Q. A solid sphere and a hollow sphere of the same mass and radius are released from rest at the same height. Which one will have a greater translational speed when they reach the ground?
A.Solid sphere
B.Hollow sphere
C.Both will have the same speed
D.Depends on the mass
Solution
The solid sphere will have a greater translational speed because it has a smaller moment of inertia.
Q. A solid sphere and a hollow sphere of the same mass and radius are released from rest at the same height. Which one will have a greater linear speed when they reach the ground?
A.Solid sphere
B.Hollow sphere
C.Both have the same speed
D.Depends on the mass
Solution
The solid sphere will have a greater linear speed because it has a smaller moment of inertia, allowing it to convert more potential energy into translational kinetic energy.
Q. A solid sphere of mass M and radius R is rolling without slipping on a horizontal surface. What is the expression for its total angular momentum about its center of mass?
A.(2/5)MR^2ω
B.MR^2ω
C.MR^2
D.0
Solution
Total angular momentum L = Iω, where I = (2/5)MR^2 for a solid sphere.
Q. A solid sphere of mass m and 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 rolling object is given by a = g sin(θ) / (1 + k^2/r^2). For a solid sphere, k^2/r^2 = 2/5, thus a = g sin(θ) / (1 + 2/5) = g sin(θ) / (7/5) = (5/7)g sin(θ).
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 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 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 without slipping down an incline. What is the ratio of its translational kinetic energy to its total kinetic energy at the bottom?
A.1:2
B.2:3
C.1:1
D.1:3
Solution
For a solid sphere, the ratio of translational kinetic energy to total kinetic energy is 2:3.
