# I really ! Torque problem

Cyannaca
I would really appreciate if anyone could help me with this problem. My exam is in 3 days and I don't understand how to do this problem!

A grinding stone of radius 10 cm ( I=0,2 kg*m^2) turns at a rate of 200 RPM. A tool is leaned against the circumference of the grinding stone with a force of 50N of radial direction. The kinetic coefficient of friction is equal to 0,6.
(A) What power is necessary to maintain the grinding stone in rotation at a constant angular velocity?

The answer is supposed to be 62,8 W and I know I have to use torque but I really don't know how to do it Cyannaca said:
A grinding stone of radius 10 cm ( I=0,2 kg*m^2) turns at a rate of 200 RPM. A tool is leaned against the circumference of the grinding stone with a force of 50N of radial direction. The kinetic coefficient of friction is equal to 0,6.
(A) What power is necessary to maintain the grinding stone in rotation at a constant angular velocity?

The answer is supposed to be 62,8 W and I know I have to use torque but I really don't know how to do it power formula: (T is torque, t is time)

$$P = \frac{T \theta}{t}$$

$$P = T\omega$$

convert 200rpm into rad/s and it should be easy from there.

I just worked the problem all the way through and the answer does work out.

Last edited:
Staff Emeritus
Gold Member
There's a real easy way to do this - almost a short cut.

Notice that if the stone must be turning at a constant angular velocity, it's Kinetic Energy must be constant. So the work done by the motor = work done by friction. Dividing by time, we have the power of motor = power removed by friction.

Also we know that power = force * velocity.
The relevant velocity here is the speed of the edge of the grinding wheel (where the friction acts) = w*R, where w is in rad/s. Lastly, the frictional force is 0.6 * 50 N = 30 N.

Plugging in numbers, you'll find that P = 63 W