object rotates is known as the axis of rotation.
- Mathematically it is written as,
τ = rFsin θ
EXPLANATION:
The relationship between the angular acceleration (α), torque (τ) and moment of inertia (I) is given by
⇒ τ = α × I
\( \Rightarrow \alpha = \frac{\tau }{I}\)
- Therefore, angular acceleration = Torque / Moment of inertia. Hence option 2 is correct.
The expression Iω2/2 represents Rotational ____________.
Answer (Detailed Solution Below)
Rotational Motion Question 11 Detailed Solution
Download Solution PDFCONCEPT:
- Rotational kinetic energy: The energy, which a body has by virtue of its rotational motion, is called rotational kinetic energy.
- A body rotating about a fixed axis possesses kinetic energy because its constituent particles are in motion, even though the body as a whole remains in place.
- Mathematically rotational kinetic energy can be written as -
\(KE = \frac{1}{2}I{ω ^2}\)
Where I = moment of inertia and ω = angular velocity.
EXPLANATION:
- From above it is clear that, \(\frac{1}{2}I{ω ^2}\) represents rotational kinetic energy.
Quantity |
Expression |
Torque |
Iα |
Angular momentum |
Iω |
Rotational kinetic energy |
\(\frac{1}{2}I{ω ^2}\) |
Work done |
τθ |
Power |
τω |
A man standing on a revolving platform spreading his hands outward. Then:
Answer (Detailed Solution Below)
Rotational Motion Question 12 Detailed Solution
Download Solution PDFCONCEPT:
Angular momentum (L):
- The angular momentum of a rigid body is defined as the product of the moment of inertia and the angular velocity i.e.,
\(⇒ L = Iω \)
Where I = moment of inertia and ω = angular velocity
Law of conservation of angular momentum:
- When the net external torque acting on a body about a given axis is zero, the total angular momentum of the body about that axis remains constant i.e.,
⇒ I1ω1 = I2ω2
EXPLANATION:
- When a man standing on a revolving platform and spreading his hands suddenly as he stretches his hand that he is increasing his moment of inertia thereby decreasing his angular velocity using the principle of conservation of angular momentum. Therefore option 2 is correct.
The product of the moment of inertia and the angular acceleration is:
Answer (Detailed Solution Below)
Rotational Motion Question 13 Detailed Solution
Download Solution PDFCONCEPT:
Torque (τ):
- It is the twisting force that tends to cause rotation.
- The point where the object rotates is known as the axis of rotation.
- Mathematically it is written as,
τ = rFsin θ
- Angular acceleration (α): It is defined as the time rate of change of angular velocity of a particle is called its angular acceleration.
- If Δω is the change in angular velocity time Δt, then average acceleration is
\(\vec α = \frac{{{\rm{\Delta }}\omega }}{{{\rm{\Delta }}t}}\)
EXPLANATION:
- Torque is the measure of the amount of force acting on an object that can cause it to rotate.
- The torque that is needed to produce angular acceleration depends on the mass distribution of the object which is described by the moment of inertia.
- Therefore, torque (\(\tau \)) is the product of the angular acceleration (a) and the moment of inertia (I) of an object. Therefore option 2 is correct.
\(\tau = I \times a\)
A thin disc and a thin ring, both have mass M and radius R. Both rotate about axes through their centre of mass and are perpendicular to their surfaces at the same angular velocity. Which of the following is true?
Answer (Detailed Solution Below)
Rotational Motion Question 14 Detailed Solution
Download Solution PDFCONCEPT:
Moment of inertia:
- The moment of inertia of a rigid body about a fixed axis is defined as the sum of the product of the masses of the particles constituting the body and the square of their respective distances from the axis of the rotation.
- The moment of inertia of a particle is
⇒ I = mr2
Where r = the perpendicular distance of the particle from the rotational axis.
- Moment of inertia of a body made up of a number of particles (discrete distribution)
⇒ I = m1r12 + m2r22 + m3r32 + m4r42 + -------
Rotational kinetic energy:
- The energy, which a body has by virtue of its rotational motion, is called rotational kinetic energy.
- A body rotating about a fixed axis possesses kinetic energy because its constituent particles are in motion, even though the body as a whole remains in place.
- Mathematically rotational kinetic energy can be written as -
⇒ KE \( = \frac{1}{2}I{\omega ^2}\)
Where I = moment of inertia and ω = angular velocity.
EXPLANATION:
- The moment of inertia of the ring about an axis passing through the center and perpendicular to its plane is given by
⇒ Iring = MR2
- Moment of inertia of the disc about an axis passing through center and perpendicular to its plane is given by -
\(⇒ {I_{disc}} = \frac{1}{2}M{R^2}\)
- As we know that mathematically rotational kinetic energy can be written as
\(⇒ KE = \frac{1}{2}I{\omega ^2}\)
- According to the question, the angular velocity of a thin disc and a thin ring is the same. Therefore, the kinetic energy depends on the moment of inertia.
- Therefore, a body having more moments of inertia will have more kinetic energy and vice - versa.
- So, from the equation, it is clear that,
⇒ Iring > Idisc
∴ Kring > Kdisc
- The ring has higher kinetic energy.
Body |
Axis of Rotation |
Moment of inertia |
Uniform circular ring of radius R |
perpendicular to its plane and through the center |
MR2 |
Uniform circular ring of radius R |
diameter |
\(\frac{MR^2}{2}\) |
Uniform circular disc of radius R | perpendicular to its plane and through the center | \(\frac{MR^2}{2}\) |
Uniform circular disc of radius R | diameter | \(\frac{MR^2}{4}\) |
A hollow cylinder of radius R | Axis of cylinder | MR2 |
A wheel has angular acceleration of 3 rad/s2 and an initial angular speed of 2 rad/s. In a time of two second it has rotated through an angle (in radians) of
Answer (Detailed Solution Below)
Rotational Motion Question 15 Detailed Solution
Download Solution PDFConcept:
The equations which are applied in linear motion can also be applied in angular motion.
|
Linear Motion |
Rotational Motion |
|
Position |
x |
θ |
Angular position |
Velocity |
v |
ω |
Angular velocity |
Acceleration |
a |
α |
Angular acceleration |
Motion equations |
x = v̅ t |
θ = ω̅t |
Motion equations |
|
v = v0 + at |
ω = ω0 + αt |
|
|
\(x = {v_0}t + \frac{1}{2}a{t^2}\) |
\(θ = {ω _0}t + \frac{1}{2}α {t^2}\) |
|
|
\({v^2} = v_0^2 + 2ax\) |
\({ω ^2} = ω _0^2 + 2α θ\) |
|
Mass (linear inertia) |
M |
I |
Moment of inertia |
Newton’s second law |
F = ma |
τ = Iα |
Newton’s second law |
Momentum |
p = mv |
L = Iω |
Angular momentum |
Work |
Fd |
τθ |
Work |
Kinetic energy |
\(\frac{1}{2}m{v^2}\) |
\(\frac{1}{2}I{ω ^2}\) |
Kinetic energy |
Power |
Fv |
τω |
Power |
Calculation:
Given:
α = 3 rad/s2, ωo = 2 rad/s, t = 2 s
We know that,
\(θ = {ω _0}t + \frac{1}{2}α {t^2}\)
\(θ = ({2}\times 2) + \frac{1}{2}\times3\times2^2\)
θ = 4 + 6 = 10 radians