JEE Main · 2019 · 09 Apr 2019 · Shift 2 · Rotational Mechanics

Moment of inertia of a body about a given axis is

1.5 kg m^{2}

Initially the body is at rest. In order to produce a rotational kinetic energy of 1200 J , the angular acceleration of

20 rad / s^{2}

must be applied about the axis of rotation for a duration of:

(A)2 s✓ Correct
(B)5 s
(C)2.5 s
(D)3 s

Answer: (A)

Solution

First find the final angular speed from the energy demanded:

\frac{1}{2} I ω^{2} = 1200

with

I = 1.5

gives

ω^{2} = \frac{2 \times 1200}{1.5} = 1600

, so

ω = 40

rad/s. The body starts from rest and is given a steady angular acceleration of

20

rad/s

^{2}

, so from

ω = α t

the time needed is

t = \frac{40}{20} = 2

JEE Main · 2019 · 09 Apr 2019 · Shift 2 · Single correct option

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