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 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).
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.
Q. A rolling object has both translational and rotational motion. Which of the following quantities remains constant for a rolling object on a flat surface?
A.Linear velocity
B.Angular velocity
C.Total energy
D.Kinetic energy
Solution
The total energy remains constant for a rolling object on a flat surface, assuming no external work is done.
Q. A solid cylinder and a hollow cylinder of the same mass and radius are released from rest at the same height. Which one will have a greater speed at the bottom?
A.Solid cylinder
B.Hollow cylinder
C.Both have the same speed
D.Depends on the mass
Solution
The solid cylinder has a smaller moment of inertia compared to the hollow cylinder, thus it will have a greater speed at the bottom.
Q. A solid cylinder of radius R rolls down a frictionless incline. What is the ratio of its translational kinetic energy to its total kinetic energy at the bottom?
A.1:1
B.2:1
C.1:2
D.3:1
Solution
At the bottom, total kinetic energy = translational + rotational. For a solid cylinder, the ratio of translational to total kinetic energy is 2:1.
Q. A solid sphere and a hollow sphere of the same mass and radius are released from rest at the same height. Which one reaches the bottom first?
A.Solid sphere
B.Hollow sphere
C.Both reach at the same time
D.Depends on the surface
Solution
The solid sphere reaches the bottom first because it has a lower moment of inertia, allowing it to convert more potential energy into translational kinetic energy.
Q. A solid sphere of radius R rolls without slipping down an inclined plane of angle θ. What is the acceleration of the center of mass of the sphere?
A.g sin(θ)
B.g sin(θ)/2
C.g sin(θ)/3
D.g sin(θ)/4
Solution
The acceleration of the center of mass of a solid sphere rolling down an incline is given by a = g sin(θ) / (1 + (2/5)) = g sin(θ) / (7/5) = (5/7) g sin(θ).
