Dynamics Help Determine Angular Acceleration of Disk?

Question

"The 100-kg cylindrical disk is at rest when the force F is applied to a cord wrapped around it. The static and kinetic coefficients of friction between the disk and the surface are 0.2. Determine the angular acceleration of the disk if (a) F=500 N and (b) F= 100 N."

They also give a hint: First solve the problem by assuming the disk does not slip, but rolls on the surface. Determine the frictional force, and find out whether it exceeds the product of the coefficient of friction and the normal force. If it does, you must rework the problem assuming that the disk slips.

The picture given shows a disk on a horizontal surface with the force moving towards the right. The disk has a radius of 300 mm. The answers in the back of the book were: (a) It doesn't slip, angular acceleration is 22.2 rad/s^2 clockwise and (b) It does slip, angular acceleration is 53.6 rad/s^2 clockwise. I got an answer close to part a but not sure if my method was right. ANy help to clear this up is great!!

Answer 1

Okay, I recently finished a dynamics course

Part a)

First make a diagram of the problem.  One of the hardest parts for me is drawing friction in the correct direction where there is rolling.  In this case friction should be pointing to the right because it is opposing the rolling motion.

Now let's do our force and moment equations:

Let f = friction force; a = acceleration (ax in x direction); N = normal force

Fx: 500 + f = 100*ax

Fy: N - 100*9.81 = 0

The moment equation is trickier: M = I*alpha

If it is rolling without slipping, alpha = a/R

I = 0.5*m*R^2

M: -0.3*500+0.3*f = 0.5*100*(0.3^2)*ax/0.3

Plugging in f to the equation from Fx:

-0.3*500 + 0.3(100ax - 500) = 50*0.3*ax

Thus, ax = 20, and f = 1500N

Since f = 1500N > 0.2*981 then the disk must be slipping.

Therefore f = friction coefficient * N and the moment equation changes to:

-150 + 0.3*0.2*981 = 0.5*100*(0.3^2)*alpha

Thus alpha = 46.4 rad/s^2 clockwise.

Part b)

Fx: 100 + f = 100ax => ax = f/100

M: -0.3*100+0.3f = 0.5*100*0.3*ax

Plugging in the equation from Fx:

-30 + 0.3f = 0.15f

f = 200, and ax = 2

Since f = 200 < 0.2*981 the cylinder is rolling

Therefore alpha = a/R = 2/0.3 = 6.67 rad/s^2 counter clockwise

I don't think the solution is right because it doesn't make sense that with a force of 500 N it would roll but with a force of 100N it would slip.  I might be wrong though.