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).
Correct Answer: A — (2/5)m(10^2)
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.
Correct Answer: B — Increases by a factor of 4
Q. A ball rolls without slipping on a flat surface. If the ball's radius is doubled, 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 for a solid sphere is (2/5)MR^2. If the radius is doubled, the moment of inertia increases by a factor of 4.
Correct Answer: B — 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.
Correct Answer: A — Increases
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.
Correct Answer: A — Increases
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.
Correct Answer: A — Increases
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.
Correct Answer: C — Remains constant
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.
Correct Answer: A — 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).
Correct Answer: B — √(2gh)
Q. A disc of radius R and mass M is rotating about its axis with an angular velocity ω. What is the kinetic energy of the disc?
A.(1/2)Iω^2
B.(1/2)Mω^2
C.Iω
D.Mω^2
Solution
Kinetic energy K = (1/2)Iω^2, where I = (1/2)MR^2 for a disc.
Correct Answer: A — (1/2)Iω^2
Q. A disk and a ring of the same mass and radius are released from rest at the same height. Which one reaches the ground first?
A.Disk
B.Ring
C.Both reach at the same time
D.Depends on the surface
Solution
The disk has a lower moment of inertia compared to the ring, thus it reaches the ground first.
Correct Answer: A — Disk
Q. A disk and a ring of the same mass and radius are rolling down an incline. Which one will have a greater translational speed at the bottom?
A.Disk
B.Ring
C.Both have the same speed
D.Cannot be determined
Solution
The disk has a lower moment of inertia than the ring, allowing it to convert more potential energy into translational kinetic energy.
Correct Answer: A — Disk
Q. A disk and a ring of the same mass and radius are rolling down an incline. Which will reach the bottom first?
A.Disk
B.Ring
C.Both will reach at the same time
D.Depends on the angle of incline
Solution
The disk has a smaller moment of inertia compared to the ring, hence it will reach the bottom first.
Correct Answer: A — Disk
Q. A disk and a ring of the same mass and radius are rolling without slipping down an incline. Which one will have a greater translational speed at the bottom?
A.Disk
B.Ring
C.Both have the same speed
D.Depends on the incline
Solution
The disk has a lower moment of inertia than the ring, allowing it to convert more potential energy into translational kinetic energy.
Correct Answer: A — Disk
Q. A disk of radius R and mass M is rotating about its axis with an angular velocity ω. What is the angular momentum of the disk?
A.(1/2)MR^2ω
B.MR^2ω
C.(1/4)MR^2ω
D.(3/2)MR^2ω
Solution
Angular momentum L = Iω = (1/2)MR^2ω, where I = (1/2)MR^2 for a disk.
Correct Answer: A — (1/2)MR^2ω
Q. A disk rolls down an incline. If the height of the incline is h, what is the speed of the disk at the bottom assuming no energy losses?
A.√(gh)
B.√(2gh)
C.√(3gh)
D.√(4gh)
Solution
Using conservation of energy, potential energy at height h converts to kinetic energy at the bottom. The speed is √(2gh).
Correct Answer: B — √(2gh)
Q. A disk rolls without slipping on a horizontal surface. If its radius is R and it rolls with a linear speed v, what is the angular speed of the disk?
A.v/R
B.R/v
C.vR
D.v^2/R
Solution
The relationship between linear speed and angular speed for rolling without slipping is given by ω = v/R.
Correct Answer: A — v/R
Q. A disk rotates about its axis with an angular velocity of ω. If its radius is doubled, what will be the new angular momentum if the mass remains the same?
A.2ω
B.4ω
C.ω
D.ω/2
Solution
Angular momentum L = Iω. If the radius is doubled, the moment of inertia increases by a factor of 4, thus L = 4Iω.
Correct Answer: B — 4ω
Q. A disk rotates about its axis with an angular velocity of ω. If its radius is doubled, what will be the new angular momentum?
A.2Iω
B.4Iω
C.Iω
D.I(2ω)
Solution
Angular momentum L = Iω, and if the radius is doubled, the moment of inertia I becomes 4I, thus L = 4Iω.
Correct Answer: B — 4Iω
Q. A disk rotates about its axis with an angular velocity of ω. If its radius is doubled, what will be the new angular velocity to maintain the same linear velocity at the edge?
A.ω/2
B.ω
C.2ω
D.4ω
Solution
The linear velocity v = rω. If the radius is doubled, to maintain the same v, the angular velocity must remain ω.
Correct Answer: B — ω
Q. A door is pushed at its edge with a force of 20 N. If the width of the door is 0.8 m, what is the torque about the hinges?
A.8 Nm
B.10 Nm
C.16 Nm
D.20 Nm
Solution
Torque (τ) = Force (F) × Distance (r) = 20 N × 0.8 m = 16 Nm.
Correct Answer: C — 16 Nm
Q. A door is pushed at its edge with a force of 20 N. If the width of the door is 1 m, what is the torque about the hinges?
A.10 Nm
B.20 Nm
C.30 Nm
D.40 Nm
Solution
Torque (τ) = F × d = 20 N × 1 m = 20 Nm.
Correct Answer: B — 20 Nm
Q. A door is pushed at its edge with a force of 50 N. If the width of the door is 1 m, what is the torque about the hinges?
A.25 Nm
B.50 Nm
C.75 Nm
D.100 Nm
Solution
Torque (τ) = Force (F) × Distance (r) = 50 N × 1 m = 50 Nm.
Correct Answer: B — 50 Nm
Q. A figure skater spins with arms extended. When she pulls her arms in, what happens to her rotational speed?
A.Increases
B.Decreases
C.Remains the same
D.Becomes zero
Solution
Pulling her arms in decreases her moment of inertia, causing her rotational speed to increase to conserve angular momentum.
Correct Answer: A — Increases
Q. A figure skater spins with arms extended. When she pulls her arms in, what happens to her angular velocity?
A.Increases
B.Decreases
C.Remains the same
D.Becomes zero
Solution
By conservation of angular momentum, pulling arms in decreases moment of inertia, thus increasing angular velocity.
Correct Answer: A — Increases
Q. A figure skater spins with arms extended. When she pulls her arms in, what happens to her angular momentum?
A.Increases
B.Decreases
C.Remains the same
D.Becomes zero
Solution
Angular momentum remains the same; however, her angular velocity increases due to a decrease in moment of inertia.
Correct Answer: C — Remains the same
Q. A flywheel is rotating with an angular speed of 20 rad/s. If it comes to rest in 5 seconds, what is the angular deceleration?
