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).
Q. A solid sphere rolls down a hill without slipping. If the height of the hill is h, what is the speed of the sphere at the bottom of the hill?
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 sphere, I = (2/5)mr^2 and ω = v/r. Solving gives v = √(2gh).
Q. A solid sphere rolls down an inclined plane without slipping. What is the ratio of its translational kinetic energy to its total kinetic energy at the bottom?
A.1:2
B.2:3
C.1:3
D.1:1
Solution
The total kinetic energy is the sum of translational and rotational kinetic energy. For a solid sphere, the ratio of translational to total kinetic energy is 2:3.
Q. A solid sphere rolls down an inclined plane without slipping. What is the ratio of its translational kinetic energy to its total kinetic energy at the bottom of the incline?
A.1:2
B.2:3
C.1:3
D.1:1
Solution
The total kinetic energy is the sum of translational and rotational kinetic energy. For a solid sphere, the ratio of translational to total kinetic energy is 2:5, which simplifies to 2:3.
Q. A wheel of radius R rolls on a flat surface. If it rolls without slipping, what is the distance traveled by the center of mass after one complete rotation?
A.2πR
B.πR
C.4πR
D.R
Solution
The distance traveled by the center of mass after one complete rotation is equal to the circumference of the wheel, which is 2πR.
Q. A wheel of radius R rolls without slipping on a horizontal surface. If it rotates with an angular velocity ω, what is the linear velocity of the center of the wheel?
A.Rω
B.2Rω
C.ω/R
D.R/ω
Solution
The linear velocity v of the center of the wheel is related to the angular velocity ω by the equation v = Rω.
Q. If a rolling object has a mass m and radius r, what is the expression for its total kinetic energy?
A.(1/2)mv^2
B.(1/2)mv^2 + (1/2)Iω^2
C.(1/2)mv^2 + (1/2)mr^2ω^2
D.(1/2)mv^2 + (1/2)(2/5)mr^2(ω^2)
Solution
The total kinetic energy of a rolling object is the sum of translational and rotational kinetic energy, which can be expressed as (1/2)mv^2 + (1/2)(2/5)mr^2(ω^2).
Correct Answer: D — (1/2)mv^2 + (1/2)(2/5)mr^2(ω^2)
Q. If a rolling object has a radius of R and rolls with a speed v, what is its kinetic energy?
A.(1/2)mv^2
B.(1/2)mv^2 + (1/2)Iω^2
C.(1/2)mv^2 + (1/2)(1/2)mR^2(v/R)^2
D.None of the above
Solution
The total kinetic energy of a rolling object is the sum of translational and rotational kinetic energy, which simplifies to (1/2)mv^2 + (1/4)mv^2 = (3/4)mv^2.
Correct Answer: C — (1/2)mv^2 + (1/2)(1/2)mR^2(v/R)^2
Q. If a solid cylinder rolls without slipping, what fraction of its total kinetic energy is translational?
A.1/3
B.1/2
C.2/3
D.1
Solution
For a solid cylinder, the total kinetic energy is KE_total = KE_translational + KE_rotational = (1/2)mv^2 + (1/2)(1/2)mR^2(ω^2). Since ω = v/R, the translational part is 2/3 of the total.
Q. If a solid cylinder rolls without slipping, what is the ratio of its translational kinetic energy to its rotational kinetic energy?
A.1:1
B.2:1
C.1:2
D.3:1
Solution
For a solid cylinder, the translational kinetic energy (KE_trans) is (1/2)mv² and the rotational kinetic energy (KE_rot) is (1/2)(Iω²). The ratio KE_trans:KE_rot is 1:2.
Q. If a solid sphere and a solid cylinder of the same mass and radius are released from rest at the same height, which will have a greater speed at the bottom?
A.Solid sphere
B.Solid cylinder
C.Both have the same speed
D.Depends on the mass
Solution
Both will have the same speed at the bottom due to conservation of energy, as they start from the same height.
