Question

A small 0.35 kg ball on the end of a light thin rod is rotated in a horizontal circle of radius 1.2 m. Calculate the moment of inertia at the center of the circle and the torque needed to keep the ball rotating if the air resistance is 0.02 N against the ball. Ignore the air resistance in the moment of inertia.

a) Moment of Inertia: 0.504 kg·m², Torque: 0.06 N·m
b) Moment of Inertia: 0.168 kg·m², Torque: 0.02 N·m
c) Moment of Inertia: 0.504 kg·m², Torque: 0.02 N·m
d) Moment of Inertia: 0.168 kg·m², Torque: 0.06 N·m

Answer 1

The moment of inertia at the center of the circle is 0.504 kg·m². The torque needed to keep the ball rotating is 0.024 N·m.

Step-by-step explanation:

To solve for the moment of inertia and torque, we need to calculate the moment of inertia at the center of the circle and the force of air resistance. The moment of inertia at the center of the circle can be calculated using the formula I = MR², where M is the mass of the ball and R is the radius of the circle. In this case, the moment of inertia is 0.35 kg * 1.2 m² = 0.504 kg·m².

The torque needed to keep the ball rotating can be calculated using the formula τ = Fr, where F is the force of air resistance and r is the radius of the circle. In this case, the torque is 0.02 N * 1.2 m = 0.024 N·m.