Question
From an inclined plane a sphere, a disc, a ring and a shell are rolled without slipping. The order of their reaching at the base will be
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Answer
When a sphere, a disc, a ring and a spherical shell are ralled down an inclined plane, their kinetic energy at the bottom will be $\Rightarrow K \cdot E=\frac{1}{2} m v^2+\frac{1}{2} I \omega^2=\frac{1}{2} m v^2\left[1+\frac{k^2}{r^2}\right]$ moment of inertia, $I=m k^2$ [ $k \rightarrow$ radius of gyration] As the bodies will have same kinetic eenergy, their velocities will depend upon ratio $(k / r)$ -
Higher value of $(k / r)$, lesser velocity $v$.
Isphere $=\frac{2}{5} m r^2 ;$ Idisc $=\frac{m r^2}{2} ; I_{r i n g}=m r^2$; Ishell $=\frac{2 m r^2}{3}$
So, $\frac{K_s}{r}=\sqrt{\frac{2}{5}} ; \frac{K_D}{r}=\frac{1}{\sqrt{2}} ; \frac{K_r}{r}=1 ; \frac{K_{s h}}{r}=\sqrt{\frac{2}{3}}$
so, $K_s
Higher value of $(k / r)$, lesser velocity $v$.
Isphere $=\frac{2}{5} m r^2 ;$ Idisc $=\frac{m r^2}{2} ; I_{r i n g}=m r^2$; Ishell $=\frac{2 m r^2}{3}$
So, $\frac{K_s}{r}=\sqrt{\frac{2}{5}} ; \frac{K_D}{r}=\frac{1}{\sqrt{2}} ; \frac{K_r}{r}=1 ; \frac{K_{s h}}{r}=\sqrt{\frac{2}{3}}$
so, $K_s
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