Question

A particle at the edge of a rotating disc speeds up at a uniform angular acceleration $\alpha$. If the radius of the disc is R, find the angular distance covered by the particle till it acquires a total acceleration $a_0$.

Answer 1

The angular displacement of the particle till it attains velocity $\omega$ is

$\theta =\frac{{\omega }^2-{\omega }_0^2}{2\alpha }$

Where $\omega =0$ (because the particle starts from rest)

Then $\theta =\frac{{\omega }^2-{\omega }_0^2}{2\alpha }$..................(i)

Let us calculate $\omega$

The total acceleration of the particle is

$a=\sqrt{a_1^2+a_r^2}$

Where $a_t=R\alpha and{a}_r=R_{\omega }^2$

If the maximum acceleration of the particle is $a_0$ we have

$a_0=\sqrt{R^2{\alpha }^2+R^2{\omega }^4}$

This given $\omega ={\left(\frac{a_0^2-R^2{\theta }^2}{2R\alpha }\right)}^{\frac{1}{4}}$