A grindstone in the shape of a solid disk with diameter 0.520 m and a mass of 50

Question

A grindstone in the shape of a solid disk with diameter 0.520 m and a mass of 50.0 kg is rotating at 850 rev/min. You press an axe against the rim with a normal force of 160 N and the grindstone comes to rest in 7.50 sec. What is the angular acceleration of the grindstone as it comes to a stop? What is the moment of inertia of the grindstone? What force is bringing the grindstone to ta stop? What is the net torque of that force? What is the size of the force? What is the coefficient of friction between the axe and the grindstone?

Explanation / Answer

normal force(Fn) = 160N.

The friction force (Ff) = u*Fn, or u*160 and is the tangential force on the grindstone.

You need to look up the rotational inertia of a solid circular disk (it's a solid disk since they don't give you the grindstone thickness), and according to the reference, this is:

I = m*r^2/2 = 50*(.52/2)^2/2 = 1.69kg*m^2

the initial angular velocity is:

850rpm*2*pi/60 = 89.01rad/s

The angular acceleration can be derived from:

omega = alpha*t

alpha = omega/t = 89.01/7.50 = 11.86rad/s^2

The torque is Ff*r but it is also I*alpha

Ff*r = u*160*r = I*alpha

u = I*alpha/(160*r) = 1.69*11.86/(160*.26)

u = 0.481

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