1. Given system

• Mass of disc $= 1\,\text{kg}$
• Radius of disc $= R$
• A body of mass $1\,\text{kg}$ is fixed at the highest point on the rim.
• Axis passes through the centre of the disc and is perpendicular to the plane of the disc.

We need the angular speed when the attached body reaches the lowest point.

2. Use conservation of mechanical energy

Initially, the attached mass is at the top of the disc. Finally, it comes to the bottom.

So the attached mass falls through a vertical distance: $2R$

Hence loss in gravitational potential energy is $\Delta U = mg(2R) = (1)(10)(2R)=20R$

This becomes rotational kinetic energy of the whole system.

3. Find total moment of inertia about the axis

(a) Disc

Moment of inertia of a disc about its central axis: $I_{\text{disc}} = \frac{1}{2}MR^2 = \frac{1}{2}(1)R^2=\frac{R^2}{2}$

(b) Attached body

It is a point mass of $1\,\text{kg}$ at distance $R$ from the axis, so $I_{\text{mass}} = mR^2 = (1)R^2 = R^2$

(c) Total

$I_{\text{total}} = \frac{R^2}{2}+R^2 = \frac{3R^2}{2}$

4. Apply energy conservation

At the lowest position, $\frac{1}{2}I_{\text{total}}\omega^2 = 20R$

Substitute $I_{\text{total}}=\frac{3R^2}{2}$: $\frac{1}{2}\left(\frac{3R^2}{2}\right)\omega^2=20R$

$\frac{3R^2}{4}\omega^2=20R$

$\omega^2=\frac{80}{3R}$

$\omega=\sqrt{\frac{80}{3R}}=4\sqrt{\frac{5}{3R}}$

5. Compare with given form

Given: $\omega = 4\sqrt{\frac{x}{3R}}$

From our result: $\omega = 4\sqrt{\frac{5}{3R}}$

So, $x=5$

6. Comparison with stored answer

Stored correct answer: $5$

Our derived answer: $5$

They match.