QJEEM20Ph66b7: JEE Main 2020 Physics - Class 11 JEEM20Ph66b7 Solution: A uniform rod of length ‘ ’ is pivoted at one of its ends on a…

Question

Verified Solution • 2025–26 Session moderate

A uniform rod of length ‘

l

’ is pivoted at one of its ends on a vertical shaft of negligible radius. When the shaft rotates at angular speed

\omega

the rod makes an angle

\theta

with it (see figure). To find

\theta

equate the rate of change of angular momentum (direction going into the paper)

{{m{l^2}} \over {12}}{\omega ^2}\sin \theta \cos \theta

about the centre of mass (CM) to the torque provided by the horizontal and vertical forces F_H and F_V about the CM. The value of

\theta

is then such that : JEE Main 2020 (Online) 3rd September Evening Slot Physics - Rotational Motion Question 129 English

Accepted Answer

{F_v} = mg

{F_H} = m{\omega ^2}{l \over 2}\sin \theta

\therefore

{\tau _{net}}\,about\,COM = {F_v}.{l \over 2}\sin \theta - {F_H}{l \over 2}\cos \theta

= {{m{l^2}} \over {12}}{\omega ^2}\sin \theta \cos \theta

\Rightarrow

mg{l \over 2}\sin \theta - m{\omega ^2}{l \over 2}\sin \theta {l \over 2}\cos \theta = {{m{l^2}} \over {12}}{\omega ^2}\sin \theta \cos \theta

\Rightarrow {{gl} \over 2} - {{{\omega ^2}{l^2}} \over 4}\cos \theta = {{{l^2}} \over {12}}{\omega ^2}\cos \theta

\Rightarrow

{{gl} \over 2} ={\omega ^2}{l^2}\cos \theta \left( {{1 \over {12}} + {1 \over 4}} \right)

\Rightarrow

{{gl} \over 2} = {{{\omega ^2}{l^2}\cos \theta } \over 3}

\Rightarrow

\cos \theta = {{3g} \over {2{\omega ^2}l}}

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