Question

A fly wheel of M.I. 6 x $10^{-2}$ kgm² is rotating with an angular velocity of 20 rads$^{-1}$. The torque required to bring it to rest in 4s is :

• A. 1.6 N m
• B. 0.6 Nm
• C. 0.8Nm
• D. 0.3 N m

Answer 1

Answer: (D)

0.3 N m

Answer 2

To bring the flywheel to rest, a torque must be applied that causes deceleration.

1. Calculate the angular acceleration ($\alpha$): We are given: Initial angular velocity ($\omega_0$) = 20 rad/s Final angular velocity ($\omega$) = 0 rad/s (since it comes to rest) Time ($t$) = 4 s

Using the first equation of rotational motion:

$$\omega = \omega_0 + \alpha t$$

Substituting the given values:

$$0 = 20 + \alpha (4)$$ $$-20 = 4\alpha$$ $$\alpha = \frac{-20}{4}$$ $$\alpha = -5 \text{ rad/s}^2$$

The negative sign indicates deceleration. For calculating the magnitude of the torque, we use the magnitude of angular acceleration, $|\alpha| = 5 \text{ rad/s}^2$.

2. Calculate the torque ($\tau$): We are given: Moment of Inertia ($I$) = 6 x $10^{-2}$ kgm² Angular acceleration ($|\alpha|$) = 5 rad/s²

Using the relationship between torque, moment of inertia, and angular acceleration:

$$\tau = I \alpha$$

Substituting the values:

$$\tau = (6 \times 10^{-2} \text{ kgm}^2) \times (5 \text{ rad/s}^2)$$ $$\tau = 30 \times 10^{-2} \text{ N m}$$ $$\tau = 0.3 \text{ N m}$$

The torque required to bring the flywheel to rest is 0.3 N m.