JEE Advance - Physics (2019 - Paper 2 Offline - No. 5)
A thin and uniform rod of mass M and length L is held vertical on a floor with large friction. The rod is released from rest so that it falls by rotating about its contact-point with the floor without slipping. Which of the following statement(s) is/are correct, when the rod makes an angle 60$$^\circ $$ with vertical? [g is the acceleration due to gravity]
The angular acceleration of the rod will be $${{2g} \over L}$$.
The normal reaction force from the floor on the rod will be $${{Mg} \over 16}$$.
The radial acceleration of the rod's center of mass will be $${{3g} \over 4}$$.
The angular speed of the rod will be $$\sqrt {{{3g} \over {2L}}} $$.
Explanation
$$\Delta $$K + $$\Delta $$U = 0
$${1 \over 2}{I_0}{\omega ^2} = - \Delta U$$
$${1 \over 2}{{m{l^2}} \over 3}{\omega ^2} = - \left( { - mg{l \over 4}} \right)$$
$$\omega = \sqrt {{{3g} \over {2l}}} $$
$$ \Rightarrow {a_{radial}} = {\omega ^2}{l \over 2} = {{3g} \over {2l}}{l \over 2} = {{3g} \over 4} \Rightarrow \tau = {I_0}\alpha $$
$$\alpha = {{mg{l \over 2}\sin 60^\circ } \over {mg{{{l^2}} \over 3}}} = {{3\sqrt 3 g} \over {4l}}$$
$$ \Rightarrow {a_v} = \left( {\alpha {l \over 2}} \right)\sin 60^\circ + {\omega ^2}{l \over 2}\cos 60^\circ $$
$${a_v} = {{3\sqrt 3 g} \over 8}{{\sqrt 3 } \over 2} + {{3g} \over 8} \Rightarrow {a_v} = {{9g} \over {16}} + {{6g} \over {16}}$$
$$mg - N = m{a_v} \Rightarrow N = {{mg} \over {16}}$$
Comments (0)
