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, $m g \sin \theta-f=m a$
$\Rightarrow f=m g \sin \theta-m a=m g \sin \theta-m \frac{5}{7} \cdot g \sin \theta$
$=\frac{2}{7} \,mg \,\sin \,\theta$
$\therefore $ Coefficient of friction $\left(\mu_{s}\right)=\frac{f}{N}=\frac{\frac{2}{7} \,m g \,\sin \theta}{m g \,\cos \,\theta}$
$=\frac{2}{7} \,\tan \theta$
To prevent slipping, $\mu_{s} \geq \frac{2}{7} \,\tan \,\theta$
$\mu_{s} \geq \frac{2}{7} \left[\because \tan 45^{\circ}=1\right]$
Hence, $\frac{3}{7}, \frac{1}{2}$ and $\frac{5}{8}$ are all greater than $\frac{2}{7}$.