Question

A uniform sphere of radius $R$ and mass $m$ is placed on an inclined plane which makes an angle $45^{\circ }$ to the horizontal. For which of the following value of coefficient of friction, the sphere rolls without slipping,

• A. $\frac{3}{7}$
• B. $\frac{1}{2}$
• C. $\frac{5}{8}$
• D. $\frac{1}{7}$

Answer 1

Answer: (A), (B), (C)

$(a,b,c)$ Given,
radius of sphere $=R$
mass of sphere $=m$
and the angle of inclination in inclined plane,
$\theta =45^{\circ }$
The linear acceleration of solid sphere is given by

$a=\frac{g\sin \theta }{1+\frac{I}{M{R}^2}}=\frac{g\sin \theta }{1+\frac{\frac{2}{5}M{R}^2}{M{R}^2}}\left[\because I=\frac{2}{5}M{R}^2\right]$
$a=\frac{5}{7}g\sin \theta$
Applying Newton's second law of motion for linear motion of the centre of mass, $mg\sin \theta -f=ma$
$\Rightarrow f=mg\sin \theta -ma=mg\sin \theta -m\frac{5}{7}\cdot g\sin \theta$
$=\frac{2}{7}mg\sin \theta$
$\therefore$ Coefficient of friction $\left({\mu }_s\right)=\frac{f}{N}=\frac{\frac{2}{7}mg\sin \theta }{mg\cos \theta }$
$=\frac{2}{7}\tan \theta$
To prevent slipping, ${\mu }_s\ge \frac{2}{7}\tan \theta$
${\mu }_s\ge \frac{2}{7}\left[\because \tan 45^{\circ }=1\right]$
Hence, $\frac{3}{7},\frac{1}{2}$ and $\frac{5}{8}$ are all greater than $\frac{2}{7}$.