Prove the result that the velocity v of translation of a rolling body (like a ring, disc, cylinder or sphere) at the bottom of an inclined plane of height h given by $\text{v}^2=\frac{2\text{gh}}{\big(1+\frac{\text{K}^2}{\text{R}^2}\big)}$ using dynamical consideration. Note K is the radius of gyration of the body about its symmetry axis, and R is the radius of the body. The body starts from rest at the top of the plane.
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Consider a mass 'm' capable of rolling down the inclined plane from a vertical height 'h'. Rolling comprises of transitory motion of centre of mass and rotation produced by frictional force. Resolving the forces acting on the mass, we have Normal reaction force $\text{N}=\text{mg}\cos\theta\text{ and }$
$=\text{mg}\sin\theta-\text{F}_{\text{f}}=\text{ma}\dots(1)$ Torque = $F_f$ r and also $\tau=\text{I}\alpha=\frac{\text{l}\alpha}{\text{r}}\dots(2)$
$\text{F}_{\text{f}\text{r}}=\frac{\text{l}\alpha}{\text{r}}$
$\Rightarrow\text{F}_{\text{f}}=\frac{\text{I}\text{a}}{\text{r}}\dots(2)$ where, r is the radius of the body
Using $F_f$_in (i), we have $\text{mg}\sin\theta=\Big(\text{m}+\frac{\text{l}}{\text{r}^2}\Big)\text{a}$
$\text{a}=\frac{\text{mg}\sin\theta}{\text{m}\Big(1+\frac{\text{K}^2}{\text{r}^2}\Big)}$
$=\frac{\text{g}\sin\theta}{\Big(1+\frac{\text{K}^2}{\text{r}^2}\Big)}[\therefore\text{I}=\text{mK}^2]$ Where K is the radius of gyration of the body about its symmetry axis using a, in $v^2 = u^2 + 2$
as, we have $\text{v}=\sqrt{\frac{2\text{g}\sin\theta\text{l}}{\Big(1+\frac{\text{K}^2}{\text{r}^2}\Big)}}$
$=\sqrt{\frac{2\text{gh}}{\Big(1+\frac{\text{K}^2}{\text{r}^2}\Big)}}[\therefore\text{i}\sin=\text{h}][\text{u}=0].$
$=\text{mg}\sin\theta-\text{F}_{\text{f}}=\text{ma}\dots(1)$ Torque = $F_f$ r and also $\tau=\text{I}\alpha=\frac{\text{l}\alpha}{\text{r}}\dots(2)$
$\text{F}_{\text{f}\text{r}}=\frac{\text{l}\alpha}{\text{r}}$
$\Rightarrow\text{F}_{\text{f}}=\frac{\text{I}\text{a}}{\text{r}}\dots(2)$ where, r is the radius of the body
Using $F_f$_in (i), we have $\text{mg}\sin\theta=\Big(\text{m}+\frac{\text{l}}{\text{r}^2}\Big)\text{a}$
$\text{a}=\frac{\text{mg}\sin\theta}{\text{m}\Big(1+\frac{\text{K}^2}{\text{r}^2}\Big)}$
$=\frac{\text{g}\sin\theta}{\Big(1+\frac{\text{K}^2}{\text{r}^2}\Big)}[\therefore\text{I}=\text{mK}^2]$ Where K is the radius of gyration of the body about its symmetry axis using a, in $v^2 = u^2 + 2$
as, we have $\text{v}=\sqrt{\frac{2\text{g}\sin\theta\text{l}}{\Big(1+\frac{\text{K}^2}{\text{r}^2}\Big)}}$
$=\sqrt{\frac{2\text{gh}}{\Big(1+\frac{\text{K}^2}{\text{r}^2}\Big)}}[\therefore\text{i}\sin=\text{h}][\text{u}=0].$
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