The acceleration of the point of contact, when a solid disc is rolling on a plank [closed]

Question

A solid disc, initially at rest has a constant force applied on it, and it moves by rolling (pure roll) on a plank of the same mass (this plank can move as well - its on a frictionless surface).

My question is, is the acceleration of the point of contact b/w the plank and disc zero (like how velocity of point of contact was zero in normal pure roll)? How would I relate the accelerations of the two bodies?

Edit : I already know that net force on the disc is F - friction, and that net force on plank is only due to friction. Also, I understand, that torque produced is only due to friction. My main trouble is in relating acceleration of the disc and the plank. I see that the disc is in a non-inertial frame of reference, does this change anything? Like when using $Torque= I*α$ how do I relate $α$ to a1 and a2 - accelerations of plank and disc.

Answer 1

The motion of the disc is governed by $m_d a_d = F - F_c$ with $F_c$ the contact force between the disc and the plank.

The motion of the plank is governed by $m_p a_p = F_c$.

Finally, the rotation of the disc is governed by $I \alpha = R F_c$.

Since there is no slipping, the point of contact has the same velocity as the plank $v_p$, so its velocity relative to the center is $v_p - v_d = - R \omega$ (negative because the bottom of the disk is moving backward). Differentiating this, we get $\alpha = \frac{a_d - a_p}{R}$.

Additionally, $I = \frac12 m_d R^2$ so we have $\frac12 m_d (a_d - a_p) = F_c.$

Substituting $a_d$ and $a_c$ using the relations above, we get $\frac12 (F - F_c - \frac{m_d}{m_p} F_c) = F_c$ and finally $F = (3 + \frac{m_d}{m_p}) F_c$.

Answer 2

Assuming the rolling slippage between the disk and the plank is zero, meaning as the disk rolls it carries the plank with it:

We know I of a disk is: $ I =1/2mR^2$

Let's call $ \ m\times a_{plank}, F_p $

the force F is shared between the torque $ \tau = _{F_b}R, \ and \ F_p$.

$ \tau = I\alpha , \ $ and $ \alpha = \frac {\Delta\omega}{\Delta t} $

But we know $ \ \omega =V/R$

Therefor:

$ F = \tau /R + F_p $

From here you should be able to handle it.