A disc of mass 2kg and diameter 2m is performing rotational motion. Find the work done, if the disc is rotating from 300rpm to 600rpm.
1479 J
147.9 J
14.79 J
1.479 J
The correct option is A) 1480 J
I = \(\frac{mR^2}{2} = 1 kg/m^2\)
\(ω_i = \frac{300 \times 2π}{60} = 10π\)
\(ω_f = \frac{600 \times2π} {60} = 20π\)
△KE = \(\frac{1}{2} I(ω_f)^2 - \frac{1}{2}(Iω_i)^2\)
= \(\frac{1}{2}((20π)^2 - (10π)^2)\)
\(2 \times300π^2= 1480 J\)
The Correct Answer is (A)

The Correct Answer is (A)
The rotational motion can be defined as the motion of the object around a circular path, that to in a fixed orbit.
Work done for rotational motion
Work done is calculated by the dot product of force and displacement of the point, where the force is applied. For the rotational motion, the force is replaced by the torque, and the displacement is replaced by the angular displacement.
Kinetic energy for rotational motion
The kinetic energy is given by ½ Iω². It is directly proportional to the rotational inertia and the square of the magnitude of the angular velocity.
Work energy theorem for rotation
For a rigid body rotating at a fixed axis, the theorem is given by:
WAB = KB - KA
Where, K= ½ Iω²
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