For the question I attached I can't seem to understand one part of it and my friend and teacher can't seem to explain it too. I've included the solution with my understanding of what's happening.

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The torque applied on a body equals the body's moment of inertia times the body's angular acceleration:

$$Fd=I\alpha=\frac{1}{2}MR^2\alpha \ \rm{(eqn 1)}$$

However, this is where I'm confused:

Since $F=Ma$ (From Newton's second law)

$$(Ma)d=\frac{1}{2}MR^2\alpha\ \rm{(eqn 2)}$$

Since $\alpha=\frac{a}{r}$

$$M(aR)d=\frac{1}{2}MR^2\alpha\ \rm{(eqn 3)}$$


The part that I'm confused is: In the solution, linear acceleration is converted to angular acceleration by the relationship $\alpha=\frac{a}{r}$; but this relationship refers to the linear acceleration at a point that is a distance $r$ from the centre of mass (since linear acceleration is different at different points of the cylinder I believe).

And so when the solution replaces the linear acceleration, $a$, to angular acceleration $\alpha$ multiplied by the radius $R$ of the cylinder in equation 3, it implies that the $a$ in $F=ma$ refers to the linear acceleration at the circumference i.e. $a_{\rm circumference}$ (since the point at circumference is a distance $R$ from centre of mass). But why is that? Why is $a_{\rm circumference}$ used for the applied force's linear acceleration instead of the linear acceleration of the cylinder at the point $d$ from the center of mass i.e. $a_{\rm d}$ (where the force is applied to provide the torque)?

If $a_{\rm d}$ is used in replacing the applied force by $F=Ma$, then the $R$ on the left side of equation 3 would not cancel with one of the $R$ in the right side of equation 3 (which will not yield the solution)

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    $\begingroup$ There is no such thing as `The Force of A Torque". $\endgroup$ – John Alexiou Oct 25 '18 at 17:31

Regardless of where you apply the force, the acceleration is $F=ma_{CM}$, where CM is center of mass. The condotion for no slipping is that the speed at the bottom is zero, so $v=\omega R$, in this way the speed at the center will cancel the speed at the bottom, and the speed at the top will be twice that speed (because the ball is rolling). Same with the acceleration: $a=\alpha R$. Notice also that the acceleration at the point of application is not even horizontal, because the cilynder is rotating.

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  • $\begingroup$ Thanks you so much, I think what I understood wrong was that I thought the centre of the cylinder was the axis of rotation (instead of the bottom of the cylinder). But could you elaborate a bit on what do you mean by "in this way the speed at the center will cancel the speed at the bottom?" $\endgroup$ – Bøbby Leung Oct 25 '18 at 16:37
  • $\begingroup$ I mean imagine a vertical line that passes through the center. The center will have velocity v, and because it is rotating without slipping, the bottom of the line will have velocity 0 (because v+(-wR)=0, and the top 2v (v+wR=2v) $\endgroup$ – Wolphram jonny Oct 25 '18 at 18:03

The equations of motion (CM) are:




$M\,\ddot{x}=F\quad\quad (2)$

$ \Rightarrow$



Roll without slip means :

$\ddot{x}=R\,\ddot{\varphi}$ $ \Rightarrow$

$\frac{F}{M}=\frac{2\,F\,d}{M\,R^2}\quad\quad (3)$

We solve equation (3) for $d$ and get:


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The confusion arises from the fact that your target is pure rolling which means the contact point has no velocity, and the velocity of the center of mass is $v = \omega r$


As a consequence (since $r$ is constant) the acceleration of the center of mass is $a = \alpha r $

The equations of motion resolved at the center of mass are thus

$$ \boxed{ \begin{aligned} F & = m (\alpha r) \\ F d &= I \alpha \end{aligned} } $$

Here $I = \tfrac{m}{2} r^2$.

which are used to solve for $\alpha$ and $d$. The acceleration of the surface of the cylinder bears no influence to the solution. It is just an illusion because it is a rolling cylinder, and the acceleration of the center of mass equals the surface acceleration of the rotating body.

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  • $\begingroup$ Thanks for the detailed answer. Just to confirm my understanding, with regard to your equation 'Fd=Ia', does Fd refers to the torque, at the point where the force is applied i.e. call it point d, relative to the centre of mass? And Ia refers to the torque, at the centre of mass, relative the ground? $\endgroup$ – Bøbby Leung Oct 25 '18 at 23:57
  • $\begingroup$ If what I said is true, then is it true to say that the rotational motion at point d relative to the centre of mass causes an equal rotational motion of the centre of mass relative to the ground? $\endgroup$ – Bøbby Leung Oct 25 '18 at 23:57
  • $\begingroup$ @BøbbyLeung - The equations of motion have to be evaluated at the center of mass, and hence $\tau = F\,d$ is the total torque applied to the body at the center of mass. The rotational motion is common throughout the body and thus there isn't such a thing as "the rotational motion at a point". $\endgroup$ – John Alexiou Oct 26 '18 at 12:52

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