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