KE = 1/2 Iω²
1200 = 1/2 × 1.5 × ω²
ω² = 1200 × 2 / 1.5 = 1600
ω = 40 rad/s
ω = ω₀ + αt
40 = 0 + 20t
t = 2 sec.

" /> KE = 1/2 Iω²
1200 = 1/2 × 1.5 × ω²
ω² = 1200 × 2 / 1.5 = 1600
ω = 40 rad/s
ω = ω₀ + αt
40 = 0 + 20t
t = 2 sec.

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Question:

Moment of inertia of a body about a given axis is 1.5kg m^2. Initially the body is at rest. In order to produce a rotational kinetic energy of 1200J, the angular acceleration of 20 rad/s^2 must be applied about the axis for a duration of?

2s

5s

2.5s

3s

Solution:

Correct option is A. 2s
Given moment of inertia 'I' = 1.5 kgm²
Angular Acc. "α" = 20 Rad/s²
KE = 1/2 Iω²
1200 = 1/2 × 1.5 × ω²
ω² = 1200 × 2 / 1.5 = 1600
ω = 40 rad/s
ω = ω₀ + αt
40 = 0 + 20t
t = 2 sec.