How Much Force is Needed to Stop a Spinning Grinding Wheel in 20 Seconds?

[tandoorichicken]

A piece of metal is pressed against the rim of a 1.6kg, 19-cm-diameter grinding wheel that is turning at 2400 rev/min. The metal has a coeffiecient of friction of .85 with respect to the wheel. When the motor is cut off, with how much force must you press to stop the wheel in 20 sec.?

Not sure I did this right, here's what I did:
2400 rev/min = 80 \pi rad/sec. = \omega_0.
t=20, and \omega_f = \omega_0 + \alpha t. After plugging everything in (w_f=0) I got \alpha = -40 \pi rad/sec^2. Since a = r \alpha, I got a = -3.4 \pi m/sec^2. Then somehow from there I jumped to an answer of 4.624 \pi N. I don't think that's right.

 

[Doc Al]

Originally posted by tandoorichicken
Not sure I did this right, here's what I did:
2400 rev/min = 80 \pi rad/sec. = \omega_0.
t=20, and \omega_f = \omega_0 + \alpha t. After plugging everything in (w_f=0) I got \alpha = -40 \pi rad/sec^2.

\alpha = -4\pi. Now that you've calculated \alpha, how do you relate this to force? Make use of \tau = I\alpha to find the torque and thus the frictional force. You will need to know the rotational inertia (I) of a disk; look it up.

 

[tandoorichicken]

So once I find the linear force of the wheel do I just multiply that by the coefficient of friction to find the force that I push with?

 

[Doc Al]

Originally posted by tandoorichicken
So once I find the linear force of the wheel do I just multiply that by the coefficient of friction to find the force that I push with?

No. The force you press with is the normal force; the force producing the torque on the wheel is the friction. F_friction=μN.