A small 520-gram ball on the end of a thin, light rod is rotated in a horizontal circle of radius 1.5 m. Calculate the moment of inertia of the ball about the center of the circle. Then calculate the torque needed to keep the ball rotating at constant angular velocity if air resistance exerts a force of 0.015 N on the ball. Ignore the rod's moment of inertia and air resistance. I don't understand how to do this problem, please help with step-by-step.

asked by Anonymous

14 years ago

2,778 views

0

0

1 answer

The moment of inertia for the mass is

I=mr^2

Now, torque to counter air friction..

airresiatance*r=torque

answered by bobpursley

2

6

Ask a New Question or answer this question.

Question

A small 520-gram ball on the end of a thin, light rod is rotated in a horizontal circle of radius 1.5 m. Calculate the moment of inertia of the ball about the center of the circle. Then calculate the torque needed to keep the ball rotating at constant angular velocity if air resistance exerts a force of 0.015 N on the ball. Ignore the rod's moment of inertia and air resistance. I don't understand how to do this problem, please help with step-by-step.

1 answer

The moment of inertia for the mass is

I=mr^2

Now, torque to counter air friction..

airresiatance*r=torque

Ask a New Question or answer this question.

bobpursley · Answer

The moment of inertia for the mass is I=mr^2 Now, torque to counter air friction.. airresiatance*r=torque