by energy conservation $$mgl = {1 \over 2}I{\omega ^2} = {1 \over 2}{{m{l^2}{\omega ^2}} \over 3}$$

$$ \Rightarrow \omega = \sqrt {{{6g} \over l}} $$

As we know the relation between the linear speed and angular speed,

$$v = \omega r = \omega l = \sqrt {6gl} $$

$$v = \sqrt {6 \times 10 \times .6} $$ = 6 m/s

Hence, the speed of the free end of the rod when it passes through its lowest position is 6 m/s.

$${1 \over 2}{I_1}{({\omega _1})^2} = {1 \over 2}{I_2}{({\omega _2})^2}$$

$${I_1}{\left( {{v \over {3R}}} \right)^2} = {I_2}{\left( {{v \over R}} \right)^2}$$

$${{{I_1}} \over {{I_2}}} = {9 \over 1}$$

$$\tau = {{\Delta L} \over {\Delta t}} = {{I({\omega _f} - {\omega _i})} \over {\Delta t}}$$

$$\tau = {{{{m{R^2}} \over 2} \times [0 - \omega ]} \over {\Delta t}}$$

$$ = {{10 \times {{(20 \times {{10}^{ - 2}})}^2}} \over 2} \times {{600 \times \pi } \over {30 \times 10}}$$

$$ = 0.4\pi = 4\pi \times {10^{ - 2}}$$

$$\overrightarrow r = 10\alpha {t^2}\widehat i + 5\beta (t - 5)\widehat j$$

$$\overrightarrow v = 20\alpha t\widehat i + 5\beta \widehat j$$

$$\overrightarrow L = m(\overrightarrow r \times \overrightarrow v )$$

$$ = m[10\alpha {t^2}\widehat i + 5\beta (t - 5)\widehat j] \times [20\alpha t\widehat i + 5\beta \widehat j]$$

$$\overrightarrow L = m[50\alpha \beta {t^2}\widehat k - 10\alpha \beta ({t^2} - 5t)\widehat k]$$

At t = 0, $$\overrightarrow L = \overrightarrow 0 $$

$$50\alpha \beta {t^2} - 100\alpha \beta ({t^2} - 5t) = 0$$

t $$-$$ 2 (t $$-$$ 5) = 0

t = 10 sec