a) all of them reach the ground at the same instant b) the sphere reaches first, the ring at last c) the sphere reaches first, the cylinder and ring reach together. d) none of the above

For pure rolling on a surface of inclination $\theta$,

$\mu \geq \large\frac{\tan \theta}{1+\large\frac{R^2}{k^2}}$

Since $k^2 $ for sphere $=\large\frac{2}{5}$$ R^2$

$k^2$ for ring $= R^2$

$k^2$ for cylinder $= \large\frac{R^2}{2}$

$\mu _s \geq \large\frac{2}{7} $$\tan \theta$

$\mu _c \geq \large\frac{1}{3} $$\tan \theta, \mu _r \geq \large\frac{\tan \theta}{2}$

$\mu _s$ is least.

Hence both cylinder and ring will slide.

$\therefore $ friction =$\mu mg \cos \theta$ will be acting on all three.

Hence all will have the same acceleration. Hence will reach the bottom simultaneously.

Ask Question

Tag:MathPhyChemBioOther

Take Test

...