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Suppose we have a cylinder on an inclined plane of mass $m$ and radius $R$ moving without sliding (so that $\varepsilon = a/R$). Why is the friction $F$ causing the circular motion sometimes lower than $F' = f \cdot N$, where $N$ is a normal force and $f$ the friction coefficient? ($F \le f \cdot N$)

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  • $\begingroup$ this sounds like a 'homework' question. $\endgroup$ – theo Dec 7 '14 at 13:18
  • $\begingroup$ But it is not. It simply wasn't sufficiently explained to us. $\endgroup$ – marmistrz Dec 7 '14 at 18:22
  • $\begingroup$ Thats OK, either way! Do you have a diagram for it? What are the parameters "$\epsilon$" and "$a$" ? $\endgroup$ – theo Dec 7 '14 at 19:37
  • $\begingroup$ You mean some drawing of the situation? As for parameters, well, as ordinarily - circular and linear acceleration. $\endgroup$ – marmistrz Dec 8 '14 at 6:07
  • $\begingroup$ Usually an $\alpha$ and not an $\varepsilon$ is used to represent angular (circular) acceleration $\endgroup$ – Steeven Sep 7 '16 at 8:34
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It's like that because the rotation without sliding is caused by static friction, not the kinetic one. Thus $T \le f N$. On the other hand, while if the cylinder is sliding, we take the kinetic friction.

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  • $\begingroup$ It doesn't specify what type of friction it is. $\endgroup$ – JMac Feb 27 '17 at 11:33
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Draw a free body diagram for the object. You will see that this friction $F$ is the only force that attacks at the edge. This is the only force that gives a torque.

Since it isn't balanced by any other torques, this is the reason for the accelerating rotation.

The size of this force $F$ doesn't matter - as long as it exists, it will cause a torque.

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