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?enter image description here

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.

  • $\begingroup$ When you say "pure roll" I'm guessing you mean "rolling without slipping", which implies the relative velocity and acceleration of the point of contact is zero. I'm curious how the force is acting on the disk - is it acting on an edge, making it rotate or at the CM, making it translate? $\endgroup$
    – levitopher
    Commented Dec 6, 2018 at 16:34
  • $\begingroup$ @levitopher I don't think the acceleration is $0$. $\endgroup$ Commented Dec 6, 2018 at 16:35
  • $\begingroup$ @AaronStevens: the answer is frame-dependent, which is kind of what I was trying to get at. One could imagine this system sliding on a frictionless surface where the point of contact is stationary in the frame of the surface. $\endgroup$
    – levitopher
    Commented Dec 6, 2018 at 16:42
  • $\begingroup$ @levitopher the force is acting at the center of mass, and there's sufficient friction b/w plank and disc to prevent slipping. $\endgroup$ Commented Dec 7, 2018 at 1:21
  • $\begingroup$ @AaronStevens : ya, I'm asking for the acceleration (in the ground frame) of a point on the disc when it touches the plank. Mainly my question is how to relate the accelerations of the plank and the disc, because they seem somewhat connected- I'm not sure how. $\endgroup$ Commented Dec 7, 2018 at 1:27

2 Answers 2


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 $F = (3 + \frac{m_d}{m_p}) F_c$.

  • $\begingroup$ Thanks, I got most of what you did.. But could you tell me how you got $α=(ad-ap)/R$ ? $\endgroup$ Commented Dec 7, 2018 at 16:10
  • $\begingroup$ Is it because $ad-ap$ is the relative acceleration of the disc relative to the plank? But, since the plank and disc move in opposite directions, wouldn't their relative acceleration be $a1+a2$ ? $\endgroup$ Commented Dec 7, 2018 at 16:19
  • $\begingroup$ @karunmathews I'll edit my post to explain that part more. I consider all accelerations in the same direction, that's why I used $a_d - a_p$. If the plank moved in the opposite direction, $a_p$ would simply be negative. However, it actually does not: it is dragged along by the disc. $\endgroup$
    – Aetol
    Commented Dec 7, 2018 at 16:29
  • $\begingroup$ No, but when the disc rolls on the plank- it seems like they move in opposite directions.. $\endgroup$ Commented Dec 7, 2018 at 16:30
  • 1
    $\begingroup$ The friction of the plank on the disc is backward, since the disc is rolling. By Newton's third law, the friction of the disc on the plank is in the opposite direction, forward. Imagine an extreme case: a very heavy, hard to turn wheel on a very light plank. If you push the wheel, it will hardly roll, instead it will slide with the plank on the frictionless ground. $\endgroup$
    – Aetol
    Commented Dec 8, 2018 at 0:18

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$


$ F = \tau /R + F_p $

From here you should be able to handle it.

  • $\begingroup$ But torque isn't a force, right? Then how can we write $F= T + Fp$ . I think Fp is the equal to the friction force acting on the disc, so that way $T= Fp*R$ . But how is F related to Fp? Is this correct: (a1 & a2 are accelerations of disc and plank respectively) because $w= v/R$ , so $α=a1/R$ => $T=I*(a1/R)$ .. So $I*(a1/R) = Fp*R$ .. And Fp=ma2, so $I*(a1/R)=m*a2$ . $\endgroup$ Commented Dec 7, 2018 at 4:30
  • $\begingroup$ @karunmathews, yes, torque is twisting force. and its work is 'tau' times 'theta'. also if the Friction force is greater than Mp times alpha, no need to consider friction, or else we substitute friction force. but it is a more interesting case if the friction is greater than Mp. a , because then the system is dynamically tied together. $\endgroup$
    – kamran
    Commented Dec 7, 2018 at 4:55
  • $\begingroup$ So, is this relation correct: $I*(a1/R)=m*a2*R$ - I forgot to add the R in my last comment. $\endgroup$ Commented Dec 7, 2018 at 5:43
  • $\begingroup$ @karunmathews, yes sounds right. so I modify my answer to express tau in terms of linear force to be at the same direction with force of plank, tau/R. $\endgroup$
    – kamran
    Commented Dec 7, 2018 at 15:25

Not the answer you're looking for? Browse other questions tagged or ask your own question.