I recently saw a Walter Lewin lecture on YouTube, and he proved how solid cylinder rolls down faster due to a smaller moment of inertia.

In one of the steps we get,

my step:

$$ma=mg\sin\theta−\mu mg\cos\theta$$

his step:


He equates friction with moment of inertia and gets a dependence relating it. (Nevermind that.)

My question is why can't we cancel mass in my step and conclude they will roll with the same acceleration downwards? (Although they do not and he also demonstrated it.)

I hope this doesn't come under check my work cause its been bugging me for way too long.

  • $\begingroup$ The second equation doesn’t seem to be dimensionally consistent. Can you check it? $\endgroup$
    – G. Smith
    Commented Oct 14, 2020 at 6:01
  • $\begingroup$ @G.Smith yep my bad , corrected. $\endgroup$
    – Linkin
    Commented Oct 14, 2020 at 6:23

2 Answers 2


In your step, you are assuming that the maximum value of static friction is acting, but that is not the case. Static friction can have variable values, with a limit that it canot cross. When a solid cylinder is rolling down, lesser frictional force acts on it than if a hollow cylinder is rolling down, while you have assumed it to be the same for both cases.

  • $\begingroup$ so when can I use μmgcosθ and when can I not sir? when it's not in pure roll? $\endgroup$
    – Linkin
    Commented Oct 14, 2020 at 5:23
  • $\begingroup$ For static friction, you can only use $f=\mu N$ if it is mentioned in the question (directly or indirectly), like find the maximum force something can tolerate before moving, etc. If it isnt mentioned, you have to calculate its value. $\endgroup$
    – dnaik
    Commented Oct 14, 2020 at 5:28
  • $\begingroup$ Here is the video, hollow verses solid starts at 10:10 youtube.com/watch?v=XPUuF_dECVI&t=200s but he says nothing about friction being a factor, which it would not. $\endgroup$ Commented Oct 14, 2020 at 6:04
  • $\begingroup$ @AdrianHoward he derived an expression using friction as a factor since they are not slipping friction won't be the same for all so my step was wrong and rest as dnaik has said. $\endgroup$
    – Linkin
    Commented Oct 14, 2020 at 6:29

Taking moments about the centre of the cylinder we have

$I\dot \omega=FR$

where $F$ is the frictional force. But $\omega = \frac v R$ so $\dot \omega = \frac {\dot v} R= \frac a R$ and so

$\displaystyle F = \frac {Ia}{R^2}$


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