A disc of mass $$1 \text{ kg}$$ and radius $$R$$ is free to rotate about a horizontal axis passing through its centre and perpendicular to the plane of disc. A body of same mass as that of disc is fixed at the highest point of the disc. Now the system is released, when the body comes to the lowest position, its angular speed will be $$4\sqrt{\frac{x}{3R}} \text{ rad s}^{-1}$$ where $$x =$$ ______.

$$Massofdisc=1kg$$

$$Massofparticle=1kg$$

$$Radius=R$$

Initially the particle is at the top, finally at the bottom

Loss in potential energy of particle:
$$ΔPE=mg(2R)=1\times g\times2R=2gR$$

Disc’s potential energy does not change (its centre stays at same height)

So total energy converted to rotation:
$$2gR=\frac{1}{2}I_{total}ω^2$$

Now moment of inertia:

Disc about centre:
$$I_{disc}=\frac{1}{2}MR^2=\frac{1}{2}R^2$$

Particle at distance R:
$$I_{particle}=MR^2=R^2$$

So:
$$I_{total}=\frac{1}{2}R^2+R^2=\frac{3}{2}R^2$$

Now substitute:

$$2gR=\frac{1}{2}\left(\frac{3}{2}R^2ω^2\right)$$

$$2gR=(3/4)R^2ω^2$$

Solve:

$$ω^2=(8g)/(3R)$$

$$ω=\sqrt{\ \frac{8g}{3r}}$$

Given answer form:
$$ω=4\sqrt{\ \frac{x}{3r}}$$

Compare:
$$x=\frac{g}{2}$$

Given g = 10:

x = 5

Final answer:

5