JEE Physics 2024 Question with Solution

Given: Radius $r = 0.5 \, \text{m}$ and initial speed $v(0) = 4 \, \text{m/s}$.

Find: The value of $\alpha$ in the expression for time of first revolution,

$$t = \frac{1}{\alpha}(1-e^{-2\pi})$$

when the normal and tangential accelerations are equal at every instant.

Since the magnitudes of tangential and normal accelerations are equal,

$$a_t = a_n$$

so

$$\frac{dv}{dt} = \frac{v^2}{r}$$

Separate the variables:

$$\frac{dv}{v^2} = \frac{dt}{r}$$

Integrating from $v = 4$ at $t = 0$ to speed $v$ at time $t$,

$$\int_4^v \frac{dv}{v^2} = \int_0^t \frac{dt}{r}$$

This gives

$$\left[-\frac{1}{v}\right]_4^v = \frac{t}{r}$$

With $r = 0.5 \, \text{m}$,

$$-\frac{1}{v} + \frac{1}{4} = 2t$$

Hence,

$$\frac{1}{v} = \frac{1}{4} - 2t$$

so

$$v(t) = \frac{4}{1-8t}$$

Comparing with the required form

$$v(t) = \frac{4}{1-\alpha t}$$

we get

$$\alpha = 8$$

Now check the revolution-time form using $v = \frac{ds}{dt}$:

$$\frac{ds}{dt} = \frac{4}{1-8t}$$

For one complete revolution, the arc length is

$$s = 2\pi r = \pi$$

Therefore,

$$\int_0^{\pi} ds = 4\int_0^t \frac{dt}{1-8t}$$

so

$$\pi = 4\left[-\frac{1}{8}\ln(1-8t)\right]_0^t = -\frac{1}{2}\ln(1-8t)$$

Thus,

$$\ln(1-8t) = -2\pi$$

which gives

$$1-8t = e^{-2\pi}$$

and hence

$$t = \frac{1-e^{-2\pi}}{8}$$

Therefore, comparing with

$$t = \frac{1}{\alpha}(1-e^{-2\pi})$$

the value of $\alpha$ is $8$.

Given: $|a_t| = |a_c|$, $r = 0.5 \, \text{m}$, and $v_0 = 4 \, \text{m/s}$.

Find: The parameter $\alpha$.

Using circular motion relations,

$$a_c = \frac{v^2}{r}, \qquad a_t = \frac{dv}{dt}$$

Given that these are equal,

$$\frac{v^2}{r} = \frac{dv}{dt}$$

Rearrange as

$$\frac{dv}{v^2} = \frac{dt}{r}$$

Integrate:

$$\int_4^v \frac{dv}{v^2} = \int_0^t \frac{dt}{r}$$ $$-\frac{1}{v} + \frac{1}{4} = \frac{t}{0.5} = 2t$$ $$\frac{1}{v} = \frac{1}{4} - 2t$$ $$v = \frac{4}{1-8t}$$

So the coefficient of $t$ in the denominator is $8$, therefore

$$\alpha = 8$$

Also, for one revolution,

$$s = 2\pi r = \pi$$

and since $v = ds/dt$,

$$\pi = \int_0^t \frac{4}{1-8t} \, dt$$ $$\pi = -\frac{1}{2}\ln(1-8t)$$ $$1-8t = e^{-2\pi}$$ $$t = \frac{1-e^{-2\pi}}{8}$$

This matches the given form exactly, so the required value is $8$.