Question

A point moves along the circumference of a circle with a speed v = αt. (αis a constant).
The acceleration of the particle when it has covered 1/nth of the circumference is​

Answer 1

Answer:

The acceleration of the particle is $\frac{\alpha}{n} \sqrt{n^2+16\pi^2}$

Explanation:

Let the radius of the circle be R

Given the tangential velocity

v = αt

or ds/dt = αt

or, ds = αt.dt

or, $\int_0^{2\pi R/n} ds =\int_0^T\alpha tdt$

$\implies \frac{2\pi R}{n} =\frac{\alpha t^2}{2} \Bigr |_0^T\\\implies \frac{2\pi R}{n} =\frac{\alpha T^2}{2}$           ................ (1)

If the tangential acceleration is $a_t$ then

$a_t=\frac{dv}{dt} \\\implies a_t=\frac{d(\alpha t)}{dt} \\\implies a_t=\alpha$

If the radial acceleration is $a_r$ then

$a_r=\frac{v^2}{R}$

In time T, the particle covers the distance 2πR/n, its velocity after time T is αT, also from eq (1)

$\frac{4\pi R}{n \alpha} =T^2$

Therefore, the radial acceleration after time T

$a_r=\frac{(\alpha T)^2}{R} \\\implies a_r=\frac{\alpha^2 T^2}{R}$

$\implies a_r=\frac{\alpha^2}{R} \times \frac{4\pi R}{n \alpha}$

$\implies a_r= \frac{4\pi \alpha}{n}$

Therefore total acceleration

$a=\sqrt{a_t^2+a_r^2}$

$\implies a=\sqrt{\alpha^2+\frac{16\pi^2 \alpha^2}{n^2}}$

$\implies a=\frac{\alpha}{n} \sqrt{n^2+16\pi^2}$