Question

A cord of negligible mass is wound round the rim of a flywheel of mass $20 \,kg$ and radius $20\, cm$. A steady pull of $25 \,N$ is applied on the cord. The work done by the pull when $2 \,m$ of the cord is unwound is, if wheel starts from rest, what is the kinetic energy of the wheel when .$2 \,m$ of the cord is unwound?

• A. $20 \,J$
• B. $25 \,J$
• C. $45 \,J$
• D. $50 \,J$

Answer 1

Answer: (D)

$50 \,J$

Answer 2

Here, $R = 20\, cm = 0.2\, m, M = 20 \,kg$ As $\tau = FR = (25 \,N) (0.2 \,m) = 5\, N m$ Moment of inertia of flywheel about its axis is $I = \frac{MR^2}{2}$ $= \frac{(20\,kg)(0.2\,m)^2}{2}$ $= 0.4\,kg\,m^2$ As $\tau = I\alpha$ $\therefore$ Angular acceleration of the wheel, $\alpha = \frac{\tau}{I} = \frac{5\,N\,m}{0.4\,kg\,m^2} = 12.5\,rad \,s^{-2}$ Angular displacement of wheel, $\theta = \frac{\text{Length of unwound string}}{\text{Radius of the wheel}}$ $= \frac{2\,m}{0.2\,m} = 10 \,rad$ Let $\omega$ be final angular velocity. As $\omega^2 = \omega_0^2 + 2\alpha \theta$ Since the wheel starts from rest, therefore $\omega_0 = 0$ $\therefore \omega^2 = 2\times (12.5\,rad \,s^{-2}) ( 10\,rad) = 250\,rad^2 \,s^{-2}$ $\therefore$ Kinetic energy gained, $K = \frac{I}{2}I\omega^2$ $K= \frac{1}{2}\times 0.4\,kg\,m^2 \times 250 \,rad^2 s^{-2} = 50\,J$