JEE MAIN - Physics (2020 - 9th January Morning Slot - No. 5)
One end of a straight uniform 1m long bar is
pivoted on horizontal table. It is released from
rest when it makes an angle 30º from the
horizontal (see figure). Its angular speed when
it hits the table is given as $$\sqrt n $$ s-1, where n is
an integer. The value of n is _________.
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Answer
15
Explanation
By energy conservation
Ui + Ki = Uf + Kf
$$ \Rightarrow $$ $$mg{l \over 2}\sin 30^\circ $$ + 0 = 0 + $${1 \over 2}I{\omega ^2}$$
$$ \Rightarrow $$ $$mg{1 \over 2} \times {1 \over 2}$$ = $${1 \over 2} \times {{m{{\left( 1 \right)}^2}} \over 3}{\omega ^2}$$
$$ \Rightarrow $$ $${\omega ^2} = {{3g} \over 2}$$
$$ \Rightarrow $$ $$\omega = \sqrt {15} $$
$$ \therefore $$ n = 15
Ui + Ki = Uf + Kf
$$ \Rightarrow $$ $$mg{l \over 2}\sin 30^\circ $$ + 0 = 0 + $${1 \over 2}I{\omega ^2}$$
$$ \Rightarrow $$ $$mg{1 \over 2} \times {1 \over 2}$$ = $${1 \over 2} \times {{m{{\left( 1 \right)}^2}} \over 3}{\omega ^2}$$
$$ \Rightarrow $$ $${\omega ^2} = {{3g} \over 2}$$
$$ \Rightarrow $$ $$\omega = \sqrt {15} $$
$$ \therefore $$ n = 15
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