**Paragraph:**
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 $\omega$, 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 vertical 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.

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 degree with x-y plane and its horizontal diameter parallel to x-axis. In both the cases, the disc is welded at point P, and the systems are rotated with constant angular speed $\omega$ about the z-axis. (IIT JEE 2012)

**Question 1:**
Which of the following statements about the instantaneous axis (passing through the centre of mass) is correct?

- It is vertical for both the case (a) and (b).
- It is vertical for case (a); and is at 45 degree to the x-z plane and lies in the plane of the disc for case (b).
- It is horizontal in case (a); and is at 45 degree to the x-z plane and is normal to the plane of the disc for case (b).
- It is vertical for case (a); and is at 45 degree to the x-z plane and is normal to the plane of the disc for case (b).

**Solution:**
Let C be the centre of the disc with position vector $\vec{r}_C$ and velocity vector $\vec{v}_{C}$. Let A be a general point on the disc having position vector $\vec{r}_A$. Let $\vec{\omega}$ be the angular velocity of the system as given in the problem and $\vec{\omega^\prime}$ be the angular velocity about instantaneous axis. The velocity of point C and point A are given by $\vec{v}_C=\vec{\omega}\times\vec{r}_C$ and $\vec{v}_A=\vec{\omega}\times\vec{r}_{A}$. Also, considering the motion about the instantaneous axis,
\begin{align}
\label{sca:eqn:1}
\vec{v}_A &=\vec{v}_C+\vec{\omega^\prime}\times \vec{r}_{A/C} \\
&=\vec{v}_C+\vec{\omega^\prime}\times (\vec{r}_A-\vec{r}_C).
\end{align}
Substitute $\vec{v}_A$ in the above equation and simplify to get
\begin{align}
\label{sca:eqn:2}
(\vec{\omega^\prime}-\vec{\omega})\times(\vec{r}_A-\vec{r}_C)=\vec{0}.
\end{align}
Since $\vec{r}_A$ is any general vector on the disc, the above equation is satisfied only when $\vec{\omega^\prime}=\vec\omega$ (Think over it!). This argument is true for both case (a) and case (b). Hence, the instantaneous axis for both the cases are parallel to the z-axis and the angular speed about these axes are equal.

**Question 2:**
Which of the following statements regarding the angular speed about the instantaneous axis (passing through the centre of mass) is correct?

- It is $\sqrt{2}\omega$ for both the cases.
- It is $\omega$ for case (a); and $\frac{\omega}{\sqrt{2}}$ for case (b).
- It is $\omega$ for case (a); and $\sqrt{2}\omega$ for case (b).
- It is $\omega$ for both the cases.

**Solution:**
It is $\omega$ for both the cases.

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