Question

At time $ \, t \, = \, 0$ , a horizontal disc starts rotating with angular acceleration $1 \, rad \, 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 $0.5 \, m$ 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.
$f=m\sqrt{\left(\alpha r\right)^{2} + \left(\alpha t\right)^{4} r^{2}} \, =\mu mg$
$m\sqrt{\left(\alpha r\right)^{2} + \left(\alpha t\right)^{4} r^{2}} \, = \, \mu mg$
$\Rightarrow \, \alpha ^{2}r^{2}+\alpha ^{4}r^{2}t^{4}=\mu ^{2}g^{2}$
$\Rightarrow t=\left(\frac{\mu^2 g^2-\alpha^2 r^2}{\alpha^4 r^2}\right)^{1 / 4}$
$\Rightarrow t=\sqrt{5} \, s$