A mass m is allowed to roll down an inclined plane $(\theta)$ If the vertical height is h, find.
- Acceleration along the inclined plane.
- Velocity down the plane.
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Consider a mass z 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 friction force. Using the forces (represented) acting on the mass, we have,
$\text{N}=\text{mg}\cos\ \theta\text{ and }\sin\ \theta-\text{F}_{\text{f}}=\text{ma}\dots(1)$ $\text{F}_{\text{f}}=\mu\text{ N}=\mu\text{ mg}\cos\ \theta$ $\therefore \text{mg}\ \sin\theta-\mu\text{ mg}\cos\ \theta$ $\text{Torque}\ \tau=\mu\text{ mg }\cos\ \theta$ $\text{r}=\text{I }\alpha=\text{I}\frac{\text{a}}{\text{r}}\dots(2)$
Using $F_f$ in (i), we have
$\text{mg }\sin\theta=\Big(\text{m}+\frac{\text{I}}{\text{r}^2}\Big)\text{a}$
$\text{a}=\frac{\text{mg }\sin\ \theta}{\text{m}\Big(1+\frac{\text{K}^2}{\text{r}^2}\Big)}=\frac{\text{g }\sin\ \theta}{\Big(1+\frac{\text{K}^2}{\text{r}^2}\Big)}$
$\text{N}=\text{mg}\cos\ \theta\text{ and }\sin\ \theta-\text{F}_{\text{f}}=\text{ma}\dots(1)$ $\text{F}_{\text{f}}=\mu\text{ N}=\mu\text{ mg}\cos\ \theta$ $\therefore \text{mg}\ \sin\theta-\mu\text{ mg}\cos\ \theta$ $\text{Torque}\ \tau=\mu\text{ mg }\cos\ \theta$ $\text{r}=\text{I }\alpha=\text{I}\frac{\text{a}}{\text{r}}\dots(2)$
- $\text{F}_{\text{f}}\text{ r}=\frac{\text{I}}{\text{r}^2}$
Using $F_f$ in (i), we have
$\text{mg }\sin\theta=\Big(\text{m}+\frac{\text{I}}{\text{r}^2}\Big)\text{a}$
$\text{a}=\frac{\text{mg }\sin\ \theta}{\text{m}\Big(1+\frac{\text{K}^2}{\text{r}^2}\Big)}=\frac{\text{g }\sin\ \theta}{\Big(1+\frac{\text{K}^2}{\text{r}^2}\Big)}$
- Using a in $v^2 = u^2 + 2as$, we have
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