Answer:
(B) 15 N.m
Explanation:
The deceleration of the wheel is first found by using the following formula:
[tex]\alpha = \frac{\omega_f - \omega_i}{t}[/tex]
where,
α = angular acceleration = ?
ωf = final angular speed = 0 rad/s
ωi = initial angular speed = 12 rad/s
t = time = 0.4 s
Therefore,
[tex]\alpha = \frac{0\ rad/s - 12\ rad/s}{0.4\ s}\\\\\alpha = -30\ rad/s^2[/tex]
here, the negative sign shows deceleration.
Now, we find the torque responsible for this deceleration:
[tex]\tau = I\alpha[/tex]
where,
τ = torque = ?
I = rotational inertia = 0.5 kg.m²
Therefore,
[tex]\tau = (0.5\ kg.m^2)(30\ rad/s^2)\\[/tex]
τ = 15 N.m
Therefore, the correct answer is:
(B) 15 N.m
