2

Image for following question

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 -

  1. 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.)
  2. As this $\omega$ might have changed, so how can we take $I_0\omega$ in the calculation ?
  3. 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)
VGisHere
  • 58

1 Answers1

1

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.

David Quinn
  • 34,121
  • David, your explanation is in coherence with the statement “Just before the point B is fixed, the disc is rotated with angular speed ”, as you cited. But the question mentioned that the initial angular velocity about A was and not centroid. My question is how the angular velocity about A and centroid same when Moment of a inertia has changed ? – VGisHere Jan 03 '21 at 10:46
  • before the axis $B$ is fixed, the angular speed about $A$ is the same as the angular speed about the centroid – David Quinn Jan 03 '21 at 19:59
  • I was of the opinion that it would change as moment of inertia about A and centroid are different. Or is it that when A is released the disc will have both translation and rotation. And the rotational energy gets converted to translational energy ? How do we prove this ? – VGisHere Jan 04 '21 at 16:13
  • You seem to be asking a number of different questions here, (1) While the body is rotating about the fixed axis $A$, the angular speed of the body is the same as the angular speed of the body with respect to the centroid. You only need to consider how the radius $OA$ is moving. (2) If the body was suddenly released from axis $A$ it would continue to spin but the centroid would also translate. Conservation of angular momentum could be used to calculate the angular speed after it it released (take moments at $A$). – David Quinn Jan 04 '21 at 19:10
  • However this is not the situation being described in your question: as soon as it is released from $A$ it is being forced to rotate about $B$ – David Quinn Jan 04 '21 at 19:11
  • Thanks David, now I get it. I was mixing two different concepts. Thanks for the clarification. :) – VGisHere Jan 06 '21 at 10:40
  • you are welcome – David Quinn Jan 06 '21 at 14:46