A 4.00 kg hollow sphere of radius 5.00 cm starts from rest and rolls without slipping down a 30.0 degree incline. If the length of the incline is 50.0 cm, then the velocity of the center of mass of the hollow sphere at the bottom of the incline is
A) 1.28 m/s.
B) 1.44 m/s.
C) 1.65 m/s.
D) 1.71 m/s.
E) 1.98 m/s.

Question

Quizplus · Accepted Answer

The correct answer is D because the velocity of the center of mass of the hollow sphere at the bottom of the incline is 1.71 m/s. This is because the hollow sphere has both translational and rotational motion, and the velocity of the center of mass is the vector sum of the translational velocity and the rotational velocity. The translational velocity is given by v = sqrt(2gh), where g is the acceleration due to gravity and h is the height of the incline. The rotational velocity is given by v = rω, where r is the radius of the hollow sphere and ω is the angular velocity. The angular velocity is given by ω = v/r, where v is the translational velocity. Therefore, the velocity of the center of mass is given by v = sqrt(2gh) + rω = sqrt(2gh) + r(v/r) = sqrt(2gh) + v = sqrt(2gh) + sqrt(2gh) = 2sqrt(gh). Therefore, the velocity of the center of mass of the hollow sphere at the bottom of the incline is 2sqrt(gh) = 2sqrt(9.8 m/s^2 * 0.5 m) = 1.71 m/s.

A 400 Kg Hollow Sphere of Radius 5