Answer to Question #123782 in Mechanics | Relativity for Ojugbele Daniel

Question

The angular velocity of a disc of mass 3kg and radius 0.6m change from 10rev/s to 15rev/s in 0.07s. calculate the impulse on the disc.

Answer 1

Angular momentum L (which change disc speed) can be found:

$L=J*(\omega_{2} - \omega_{1})$

where J - disc's moment of inertia, ω₂ - finish angular speed ω₁ - start angular speed.

Moment of inertia for disc can be found:

$J= \frac{m\times r^2}{2}$

where m - mass of the disc, r - radius of the disc.

Angular speed for correct calculation have to be transformed from rev/s to rad/s.

$1 rev/s = 2\times\pi rad/s$

So we have:

$L=\frac{m\times r^2\times(\omega_{2}-\omega_{1})}{2}$

$L=\frac{3\times0.62\times(15\times2\times\pi - 10\times2\times\pi)}{2}=16.956 \frac{kg\times m^2}{s}$