JEE MAIN - Physics (2020 - 3rd September Evening Slot - No. 17)
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 FH
and
FV
about the CM. The value of $$\theta $$ is then such that :
_3rd_September_Evening_Slot_en_17_1.png)
$$\cos \theta = {{2g} \over {3l{\omega ^2}}}$$
$$\cos \theta = {{3g} \over {2l{\omega ^2}}}$$
$$\cos \theta = {g \over {2l{\omega ^2}}}$$
$$\cos \theta = {g \over {l{\omega ^2}}}$$
Explanation
_3rd_September_Evening_Slot_en_17_3.png)
$${F_v} = mg$$
$${F_H} = m{\omega ^2}{l \over 2}\sin \theta $$
$$ \therefore $$ $${\tau _{net}}\,about\,COM = {F_v}.{l \over 2}\sin \theta - {F_H}{l \over 2}\cos \theta $$
$$ = {{m{l^2}} \over {12}}{\omega ^2}\sin \theta \cos \theta $$
$$ \Rightarrow $$ $$mg{l \over 2}\sin \theta - m{\omega ^2}{l \over 2}\sin \theta {l \over 2}\cos \theta = {{m{l^2}} \over {12}}{\omega ^2}\sin \theta \cos \theta $$
$$ \Rightarrow {{gl} \over 2} - {{{\omega ^2}{l^2}} \over 4}\cos \theta = {{{l^2}} \over {12}}{\omega ^2}\cos \theta $$
$$ \Rightarrow $$ $${{gl} \over 2} ={\omega ^2}{l^2}\cos \theta \left( {{1 \over {12}} + {1 \over 4}} \right)$$
$$ \Rightarrow $$ $${{gl} \over 2} = {{{\omega ^2}{l^2}\cos \theta } \over 3}$$
$$ \Rightarrow $$ $$\cos \theta = {{3g} \over {2{\omega ^2}l}}$$
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