Question

A ball is placed on a smooth inclined plane of inclination $\theta ={30}^{\circ }$ to the horizontal, the inclined plane is rotating at frequency $0.5$ $\text{Hz}$ about a vertical axis passing through its lower end. At what distance from the lower end does the ball remains at rest?

• A. $0.87\text{m}$
• B. $0.33\text{m}$
• C. $0.5\text{m}$
• D. $0.67\text{m}$

Answer 1

The correct option is D $0.67\text{m}$

Let the ball remains at rest at a distance '$d$' from lower end.
The radius of horizontal circle described is given as,
$R=d\cos \theta ....\left(1\right)$


Applying equilibrium condition in vertical direction because the body is performing circular motion in a horizontal plane;

$N\cos \theta =mg.....\left(2\right)$

Equation of circular dynamics gives;

$N\sin \theta =m{\omega }^{2}R.....\left(3\right)$

Dividing Eq.$\left(i\right)$ and $\left(ii\right)$ we get,

$\Rightarrow \tan \theta =\frac{R{\omega }^2}{g}$

$\Rightarrow R=\frac{g\tan \theta }{{\omega }^2}$

Also we can write,

$\omega =2\pi f=2\pi \left(0.5\right)=\pi \text{rad/s}$

$\Rightarrow R=\frac{g\tan {30}^{\circ }}{(\pi {)}^2}=\frac{1}{\sqrt{3}}$

$(\because {\pi }^2\approx 10)$

Substituting the value of $R$ in Eq.$\left(i\right)$ we get,

$d=\frac{R}{\cos \theta }$

$\Rightarrow d=\frac{\frac{1}{\sqrt{3}}}{\cos {30}^{\circ }}=\frac{\frac{1}{\sqrt{3}}}{\frac{\sqrt{3}}{2}}$

$\therefore d=\frac{2}{3}\simeq 0.67\text{m}$