Question

A particle of mass is kept on the top of a smooth sphere of radius. It is given a sharp impulse which imparts it a horizontal speed V.
(a) Find the normal force between the sphere and the particle just after the impulse.
(b) What should be the minimum value of V for which the particle does not slip on the surface?
(c) Assuming the velocity v to be half the minimum calculate in part,
(d) Find the angle made by the radius.

Answer 1

Hint: The minimum velocity v to which it prevent slipping can be given by;
 $ m\dfrac{{{v_{\min }}^2}}{R} = mg \\
   \Rightarrow {v_{\min }} = \sqrt {Rg} \\ $
The particle descends by an angle $ \theta $ and $ h $ be the height of the initial point when it leaves the contact.
After application of conservation of energy:
 $ \dfrac{1}{2}m{v^2} + mgh = \dfrac{1}{2}m{v^2}_f $
Also height can be represented by;
 $ h = R(1 - \cos \theta ) \\$

Complete step by step solution:
In the given solution, in part (a); here at the instant right after the implementation of the impulse, the particle comes in the circular motion.
Hence, it will require the centripetal force, which will be provided by the resultant of $ mg $ and $ N $ .
Now,
 $ m\dfrac{{{v^2}}}{R} = mg - N \\
   \Rightarrow N = mg - m\dfrac{{{v^2}}}{R} \\ $
 (b) The minimum velocity v to which it prevent slipping can be given by;
 $ m\dfrac{{{v_{\min }}^2}}{R} = mg \\
   \Rightarrow {v_{\min }} = \sqrt {Rg} \\ $
The normal reaction N is zero.
(c) Suppose $ v = \dfrac{{{v_{\min }}}}{2} $
Final velocity here be $ {v_f} $ .
Let the particle descend by an angle $ \theta $ and $ h $ be the height of the initial point when it leaves the contact.
After application of conservation of energy:
 $ \dfrac{1}{2}m{v^2} + mgh = \dfrac{1}{2}m{v^2}_f $
Also,
 $ h = R(1 - \cos \theta ) \\
    \\ $
Now, $ v = \dfrac{{{v_{\min }}}}{2} = \dfrac{{\sqrt {Rg} }}{2} \\
  \dfrac{1}{2}m\dfrac{{Rg}}{4} + mgR(1 - \cos \theta ) = \dfrac{1}{2}m{v^2}_f \\ $
 $ \Rightarrow {v^2}_f = \dfrac{{Rg}}{4}(9 - 8\cos \theta ) $
(d) Applying the newton's second law in the radial direction,
We get;
 $ mg\cos \theta - {\rm N} = m\dfrac{{{v^2}_f}}{R} $
Also $ N = 0 $
Now;
 $ {v^2}_f = Rg\cos \theta $
Substituting this value in the obtained equation,
 $ \dfrac{{Rg}}{4}(9 - 8\cos \theta ) = Rg\cos \theta \\
   \Rightarrow \cos \theta = \dfrac{3}{4} \\
  \theta = {\cos ^{ - 1}}(\dfrac{3}{4}) \\ $

Note:
At the instant right after the implementation of the impulse, the particle comes in the circular motion. The minimum velocity v to which it prevent slipping can be given by;
 $ m\dfrac{{{v_{\min }}^2}}{R} = mg \\
   \Rightarrow {v_{\min }} = \sqrt {Rg} \\ $
The particle descends by an angle $ \theta $ and $ h $ be the height of the initial point when it leaves the contact.
After application of conservation of energy:
 $ \dfrac{1}{2}m{v^2} + mgh = \dfrac{1}{2}m{v^2}_f $
Also height can be represented by;
$ h = R(1 - \cos \theta ) \\
    \\ $