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

Work done, W = 255.21 J

Step-by-step 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 =(\omega_f^2-\omega_i^2)/(2\theta)`

`\alpha =(9^2-7^2)/(2* 12.56)`

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

Work done by the torque is given by :

`W=\tau* \theta`

`W=I* \alpha * \theta`

`W=16* 1.27 * 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.