Question

A ring of mass $m$ and radius $r$ is allowed to roll down without slipping on an inclined plane having inclination ${\sin }^{-1}\left(\frac{1}{10}\right)$ from the horizontal. The magnitude of acceleration of it's centre of mass is

• A. $9.8{\text{ms}}^{-2}$
• B. $4.9{\text{ms}}^{-2}$
• C. $0.98{\text{ms}}^{-2}$
• D. $0.49{\text{ms}}^{-2}$

Answer 1

The correct option is D $0.49{\text{ms}}^{-2}$

Magnitude of linear acceleration of centre of mass of a body rolling down an inclined plane is given by $a$, where $a=a_{CM}$:

$\Rightarrow a=\frac{g\sin \theta }{1+\frac{K^2}{r^2}}=\frac{g\sin \theta }{1+\frac{I}{m{r}^2}}$
$K=$ Radius of gyration
$\Rightarrow a=\frac{g\sin \theta }{(1+1)}$
$\because I=m{r}^2$ for ring, about an axis passing through it's $\text{COM}$
$\therefore a=\frac{(9.8\times \frac{1}{10})}{(1+1)}{\text{ms}}^{-2}=0.49{\text{ms}}^{-2}$