Answer to Question #118177 in Mechanics | Relativity for rishal deo

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

Given, Moment of inertia $I=1.6 × 10^{−3} kg.m^2$ ,

Initial angular speed $\omega_0=0$ ,final angular speed $\omega=1200\times \frac{2\pi}{60}=40\pi \:rad/s$ ,time $t=15s$ ,

i). Angular acceleration

$\alpha=\frac{\omega-\omega_0}{t}\implies \alpha=\frac{40\pi}{15}=\frac{8\pi}{3} \: rad/s^2$

ii).

Unbalanced torque

$\tau=I\alpha\\ \implies \tau=1.6\times 10^{-3}\times \frac{8\pi}{3}=13.40\times 10^{-3}\:Nm$

iii). Let the required angle is $\theta$ , thus

$\omega^2=\omega_0^2+2\alpha \theta\\ \implies \theta=\frac{\omega^2-\omega_0^2}{2\alpha}\\ \implies \theta=\frac{(40\pi)^2}{2\frac{8\pi}{3}}=942.47 \: rad$

Alternatively, we can apply

$\theta=\omega_0t+\frac{1}{2}\alpha t^2$

and eventually we will get the same result.

iv).

By applying work energy theorem for rotational motion we get,

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

Thus we get,

$W_{rot}=\frac{1}{2}\times 1.6\times 10^{-3} (40\pi)^2=12.63J$

Hence we done.