A circular disc reaches from top to bottom of an inclined plane o… — Solution | JEE Physics

MCQMediumJEE 2024Rolling Motion & Rotational Kinematics

A circular disc reaches from top to bottom of an inclined plane of length $l$ . When it slips down, it takes $t$ seconds. When it rolls down, it takes $\left(\alpha/2\right)^{1/2} \times t$ seconds. Find $\alpha$ .

A
$2$
B
$3$
C
$4$
D
$5$

Answer

Correct answer:B

Step-by-step solution

Standard Method

Given: A circular disc moves down an inclined plane of length $l$ . Time for slipping is $t$ , and time for rolling is $\sqrt{\alpha/2}\, t$ .

Find: The value of $\alpha$ .

For slipping down the incline, the acceleration along the plane is

a_{\text{slip}} = g \sin\theta

Using

l = \frac{1}{2} a_{\text{slip}} t^2

we get

t = \sqrt{\frac{2l}{g\sin\theta}}

For rolling motion,

a_{\text{roll}} = \frac{g\sin\theta}{1 + \frac{k^2}{r^2}}

For a disc,

k = \frac{r}{\sqrt{2}}

So,

a_{\text{roll}} = \frac{g\sin\theta}{1 + \frac{\left(r/\sqrt{2}\right)^2}{r^2}} = \frac{g\sin\theta}{1 + 1/2} = \frac{2g\sin\theta}{3}

Now the rolling time is

t_{\text{roll}} = \sqrt{\frac{2l}{a_{\text{roll}}}} = \sqrt{\frac{2l}{2g\sin\theta/3}} = \sqrt{\frac{3l}{g\sin\theta}}

Hence,

\frac{t_{\text{roll}}}{t} = \sqrt{\frac{a_{\text{slip}}}{a_{\text{roll}}}} = \sqrt{\frac{g\sin\theta}{2g\sin\theta/3}} = \sqrt{\frac{3}{2}}

But given,

t_{\text{roll}} = \sqrt{\frac{\alpha}{2}}\, t

Therefore,

\sqrt{\frac{\alpha}{2}} = \sqrt{\frac{3}{2}}

Squaring both sides,

\alpha = 3

Therefore, the value of $\alpha$ is $3$ . The correct option is B.

Time Ratio Method

Given: Slipping acceleration is compared with rolling acceleration for the same distance.

Find: $\alpha$ .

For the same displacement from rest, time is inversely proportional to the square root of acceleration:

t \propto \frac{1}{\sqrt{a}}

So,

\frac{t_{\text{roll}}}{t_{\text{slip}}} = \sqrt{\frac{a_{\text{slip}}}{a_{\text{roll}}}}

Now,

a_{\text{slip}} = g\sin\theta

and for a disc,

a_{\text{roll}} = \frac{g\sin\theta}{1 + k^2/r^2} = \frac{g\sin\theta}{1 + 1/2} = \frac{2g\sin\theta}{3}

Thus,

\frac{t_{\text{roll}}}{t_{\text{slip}}} = \sqrt{\frac{g\sin\theta}{2g\sin\theta/3}} = \sqrt{\frac{3}{2}}

Given,

\frac{t_{\text{roll}}}{t_{\text{slip}}} = \sqrt{\frac{\alpha}{2}}

So,

\sqrt{\frac{\alpha}{2}} = \sqrt{\frac{3}{2}} \Rightarrow \alpha = 3

Therefore, the correct option is B.

Common mistakes

Using the rolling acceleration formula incorrectly by taking $I = mr^2$ instead of the disc value $I = \frac{1}{2}mr^2$ . This gives the wrong effective acceleration. For a disc, use $k = r/\sqrt{2}$ or equivalently $I = \frac{1}{2}mr^2$ .
Assuming the time ratio is directly proportional to acceleration. Time for motion from rest over the same distance varies as $1/\sqrt{a}$ , not as $1/a$ . Always use $t = \sqrt{2l/a}$ before comparing times.
Equating $\sqrt{\alpha/2}$ with $3/2$ instead of $\sqrt{3/2}$ . The ratio comes from square roots of accelerations, so the square root must be retained until the final squaring step.

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