Question 451: Centre of Mass and Rotational Motion

Question

A 2 kg steel rod of length 0.6 m is clamped on a table vertically at its lower end and is free to rotate in vertical plane. The upper end is pushed so that the rod falls under gravity, ignoring the friction due to clamping at its lower end, the speed of the free end of rod when it passes through its lowest position is ____________ ms
$$-$$1
. (Take g = 10 ms
$$-$$2
)

Accepted Answer

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,
$$v = \omega r = \omega l = \sqrt {6gl} $$
$$v = \sqrt {6 	imes 10 	imes .6} $$ = 6 m/s
Hence, the speed of the free end of the rod when it passes through its lowest position is 6 m/s.

Centre of Mass and Rotational Motion

Integer / Numeric Answer Type