Rotational motion is a crucial topic in physics that often appears in school and competitive exams. Understanding this concept is essential for students aiming to excel in their exams. Practicing MCQs and objective questions on rotational motion not only enhances conceptual clarity but also boosts confidence, helping students score better in their assessments.
What You Will Practise Here
Fundamental concepts of rotational motion and angular displacement
Key formulas related to angular velocity and angular acceleration
Understanding torque and its applications in various scenarios
Moment of inertia and its significance in rotational dynamics
Equations of motion for rotating bodies
Conservation of angular momentum and its implications
Real-world applications of rotational motion in engineering and daily life
Exam Relevance
Rotational motion is a significant part of the physics syllabus for CBSE, State Boards, NEET, and JEE. Students can expect questions that test their understanding of concepts, calculations involving formulas, and application-based scenarios. Common question patterns include numerical problems, conceptual questions, and diagram-based queries, making it essential for students to practice thoroughly.
Common Mistakes Students Make
Confusing linear motion concepts with rotational motion principles
Miscalculating torque due to incorrect application of the lever arm
Overlooking the importance of units in angular measurements
Failing to apply the parallel axis theorem correctly
Neglecting to visualize problems involving rotating objects
FAQs
Question: What is the difference between angular velocity and linear velocity? Answer: Angular velocity refers to the rate of change of angular displacement, while linear velocity is the rate of change of linear displacement. They are related through the radius of the circular path.
Question: How is torque calculated? Answer: Torque is calculated using the formula τ = r × F, where τ is torque, r is the distance from the pivot point to the point of force application, and F is the force applied.
Now is the time to enhance your understanding of rotational motion! Dive into our practice MCQs and test your knowledge to ensure you are well-prepared for your exams. Every question you solve brings you one step closer to success!
Q. A ball rolls down a ramp and reaches a speed of 10 m/s at the bottom. If the ramp is 5 m high, what is the ball's moment of inertia if it is a solid sphere?
A.
(2/5)m(10^2)
B.
(1/2)m(10^2)
C.
(1/3)m(10^2)
D.
(5/2)m(10^2)
Solution
Using conservation of energy, mgh = (1/2)mv^2 + (1/2)(2/5)mv^2. Solving gives the moment of inertia I = (2/5)m(10^2).
Q. A ball rolls down a ramp. If it starts from rest and rolls without slipping, what is the relationship between its linear speed and angular speed at the bottom?
A.
v = Rω
B.
v = 2Rω
C.
v = R/2ω
D.
v = 3Rω
Solution
The relationship is given by v = Rω, where v is the linear speed, R is the radius, and ω is the angular speed.
Q. A ball rolls without slipping on a flat surface. If the ball's radius is doubled while keeping its mass constant, how does its moment of inertia change?
A.
Increases by a factor of 2
B.
Increases by a factor of 4
C.
Increases by a factor of 8
D.
Remains the same
Solution
The moment of inertia of a solid sphere is (2/5)MR^2. If the radius is doubled, the moment of inertia increases by a factor of 4.
Q. A child is sitting on a merry-go-round that is rotating. If the child moves towards the center, what happens to the rotational speed of the merry-go-round?
A.
Increases
B.
Decreases
C.
Remains the same
D.
Becomes zero
Solution
As the child moves towards the center, the moment of inertia decreases, causing the rotational speed to increase to conserve angular momentum.
Q. A child is sitting on a merry-go-round that is spinning. If the child moves closer to the center, what happens to the angular velocity of the merry-go-round?
A.
Increases
B.
Decreases
C.
Remains the same
D.
Becomes zero
Solution
As the child moves closer to the center, the moment of inertia decreases, causing the angular velocity to increase to conserve angular momentum.
Q. A child is sitting on a merry-go-round that is spinning. If the child moves towards the center of the merry-go-round, what happens to the angular velocity of the system?
A.
Increases
B.
Decreases
C.
Remains the same
D.
Becomes zero
Solution
As the child moves towards the center, the moment of inertia decreases, thus the angular velocity increases to conserve angular momentum.
Q. A child is sitting on a merry-go-round that is spinning. If the child moves towards the center, what happens to the angular velocity of the merry-go-round?
A.
Increases
B.
Decreases
C.
Remains the same
D.
Becomes zero
Solution
As the child moves towards the center, the moment of inertia decreases, and to conserve angular momentum, the angular velocity must increase.
Q. A child sitting at the edge of a merry-go-round throws a ball tangentially. What happens to the angular momentum of the system (merry-go-round + child + ball)?
A.
Increases
B.
Decreases
C.
Remains constant
D.
Becomes zero
Solution
Angular momentum of the system remains constant due to conservation of angular momentum.
Q. A composite body consists of a solid cylinder and a solid sphere, both of mass M and radius R. What is the total moment of inertia about the same axis?
A.
(7/10) MR^2
B.
(9/10) MR^2
C.
(11/10) MR^2
D.
(13/10) MR^2
Solution
The total moment of inertia is I_cylinder + I_sphere = (1/2 MR^2) + (2/5 MR^2) = (7/10) MR^2.
Q. A cylinder rolls down a hill of height h. What is the speed of the center of mass when it reaches the bottom?
A.
√(2gh)
B.
√(3gh)
C.
√(4gh)
D.
√(5gh)
Solution
Using conservation of energy, potential energy at the top (mgh) converts to kinetic energy (1/2 mv^2 + 1/2 Iω^2). For a solid cylinder, I = (1/2)mR^2 and ω = v/R. Solving gives v = √(3gh).
Q. A cylinder rolls down a hill. If it has a radius R and rolls without slipping, what is the relationship between its linear velocity v and its angular velocity ω?
A.
v = Rω
B.
v = 2Rω
C.
v = ω/R
D.
v = R^2ω
Solution
For rolling without slipping, the relationship is v = Rω.
Q. A cylinder rolls down a hill. If the height of the hill is h, what is the speed of the center of mass of the cylinder at the bottom of the hill?
A.
√(gh)
B.
√(2gh)
C.
√(3gh)
D.
√(4gh)
Solution
Using conservation of energy, potential energy at the top (mgh) converts to kinetic energy (1/2 mv^2 + 1/2 Iω^2). For a solid cylinder, I = 1/2 mr^2, leading to v = √(2gh).
