Question

A solid sphere of mass $5\text{kg}$ rolls without slipping on an inclined plane of inclination ${30}^{\circ }$. The linear acceleration (in $\text{m}/{\text{s}}^2$) & force of friction (in $\text{N}$) acting on the sphere are :-

• A. $\frac{5g}{14},\frac{5g}{7}$
• B. $\frac{5g}{7},\frac{5g}{7}$
• C. $5g,\frac{2g}{3}$
• D. $2g,\frac{4g}{3}$

Answer 1

The correct option is A $\frac{5g}{14},\frac{5g}{7}$


Given
$m=5\text{kg}$, $\theta ={30}^{\circ }$
$\because$ Sphere is rolling without slipping.
$a=R\alpha ...\left(i\right)$

$\sum {\overrightarrow{\tau }}_{com}=I_{com}\overrightarrow{\alpha }$
$\Rightarrow -fR=-\left(\frac{2}{5}m{R}^2\right)\alpha$
$\Rightarrow f=\frac{2}{5}mR\alpha$
Substituting from $\left(i\right)$,
$f=\frac{2}{5}mR\left(\frac{a}{R}\right)=\frac{2}{5}ma...\left(ii\right)$
Applying Newton's 2nd law :-
$\sum {\overrightarrow{F}}_{ext}=m{\overrightarrow{a}}_{com}$
$\Rightarrow mg\sin \theta -f=ma...\left(iii\right)$

Putting (ii) in (iii),
$mg\sin \theta -\frac{2}{5}ma=ma$
$\Rightarrow a=\frac{5g\sin \theta }{7}$ & $f=\frac{2mg\sin \theta }{7}$
Putting values of $m=5\text{kg}$ & $\theta ={30}^{\circ }$
$a=\frac{5}{14}g{\text{m/s}}^2$ & $f=\frac{5g}{7}\text{N}$