Answer:
Work done, W = 255.21 J
Explanation:
It is given that,
Rotational inertia of the wheel, [tex]I=16\ kg-m^2[/tex]
Angular displacement, [tex]\theta=2\ rev=12.56\ rad[/tex]
Initial angular velocity, [tex]\omega_i=7\ rad/s[/tex]
Final angular velocity, [tex]\omega_f=9\ rad/s[/tex]
Firstly finding the angular acceleration of the wheel using the equation of rotational kinematics as :
[tex]\alpha =\dfrac{\omega_f^2-\omega_i^2}{2\theta}[/tex]
[tex]\alpha =\dfrac{9^2-7^2}{2\times 12.56}[/tex]
[tex]\alpha =1.27\ rad/s^2[/tex]
Work done by the torque is given by :
[tex]W=\tau\times \theta[/tex]
[tex]W=I\times \alpha \times \theta[/tex]
[tex]W=16\times 1.27 \times 12.56[/tex]
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.
