JEE MAIN - Physics (2019 - 9th January Evening Slot - No. 1)
A rod of length 50 cm is pivoted at one end. It is raised such that if makes an angle of 30o from the horizontal as shown and released from rest. Its angular speed when it passes through the horizontal (in rad s$$-$$1) will be (g = 10 ms$$-$$2)
_9th_January_Evening_Slot_en_1_1.png)
$$\sqrt {{{30} \over 7}} $$
$$\sqrt {30} $$
$${{\sqrt {20} } \over 3}$$
$${{\sqrt {30} } \over 2}$$
Explanation
_9th_January_Evening_Slot_en_1_2.png)
When this rod move from initial position to final position then,
Gain in kinetic energy = loss in potential energy
$$ \therefore $$ $${1 \over 2}I$$$$\omega $$2 = mgh
$${1 \over 2}\left( {{{m{l^3}} \over 3}} \right)$$ $$\omega $$2 = mg$$\left( {{l \over 2}\sin {{30}^ \circ }} \right)$$
$$ \Rightarrow $$ $$\omega $$2 = $${{3g} \over {2l}}$$
$$ \Rightarrow $$ $$\omega $$ = $$\sqrt {{{3g} \over {2l}}} $$
$$ \Rightarrow $$ $$\omega $$ = $$\sqrt {{{3 \times 10} \over {2 \times 0.5}}} $$
$$ \Rightarrow $$ $$\omega $$ = $$\sqrt {30} $$ rad/sec
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