A circular disc reaches from top to bottom of an inclined plane of length $$l$$. When it slips down the plane, it takes $$t$$ s. When it rolls down the plane then it takes $$\left(\frac{\alpha}{2}\right)^{1/2} t$$ s, where $$\alpha$$ is ______

A circular disc rolls and slides down an inclined plane of length $$l$$. When it slides (no rolling), it takes $$t$$ seconds. When it rolls without slipping, it takes $$\left(\frac{\alpha}{2}\right)^{1/2} t$$ seconds. Find $$\alpha$$.

When the disc slides without friction, the acceleration along the incline is:

$$a_{\text{slide}} = g\sin\theta$$

Using $$l = \frac{1}{2}a_{\text{slide}} t^2$$:

$$t^2 = \frac{2l}{g\sin\theta} \quad \cdots (i)$$

When a disc rolls without slipping, the moment of inertia about the centre is $$I = \frac{1}{2}mR^2$$, so $$\frac{I}{mR^2} = \frac{1}{2}$$.

The translational acceleration along the incline is:

$$a_{\text{roll}} = \frac{g\sin\theta}{1 + I/(mR^2)} = \frac{g\sin\theta}{1 + 1/2} = \frac{2g\sin\theta}{3}$$

Using $$l = \frac{1}{2}a_{\text{roll}} t_{\text{roll}}^2$$:

$$t_{\text{roll}}^2 = \frac{2l}{a_{\text{roll}}} = \frac{2l \cdot 3}{2g\sin\theta} = \frac{3l}{g\sin\theta} \quad \cdots (ii)$$

$$\frac{t_{\text{roll}}^2}{t^2} = \frac{3l/(g\sin\theta)}{2l/(g\sin\theta)} = \frac{3}{2}$$

$$t_{\text{roll}} = t\sqrt{\frac{3}{2}} = \left(\frac{3}{2}\right)^{1/2} t$$

We are given $$t_{\text{roll}} = \left(\frac{\alpha}{2}\right)^{1/2} t$$.

Comparing: $$\frac{\alpha}{2} = \frac{3}{2} \implies \alpha = 3$$.

The answer is 3.