Question

A ball is placed on a smooth inclined plane of inclination $\theta ={45}^{\circ }$ to the horizontal, the inclined plane is rotating at frequency $1\text{Hz}$ about a vertical axis passing through its lower end. At what distance from the lower end does the ball remain at rest? $[$Take $g=10{\text{m/s}}^2$ and ${\pi }^2\approx 10$$]$

• A. $\frac{1}{\sqrt{2}}\text{m}$
  
• B. $\frac{1}{2}\text{m}$
  
• C. $1\text{m}$
  
• D. $\frac{1}{2\sqrt{2}}\text{m}$
  

Answer 1

The correct option is D $\frac{1}{2\sqrt{2}}\text{m}$


Let the ball remains at rest at a distance '$d$' from lower end.
The radius of the 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.$\left(1\right)$ and $\left(2\right)$ we get,

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

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

Also we can write,

$\omega =2\pi f=2\pi \left(1\right)=2\pi \text{rad/s}$

$\Rightarrow R=\frac{g\tan {45}^{\circ }}{(2\pi {)}^2}=\frac{1}{4}\text{m}$

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

Substituting the value of $R$ in Eq.$\left(1\right)$ we get,

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

$\Rightarrow d=\frac{\frac{1}{4}}{\cos {45}^{\circ }}=\frac{\frac{1}{4}}{\frac{1}{\sqrt{2}}}$

$\therefore d=\frac{1}{2\sqrt{2}}\text{m}$

Hence, option (b) is the correct answer.