Question

A rigid body of mass $M$ and radius $R$ rolls without slipping on an inclined plane of inclination $\theta$, under gravity. Match the type of body Column I with magnitude of the force of friction Column II.

Column I		Column II
A	For ring	I	$\frac{Mg\sin \theta }{2.5}$
B	For solid sphere	II	$\frac{Mg\sin \theta }{3}$
C	For solid cylinder	III	$\frac{Mg\sin \theta }{3.5}$
D	For hallow cylinder	IV	$\frac{Mg\sin \theta }{2}$

• A. A=IV, B=III ,C= II,D=IV
• B. A=I, B=II ,C=IV ,D= IV
• C. A=II, B=I ,C= IV,D= III
• D. A=II, B=IV ,C=I ,D= III

Answer 1

Answer: (A)

Given, mass, radius of rigid body be $m$ and $R$ respectively,
$g$ is acceleration due to gravity and $\theta$ is inclination.
Let $I$ be the moment of inertia, then
$I=nm{R}^2$
Taking force along inclined
$mg\sin \theta -f=ma\dots \text{ (i) }$

where, $f$ is friction force and $a$ is acceleration.
$\tau =I\alpha =fR\dots \text{ (ii) }$
where, $\tau =$ torque,
$I=$ moment of inertia
and $\alpha =$ angular acceleration $=\frac{a}{R}$.
Substituting in Eq. (ii), we get
$\Rightarrow mg\sin \theta -nma=ma$
$\Rightarrow g\sin \theta =a(n+1)$
$\Rightarrow a=\frac{g\sin \theta }{n+1}$
Substituting in Eq. (iii), we get
$f=nma$
$=nm\frac{g\sin \theta }{n+1}\dots$
(A) For ring, $I=m{R}^2$
$\therefore n=1$
$\text{ and }{f}_{\text{ring }}=\frac{mg\sin \theta }{1+1}=\frac{mg\sin \theta }{2}$
(B) For solid sphere, $I=\frac{2}{5}m{R}^2$
$\therefore n=\frac{2}{5}$
$f_{\text{solid sphere }}=\frac{\frac{2}{5}mg\sin \theta }{\frac{2}{5}+1}$
$=\frac{2mg\sin \theta }{7}$
$=\frac{mg\sin \theta }{3.5}$
$\therefore n=\frac{2}{5}$
$f_{\text{solid sphere }}=\frac{\frac{2}{5}mg\sin \theta }{\frac{2}{5}+1}$
$=\frac{mg\sin \theta }{3}$
$\therefore n=\frac{1}{2}$
$f_{\text{solid cylinder }}=\frac{\frac{1}{2}mg\sin \theta }{\frac{1}{2}+1}$ $=\frac{mg\sin \theta }{3}$
(D) For hollow cylinder $I=M{R}^2$
$\therefore n=1$
$f_{\text{hollow cylinder }}=\frac{mg\sin \theta }{1+1}$
$=\frac{mg\sin \theta }{2}$
$\therefore$ Hence, $A\to 4,B\to 3,C\to 2,D\to 4$ is the correct.