Initially a disc is rotating about a point A on circumference with angular velocity $\omega$ . The point is then released and almost immediately point B is fixed on circumference. Show that the disc starts to rotate about B with angular velocity $\omega/3*(1+2*\cos \alpha)$ where $\alpha$ is angle subtended by AB at centre of disc.
This question is solved by conservation of angular momentum. $I\omega’ = I_0\omega + MV_{cm}*r*cos\alpha$
My queries are as follows –
- When point A is released, the disc rotation would shift to its centre of mass, and because of this shift wouldn’t the $\omega$ change as Moment of Inertia has changed? (new MoI is 3x less.)
- As this $\omega$ might have changed, so how can we take $I_0\omega$ in the calculation ?
- Why this can’t be achieved by conserving energy ? (Is it because we are trying to minimise, rather ignore, effects of any reaction forces that might apply)
Best Answer
The total angular momentum of the body about an axis through $B$ perpendicular to the disc comprises the angular momentum of the body about its centroid plus the moment of momentum, about that axis, of a particle of the same mass as the disc situated at the centroid.
Just before the point B is fixed, the disc is rotated with angular speed $\omega$ about the centroid, and the centroid has a velocity component perpendicular to the radius through B which is $r\omega\cos\alpha$.
Therefore, relative to axis $B$ the total angular momentum is $$I_o\omega+M(r\omega\cos\alpha)r$$
By conservation of angular momentum, this must equal $$I_B\omega'$$
Therefore we have$$\frac{3}{2}Mr^2\omega'=\frac{1}{2}Mr^2\omega+Mr^2\omega\cos\alpha$$ $$\implies\omega'=\frac{1}{3}\omega(1+2cos\alpha)$$
Note that conservation of energy does not apply because the act of fixing the axis at $B$ imparts an impulse to the disc.