Question

A small sphere of mass ‘m’ is tied to the top of the smooth inclined plane with the help of a string of length ‘L’. The string and the inclined plane makes an angle ‘$\theta$’ with the horizontal. The inclined plane is rotated with a constant angular velocity ‘$\omega$’ about the vertical axis passing through the end of the string fixed to the plane as shown in the figure.

The maximum value of $\omega$ so that the sphere maintains contact with the inclined plane is:

• A. $\sqrt{\frac{gsin\theta }{L}}$
• B. $\sqrt{\frac{g}{Lsin\theta }}$
• C. $\sqrt{\frac{2g}{Lsin\theta }}$
• D. $\sqrt{\frac{2gsin\theta }{L}}$

Answer 1

The correct option is B $\sqrt{\frac{g}{Lsin\theta }}$


Along vertical axis,
$Tsi{n}^{\theta }+Ncos\theta =mg............\left(i\right)$
Along radial axis,
$Tcos\theta -Nsin\theta =m{\omega }^2\left(Lcos\theta \right)$..........(ii)
From (i) and (ii), we get
$N=mgcos\theta -m{\omega }^{2}Lcos\theta sin\theta$

To maintain contact,
N > 0
$\Rightarrow mgcos\theta -m{\omega }^{2}Lcos\theta sin\theta >0\Rightarrow \omega <\sqrt{\frac{g}{Lsin\theta }}$