A 2.0-kg hoop rolls without slipping on a horizontal surface so that its center proceeds to the right with a constant linear speed of 6.0 m/s.
-What is the total kinetic energy of the hoop?
A)36 J
B)54 J
C)72 J
D)96 J
E)140 J

Question

A 2.0-kg hoop rolls without slipping on a horizontal surface so that its center proceeds to the right with a constant linear speed of 6.0 m/s. 
-What is the total kinetic energy of the hoop?
A)36 J 
B)54 J 
C)72 J 
D)96 J 
E)140 J

Quizplus · Accepted Answer

Since the hoop is moving with a constant linear speed, its kinetic energy is only due to its rotational KE. The equation for rotational KE is:
KE(rot) = (1/2)*I*w^2
where I is the moment of inertia and w is the angular velocity.
For a hoop, I = MR^2, where M is the mass and R is the radius.
Since the hoop is rolling without slipping, w = v/R, where v is the linear speed.
Substituting the values given in the problem:
KE(rot) = (1/2)*MR^2*(v/R)^2 = (1/2)*M*v^2
= (1/2)*2.0 kg*(6.0 m/s)^2 = 72 J
Therefore, the correct answer is (C)

A 20-Kg Hoop Rolls Without Slipping on a Horizontal Surface So