JEE MAIN - Physics (2019 - 9th January Morning Slot - No. 12)
Two masses m and $${m \over 2}$$ are connected at the two ends of a massless rigid rod of length l. The rod is suspended by a thin wire of torsional constant k, at the centre of mass of the rod-mass system(see figure). Because of torsional constant k, the restoring torque is $$\tau $$ = k$$\theta $$ for angular displacement $$\theta $$. If the rod is rotated by $$\theta $$0 and released, the tension in it when it passes through its mean position will be :
_9th_January_Morning_Slot_en_12_1.png)
$${{3k{\theta _0}^2} \over l}$$
$${{2k{\theta _0}^2} \over l}$$
$${{k{\theta _0}^2} \over l}$$
$${{k{\theta _0}^2} \over {2l}}$$
Explanation
_9th_January_Morning_Slot_en_12_2.png)
$$ \therefore $$ $${r_1} = \left( {{{{n \over 2}} \over {m + {m \over 2}}}} \right)l$$
$${r_1} = {l \over 3}$$
When rod is rotated by $$\theta $$0 and released, then it look this.
_9th_January_Morning_Slot_en_12_3.png)
After roating $$\theta $$0 and released, when the rod go through mean position it will have maximum speed.
Vmax $$=$$ A$$\omega $$
$$=$$ r1$$\theta $$0$$\omega $$
$$=$$ $$\left( {{l \over 3}{\theta _0}} \right)\omega $$
We know,
$$\omega $$ $$ = \sqrt {{K \over {\rm I}}} $$
and $${\rm I} = m{\left( {{l \over 3}} \right)^2} + {m \over 2}{\left( {{{2l} \over 3}} \right)^2}$$
$$ = {{m{l^2}} \over 9} + {{4m{l^2}} \over {18}}$$
$$ = {{2m{l^2} + 4m{l^2}} \over {18}}$$
$$ = {{m{l^2}} \over 3}$$
$$ \therefore $$ $$\omega = \sqrt {{{3K} \over {m{l^2}}}} $$
$$ \therefore $$ Tension in the rod when it passes through the mean position is,
T = $${{mv_{\max }^2} \over {{r_1}}}$$
$$ = {{m{{\left( {{l \over 3}{\theta _0}\omega } \right)}^2}} \over {{l \over 3}}}$$
$$ = {{m{{\left( {{l \over 3}} \right)}^2}\theta _0^2{\omega ^2}} \over {{\ell \over 3}}}$$
$$ = m\left( {{l \over 3}} \right)\theta _0^2\left( {{{3K} \over {m{l^2}}}} \right)$$
$$ = {{K\theta _0^2} \over l}$$
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