Question

At time $t=0$ , a horizontal disc starts rotating with angular acceleration $1rad{s}^{-2}$ about an axis perpendicular to its plane and passing through its Centre. A small block is lying on this disc at a distance $0.5m$ from the centre, coefficient of friction between the surface of block and disc is $0.255$ . The block will start slipping on the disc at time $t$ , which is approximately equal to

• A. $2\sqrt{3}s$
• B. $2\sqrt{2}s$
• C. $6s$
• D. $\sqrt{5}s$

Answer 1

Answer: (D)

The block will start slipping on the table when friction will reach its limiting value.
$f=m\sqrt{{\left(\alpha r\right)}^2+{\left(\alpha t\right)}^4{r}^2}=\mu mg$
$m\sqrt{{\left(\alpha r\right)}^2+{\left(\alpha t\right)}^4{r}^2}=\mu mg$
$\Rightarrow {\alpha }^2{r}^2+{\alpha }^4{r}^2{t}^4={\mu }^2{g}^2$
$\Rightarrow t={\left(\frac{{\mu }^2{g}^2-{\alpha }^2{r}^2}{{\alpha }^4{r}^2}\right)}^{1/4}$
$\Rightarrow t=\sqrt{5}s$