JEE MAIN - Physics (2019 - 9th April Evening Slot - No. 18)
Moment of inertia of a body about a given axis
is 1.5 kg m2. Initially the body is at rest. In order
to produce a rotational kinetic energy of
1200 J, the angular accleration of 20 rad/s2
must be applied about the axis for a
duration of :-
2.5 s
3 s
5s
2 s
Explanation
KE = $${1 \over 2}I{\omega ^2} = 1200$$ (given)
$$ \Rightarrow \omega = 40\,rad/s$$
$$ \Rightarrow \omega = {\omega _0} + \alpha t$$
$$ \Rightarrow 40 = 0 + (20)t$$
$$ \Rightarrow t = 2\,\sec $$
$$ \Rightarrow \omega = 40\,rad/s$$
$$ \Rightarrow \omega = {\omega _0} + \alpha t$$
$$ \Rightarrow 40 = 0 + (20)t$$
$$ \Rightarrow t = 2\,\sec $$
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