Question

A wheel initially rotating at 12 rad/s decelerates uniformly to rest in 0.4 s. If the wheel has a rotational inertia of 0.5 kg.m², what is the magnitude of the torque causing this deceleration? (A) 1.5 N.m (B) 15 N.m (C) 30 Nm (D) 38 Nm​

Answer 1

Answer:

(B) 15 N.m

Explanation:

The deceleration of the wheel is first found by using the following formula:

$\alpha = \frac{\omega_f - \omega_i}{t}$

where,

α = angular acceleration = ?

ωf = final angular speed = 0 rad/s

ωi = initial angular speed = 12 rad/s

t = time = 0.4 s

Therefore,

$\alpha = \frac{0\ rad/s - 12\ rad/s}{0.4\ s}\\\\\alpha = -30\ rad/s^2$

here, the negative sign shows deceleration.

Now, we find the torque responsible for this deceleration:

$\tau = I\alpha$

where,

τ = torque = ?

I = rotational inertia = 0.5 kg.m²

Therefore,

$\tau = (0.5\ kg.m^2)(30\ rad/s^2)$

τ = 15 N.m

Therefore, the correct answer is:

(B) 15 N.m