For a hoop (ring)of mass M and radius R that is rolling without slipping,which is greater,its translational or its rotational kinetic energy?
A)Its translational kinetic energy is greater.
B)Its rotational kinetic energy is greater.
C)They are equal.
D)The answer depends on the radius.
E)The answer depends on the mass.

Question

Quizplus · Accepted Answer

For a hoop rolling without slipping, its translational and rotational kinetic energies are related as follows: 
KE_translational = (1/2) * M * v^2 
KE_rotational = (1/2) * I * w^2 
where v is the linear speed, w is the angular speed, and I is the moment of inertia of the hoop. 
For a hoop rolling without slipping, we know that v = R*w (where R is the radius of the hoop), and I = (1/2) * M * R^2. 
Substituting these relations in the expressions for KE_translational and KE_rotational, we get: 
KE_translational = (1/2) * M * (R*w)^2 = (1/2) * M * R^2 * w^2 = (1/2) * I * w^2 = KE_rotational

Therefore, the translational and rotational kinetic energies are equal.

For a Hoop (Ring)of Mass M and Radius R That