Question

A solid sphere of radius $R$ makes a perfect rolling down on a plane which is inclined to the horizontal axis at an angle $\theta .$ If the radius of gyration is $k$, then its acceleration is

• A. $\frac{g\sin \theta }{\left(1+\frac{k^2}{R^2}\right)}$
• B. $\frac{g\sin \theta }{\left(R^2+k^2\right)}$
• C. $\frac{g\sin \theta }{2\left(R^2+k^2\right)}$
• D. $\frac{g\sin \theta }{2\left(1+\frac{k^2}{R^2}\right)}$

Answer 1

Answer: (A)

By forces balancing perpendicular to inclined plane,
$N=mg\cos \theta$
By forces balancing parallel to inclined plane,
$mg\sin ,\theta -f_r=m\frac{dv}{dt}$
$\Rightarrow \frac{dv}{dt}=\frac{mg\sin \theta -f_r}{m}\Rightarrow v(t)=\int g\sin \theta dt-\frac{1}{m}\int f_{r}dt$
Now, torque, $\tau =I\frac{d\omega }{dt}=r\times F_r$ and $I=m{k}^2$
$\therefore m{k}^2\frac{d\omega }{dt}=r{F}_r$ r $\frac{d\omega }{dt}=\frac{r{F}_r}{m{k}^2}$
$\Rightarrow \omega (t)=\frac{R}{m{k}^2}\int f_{r}dt$
or $\frac{1}{m}\int f_{r}dt=\frac{k^2}{R}\omega (t)=\frac{k^2}{R^2}v(t)($ using $\omega =v/R)$
$\Rightarrow v(t)=\int g\sin \theta dt-\frac{k^2}{R^2}v(t)$
or $v(t)=\frac{1}{\left(1+\frac{k^2}{R^2}\right)}\int g\sin \theta dt$
So, acceleration,
$a(t)=\frac{dv(t)}{dt}=\frac{g\sin \theta }{\left(1+\frac{k^2}{R^2}\right)}$