A solid sphere of radius r = 5 cm and mass m = 0.2 kg rolls in a groove as shown. The angle, $\theta$, between the points of contact with the groove is 120 $\circ$ . If the sphere has a linear velocity of 10 mm/s, what is the rotational kinetic energy?
A) 0.8 mJ
B) 1.0 mJ
C) 1.6 mJ
D) 2.0 mJ
E) undetermined.

Question

A solid sphere of radius r = 5 cm and mass m = 0.2 kg rolls in a groove as shown. The angle, $\theta$, between the points of contact with the groove is 120 $\circ$ . If the sphere has a linear velocity of 10 mm/s, what is the rotational kinetic energy? 
A) 0.8 mJ 
B) 1.0 mJ 
C) 1.6 mJ 
D) 2.0 mJ 
E) undetermined.

Quizplus · Accepted Answer

The rotational kinetic energy of a solid sphere is given by:

KE_rot = (2/5) * m * r^2 * w^2

where m is mass of the sphere, r is its radius, and w is its angular velocity.

To find the angular velocity, we need to use the fact that the sphere is rolling without slipping. In this case, the velocity of the center of mass of the sphere, v_cm, is related to its angular velocity by:

v_cm = r * w

We are given that the linear velocity of the sphere is 10 mm/s, which is equal to 0.01 m/s. Since the sphere is rolling without slipping, its angular velocity is w = v_cm/r = 0.01/0.05 = 0.2 rad/s.

Now we can plug in the values for m, r, and w into the expression for KE_rot:

KE_rot = (2/5) * m * r^2 * w^2 = (2/5) * 0.2 * 0.05^2 * 0.2^2 = 0.00016 J = 1.6 mJ

Therefore, the correct answer is C.

A Solid Sphere of Radius R = 5 Cm and Mass