A 2.0-kg stone is tied to a 0.50 m long string and swung around a circle at a constant angular velocity of 12 rad/s. The circle is parallel to the xy plane and is centered on the z axis, 0.75 m from the origin. The magnitude of the torque about the origin is:
A) 0
B) 6.0 N . m
C) 14 N .m
D) 72 N. m
E) 108 N. m

Question

Quizplus · Accepted Answer

The torque is calculated using τ = r × F, where F is the centripetal force (mω²r) and r is the perpendicular distance from the axis of rotation. Here, r = 0.75 m, m = 2.0 kg, and ω = 12 rad/s. Thus, τ = 0.75 m × (2.0 kg × (12 rad/s)² × 0.5 m) = 108 N·m.

A 20-Kg Stone Is Tied to a 0