## Rolling without slip

andres13_x
Posts: 1
Joined: Fri Jan 26, 2018 2:23 pm

### Rolling without slip

Hello, I have a question about this problem, I am not sure how to approach it. This is the problem:

When a disk rolls in a surface whitout slip, the velocity of the disk's edge (where it contacts the surface) is zero with respect the surface and the friction force is less than the maximum allowable of Us.N, where Us is the coefficient of static friction and N is the normal force exerted by the surface against the disk. Determine:

A)the maximum value of the force P such that the disk rolls without slip
B) angular acceleration of the disk for this maximum value of P

DATA:
m=1.8 kg
R=20cm
Us=0.25
The force P is placed in the center of disk and it goes to the right.

My attempt:

I know that when a disk rolls without slip, the condition Vcm= w.r must be satisfied, and also I read that in that condition, the friction force must be zero (But I am not sure about that last thing).

So setting my second newton's law we have the next (Ignoring the last thing I wrote):

P-fr= m.Acm (Where Acm is the acceleration in the mass center)

Now, I use the next equation:
Torque= I.α

Where Torque= Fr. R

So I would have: Fr. R = I.α

the moment of inertia for a disk is I=1/2 MR^2
So replacing I have:

Fr. R = 1/2MR^2.α

Canceling one R in each side I have:

Fr = 1/2MR.α

I also know α= Acm/R (Due the condition of rolling without slip)

So I have

Fr =1/2 MR.Acm /R (Where again I can cancel out both R's)

Fr =1/2 M.Acm

Solving for Acm: Acm= 2Fr/M
Now replacing this thing in the second newton's law I have this:

P- Fr = M. (2Fr/M)

Canceling both M and solving por P i have this:
P=3FR
Obiously I can find Fr with Fr= N.Us, But I am not sure if all that process is okay.

And also, I don't why angular acceleration depends of P, because I think I could find it with

Fr = 1/2MR.α

I would be happy if someone could help me with that Vimal
Posts: 2
Joined: Mon Jan 29, 2018 6:05 am

### Re: Rolling without slip

The first one you did is correct .
For the 2nd one we have to just place the value of fr. in the eqn. : P = 3fr.

and then the answer becomes:
alpha = P × 10^3/108

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