Question

A solid body starts rotating about a stationary axis with an angular acceleration = $at$. How soon after the beginning of rotation will the total acceleration vector of an arbitrary point of the body form an angle = ${60}^{\circ }$ with its velocity vector?

• A.
• B.
• C.
• D.

Answer 1

The correct option is C

Total acceleration making 60⁰ with the velocity vector is as good as it making 60⁰ with tangential acceleration since the direction of both v & a_t is same.

Now from the figure we know
$Tan{60}^o=\frac{a_N}{a_t}=\sqrt{3}$ .......(i)

A = a x t
Tangential acceleration $a_t=r\alpha =rat$ ........(ii)

Normal acceleration $a_N=\frac{V^2}{R}$

$V=r\omega$

$\Rightarrow a_N=r{\omega }^2$

So we need $\omega$ or v

Hmmm................... (we cannot use equations of motion as you know $\alpha$ is variable)

$\alpha =\frac{d\omega }{dt}=at$
$\Rightarrow {\int }_0^{\omega }d\omega ={\int }_0^{t}at.dt$
$\Rightarrow \omega =\frac{a{t}^2}{2}$
$\Rightarrow a_N=\frac{r{a}^2{t}^4}{4}$
Now substituting (ii) and (iii) in (i) we get
$\frac{\frac{r{a}^2{t}^4}{4}}{rat}=\sqrt{3}$
$\frac{a{t}^3}{4}=\sqrt{3}$
$t={\left(\frac{4\sqrt{3}}{a}\right)}^{\frac{1}{3}}$