Question

A particle slides down from the top outside smooth surface of a fixed sphere of radius a = $10$ m. The initial horizontal velocity to be imparted to the particle 'at the top' is $5Km/s$ if it leaves the surface at a point whose vertical height above the center of sphere is $3a/4$. Find the value of K.

Answer 1

Let the particle be at a point $P$ on the sphere having center $O$, $OP$ forms an angle $\theta$ with the vertical

Net centripetal force at $P=mgcos\theta -N-\left(1\right)$

From work-energy theorem:

$\frac{1}{2}m{v}_p^2-\frac{1}{2}m{v}_{initial}^2=W$

Since normal force is perpendicular to motion, it does no work. Work done by gravity

$=mga(1-cos\theta )$

$\therefore \frac{1}{2}m{v}_p^2-\frac{1}{2}m{v}_{initial}^2=mga(1-\cos \theta )\Rightarrow v_p^2=v_{initial}^2+2ga(1-\cos \theta )-\left(2\right)$

Also, centripetal acceleration:

$a_c=\frac{v_p^2}{a}-\left(3\right)$

using $\left(1\right)$,(2)$and$(3)$:

$mg\cos \theta -N=\frac{m}{a}[v_{initial}^2+2ga(1-\cos \theta )]$

when particle loses contact $N=0$ Also height$=\frac{3}{4}=a\cos \theta$

$\Rightarrow \cos \theta =\frac{3}{4}$ (when particle loses contact)

$\therefore \frac{3}{4}mg=\frac{m}{a}[v_{initial}^2-2ga(1-\frac{3}{4})]\therefore v_{initial}=\frac{1}{2}\sqrt{ag}=5K\Rightarrow K=\frac{1}{10}\sqrt{ag}K=\frac{10}{10}=1$