Discussion Forum



1)

The general motion of a rigid body can be considered to be a combination of (i) a motion of its centre of mass about an axis, and (ii) its motion about an instantaneous axis passing through the centre of mass. These axes need not be stationary. Consider, for example, a thin uniform disc welded (rigidly fixed) horizontally at its rim to a massless stick, as shown in the figure. When the disc-stick system is rotated about the origin on a horizontal frictionless plane with angular speed ω, the motion at any instant can be taken as a combination of (i) a rotation of the centre of mass of the disc about the z-axis, and (ii) a rotation of the disc through an instantaneous .vertlcal axis passing through its centre of mass (as is seen from the changed orientation of points P and Q). Both these motions have the same angular speed $\omega$ in this case.

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Now consider two similar systems as shown in the figure: Case (a) the disc with its face vertical and parallel to x-z plane; Case (b) the disc with its face making an angle of $45^{0}$ with the x-y plane and its horizontal diameter parallel to the x-axis. In both cases, the disc is welded at point P, and the systems are rotated with constant angular speed to about the z-axis.

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Which of the following statements regarding the angular speed about the instantaneous axis (passing through the centre of mass) is correct?


A) It is $\sqrt{2} \omega$ for both cases

B) It is $\omega$ for case (a); and $\frac{\omega}{\sqrt{2}}$ for case (b)

C) It is $\omega$ for case (a); and $\sqrt{2}\omega$ for case (b)

D) It is $\omega$ for both the cases

Answer:

Option D

Explanation:

(i) Every particle of the disc rotating in a horizontal circle.
(ii) Actual velocity of any particle horizontal.
(iii) Magnitude of the velocity of any particle is  $v=r \omega$ where r is the perpendicular distance of that particle from the actual axis of rotation (z-axis)

(iv) When it is broken into two parts then the actual velocity of any particle is the resultant of two velocities

 $v_{1}=r_{1}\omega_{1}$  and $v_{2}=r_{2}\omega_{2}$

Here,

$r_{1}=$perpendicular distance of the centre of mass from z-axis.

$\omega_{1}$=angular speed of rotation of centre of mass from z-axis.

$r_{2}$= distance of the particle from centre of mass and

$\omega_{2}$=angular speed of rotation of the disc about the axis passing through the centre of mass. (v) Net v will be horizontal, if $v_{1}$ and $v_{2}$ both are horizontal. Further, v, is already horizontal, because the centre of mass is rotating about the vertical z-axis.

To make v2, also horizontal, the second axis should also be vertical.