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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?
(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?
Solution:
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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.
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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