Answer to Question #118722 in Classical Mechanics for nikil

Question

A flywheel has a moment of inertia of 1.6 × 10−3 kg.m2. When a constant torque is applied, it reaches an angular velocity of 1200 rev/min in 15 s. Assuming it started from rest, find: i) the angular acceleration ii) the unbalanced torque applied iii) the angle turned through 15 s iv) the work done on the flywheel by the torque

Answer 1

Let's write the given information first below,

Moment of Inertia of the Flywheel,$I=1.6\times10^{-3}\: Kg\:m^2$ .

Final angular speed $\omega=1200rev/min=1200\times \frac{2\pi}{60}=40\pi \:rad/s$ .

Initial angular speed $\omega_0=0$ .

Time $t=15s$ .

i).Since,

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

On putting the given value to above equation we get,

$$40\pi=0+15\alpha\implies \alpha=8.38 rad/s^2$$

ii).Torque is given by,

$$\tau=I\alpha\\\implies \tau=1.6\times 10^{-3}\times 8.38=13.408\times 10^{-3}Nm$$

iii).We know that,

$$\theta=\omega_0t+\frac{1}{2}\alpha t^2\\ \implies \theta=\frac{1}{2}\times13.408\times10^{-3}\times 225=942.47 \:rad$$

iv).From, work energy-principle,

$$W=\frac{1}{2}I\omega^2-\frac{1}{2}I\omega_0^2$$

Thus, work done on the Flywheel by given torque is,

$$W=\frac{1}{2}\times1.6\times10^{-3}(40\pi)^2\\\implies W=12.63J$$