Question

A wheel has a rotational inertia of 16 kgm2. As it turns through 2.0 rev its angular velocity increases from 7.0 rad/s to 9.0 rad/s. Assuming the net torque is constant, how much work does that torque do?A 510 J B 130 J C 1600 J D 260 J E 400 J

Answer 1

Answer:

Work done, W = 255.21 J

Explanation:

It is given that,

Rotational inertia of the wheel, $I=16\ kg-m^2$

Angular displacement, $\theta=2\ rev=12.56\ rad$

Initial angular velocity, $\omega_i=7\ rad/s$

Final angular velocity, $\omega_f=9\ rad/s$

Firstly finding the angular acceleration of the wheel using the equation of rotational kinematics as :

$\alpha =\dfrac{\omega_f^2-\omega_i^2}{2\theta}$

$\alpha =\dfrac{9^2-7^2}{2\times 12.56}$

$\alpha =1.27\ rad/s^2$

Work done by the torque is given by :

$W=\tau\times \theta$

$W=I\times \alpha \times \theta$

$W=16\times 1.27 \times 12.56$

W = 255.21 J

Out of given options, the correct option for the work done by the torque is 255.21 J. Hence, this is the required solution.