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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

1 Answer

1 vote

Answer:

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.

User Matt Foxx Duncan
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