Physics - Rotation Question with Solution | TestHub

Let$\mathrm{m}$be the mass of the sphere.

Since, it is a case of backward slipping, force of friction is in forward direction. Limiting friction will act in this case.



$\text{Linear acceleration}\text{a}=\frac{\text{f}}{\text{m}}=\frac{\text{μ}mg}{\text{m}}=\text{μ}\text{g}$

$\text{Angular retardation}\text{α}=\frac{\text{τ}}{\text{I}}=\frac{\text{f}·\text{r}}{\frac{2}{5}{\text{mr}}^2}=\frac{5}{2}\frac{\text{μ}\text{g}}{\text{r}}$

Slipping will be stopped when$\mathrm{v}=\mathrm{r\omega }$

$\mathrm{at} = \mathrm{r}({\mathrm{\omega }}_0 - \mathrm{\alpha t})$

$\text{μ}\text{gt}=\text{r}({\text{ω}}_0-\frac{5}{2}\frac{\text{μ}\text{g}\text{t}}{\text{r}})$

$\frac{7}{2}\text{μ}\text{gt}=\text{r}{\text{ω}}_0$

$\therefore \text{t}=\frac{2}{7}\frac{\text{r}{\text{ω}}_0}{\text{μ}\text{g}}$