Answer to Question #118743 in Classical Mechanics for Donovan

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

As per the question,

Moment of inertia of the flywheel $(I)=1.6 × 10^{−3} kg.m^2$

Angular velocity $(\omega)=1200\times 2\pi rad/min =40 \pi rad/sec$

time (t)=15 sec

i) Angular acceleration $(\alpha)=\dfrac{40 \pi}{15}rad/sec^2=8 \pi/3 rad/sec^2$

ii) Torque $(\tau)=I\alpha=1.6\times 10^{-3}\times \dfrac{8 \pi}{3}=\dfrac{12.8\pi}{3}\times 10^{-3}N-m$

iii) $(\theta)=\dfrac{\omega^2}{2\alpha }=\dfrac{(40\pi)^2\times 3}{2\times 12.8\times \pi}$

$=4.68\pi rad$

iv) Work done $=\dfrac{I\omega^2}{2}=\dfrac{1.6\times 10^{-3}\times(40 \pi)^2}{2}$

$=0.64 \pi^2 J$