Question

The angular momentum of a flywheel having a rotational inertia of 0.225 kg·m² about its central axis decreases from 9.10 to 0.960 kg·m²/s in 3.60 s.
(a) What is the magnitude of the average torque acting on the flywheel about its central axis during this period?
(b) Assuming a constant angular acceleration, through what angle does the flywheel turn?
(c) How much work is done on the wheel?
(d) What is the magnitude of the average power done on the flywheel?

Answer 1

Answer:

Part a)

$\tau = -2.26 Nm$

Part b)

$\theta = 80.48 rad$

Part c)

$W = -181.97 J$

Part d)

$P = 50.5 W$

Explanation:

Part a)

As we know that torque on the rotating wheel is given as

$\tau = I\alpha$

also we can write this in terms of angular momentum

$\tau = \frac{\Delta L}{\Delat t}$

so we have

$\tau = \frac{L_f - L_i}{\Delta t}$

$\tau = \frac{0.960 - 9.10}{3.60}$

$\tau = -2.26 Nm$

Part b)

Angular displacement of the wheel at constant angular acceleration is given as

$\theta = \frac{\omega_f + \omega_i}{2}\Delta t$

$\theta = \frac{\frac{L_f}{I} + \frac{L_i}{I}}{2}\Delta t$

$\theta = \frac{0.960 + 9.10}{2 \times 0.225}(3.60)$

$\theta = 80.48 rad$

Part c)

Work done on the wheel is equal to the change in kinetic energy of the wheel

so we have

$W = \frac{L_f^2}{2I} - \frac{L_i^2}{2I}$

so we have

$W = \frac{0.960^2 - 9.10^2}{2(0.225)}$

$W = -181.97 J$

Part d)

Average power is defined as the rate of work done

so it is given as

$P = \frac{W}{t}$

$P = \frac{-181.97}{3.60}$

$P = 50.5 W$