Question

Diameter of a disc is 0.5 m and its mass is 16 kg. What torque will be required to increase the angular velocity from 0 (zero) to 120 c min^-1 in 8 s? At what rate the work will done after 8s?

Answer 1

Given: Mass of the disc M = 16 kg

Radius R = \(\frac{0.5}{2}\) = 0.25 m; t = 8 s

ω₁ = 0; ω₂ = 2πn² = 2π x \(\frac{120}{60}\) = 4π rad.s^-1

From first equation of rotational motion,

ω = ω₀ + α.t

ω₂ = ω₁ + αt

or ω₂ = 0 + α.t

\(\therefore\ \alpha = \frac{\omega^2}{t} = \frac{4\pi}{8} = \frac{\pi}{2} = \frac{3.14}{2} = 1.57\ rad.s^{-1}\) 

Moment of inertia of the disc,

\(i = \frac{1}{2}MR^2 = \frac{1}{2}\times16\times(0.25)^2 = 8\times0.0625\)

= 0.5 kgm²

^{\(\therefore\ \tau = I\alpha = 0.5\frac{\pi}{2} = \frac{\pi}{4}\ Nm\)}

Rate of doing work i.e., power

\(P = \tau\omega = \frac{\pi}{4}\times4\pi = \pi^2\) 

or P = π² watt