Question

A disc of mass 2 kg and radius 0.2 m is rotating with an angular velocity of 30 rad/s. What is the angular velocity if the mass of 0.25 kg is put on the periphery of the disc?

Answer 1

Angular momentum in the absence of any external torque remains constant,
i.e. $L=I\omega =constant$
$\Rightarrow I_1{\omega }_1=I_2{\omega }_2$
$\therefore \frac{1}{2}M{r}^2{\omega }_1=\frac{1}{2}(M+2m)r^2{\omega }_2\dots \dots \left(i\right)$
Here, $M=2kg,m=0.25kg,r=0.2m$
${\omega }_1=30rad/s$
Hence, we get after putting the given values in Eq. (i)
=$\frac{1}{2}\times 2\times (0.2{)}^2\times 30=\frac{1}{2}\times (2+2\times 0.25\left)\right(0.2{)}^2\times {\omega }_2\Rightarrow 1.2=0.05{\omega }_2\therefore {\omega }_2=24rad/s$
Because we have to balance the weight, we cannot simply put 0.25 kg mass on one side of the disk we have to either distribute the mass equally on the entire disk or we have to add equal mass on the opposite end of the disk.