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Question: Answered & Verified by Expert
A cylinder of mass $10 \mathrm{~kg}$ and radius $15 \mathrm{~cm}$ is rolling perfectly on a plane of inclination $30^{\circ}$. The coefficient of static friction is $\mu_s=\mathbf{0 . 2 5}$.
(a) How much is the force of friction acting on the cylinder?
(b) What is the work done against friction during rolling?
(c) If the inclination $\theta$ of the plane is increased, at what value of $\boldsymbol{\theta}$ does the cylinder begin to skid and not roll perfectly?
PhysicsSystem of Particles and Rotational Motion
Solution:
2824 Upvotes Verified Answer
Given: $m=10 \mathrm{~kg}, r=15 \mathrm{~cm}=0.15 \mathrm{~m}, \theta=30^{\circ}, \mu_s=0.25$
Acceleration of the cylinder when rolling down the inclined plane $=a$
$$
=\frac{2}{3} \mathrm{~g} \sin \theta=\frac{2}{3} 9.8 \times \sin 30^{\circ}=\frac{9.8}{3} \mathrm{~m} / \mathrm{s}^2
$$
(a) Force of friction $=f=m g \sin \theta-m a$
$$
\begin{aligned}
&=m(g \sin \theta-a)=10\left(9.8 \sin 30^{\circ}-\frac{9.8}{3}\right) \\
&=16.4 \mathrm{~N}
\end{aligned}
$$
(b) Work done against the friction during rolling is zero as the point of contact is at rest.
(c) For rolling without slipping, $\mu=\frac{1}{3} \tan \theta$ $\Rightarrow \tan \theta=3 \times 0.25=0.75 \Rightarrow \theta=37^{\circ}$
Beyond this $\theta$, it will not roll perfectly.

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