Calculating Power and Torque for Rotational Motion

[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

 

[ShawnD]

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)

[tex]P = \frac{T \theta}{t}[/tex]

[tex]P = T\omega[/tex]


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.

 

[Gokul43201]

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

 

[ShawnD]

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.

You totally stole that from my post :tongue:

 

[Theelectricchild]

Shawn how did he steal that from your post?