Question

A small glass marble of mass $m$oscillates between the two edges, inside a hemispherical glass bowl of radius $r$. If V is the speed of the marble at the lowest position, the normal reaction at that position is:
A) $\dfrac{{m{V^2}}}{r}$
B) $\dfrac{{2m{V^2}}}{r}$
C) $\dfrac{{3m{V^2}}}{r}$
D) $\dfrac{{3m{V^2}}}{{2r}}$

Answer 1

Hint: Recall that the kinetic energy of a particle is the energy possessed by it due to its motion. It is the work that is done to accelerate any particle from rest to the required velocity. On the other hand, potential energy of a particle is defined as the energy stored in an object due to its position or height.

Complete step by step solution:
Given that the mass of the marble is $ = m$
The radius of hemispherical glass bowl is $ = r$
Speed of the marble at the lowest position is $ = V$
At the highest point the kinetic energy of the marble will be zero.
$ \Rightarrow K.E. = 0$
And the potential energy of the system will be
$ \Rightarrow P.E. = mgr$
When the marble is at some middle point between the edges making an angle $\theta $ with the normal, the marble will be moving in a hemispherical path. So there will be a centripetal force acting on the marble. The kinetic energy of the marble will be given by
$ \Rightarrow K.E. = \dfrac{{m{V^2}}}{2}$
Similarly, the potential energy of the system is given by
$ \Rightarrow P.E. = mgr(1 - \sin \theta )$
Since it is known that the energy of an isolated system is always conserved, so the total energy of the system at the highest point and the total energy of the system when it is in the middle will be equal.
Therefore, it can be written that
${(K.E. + P.E.)_{middle}} = {(K.E. + P.E.)_{HighestPo\operatorname{int} }}$
$ \Rightarrow \dfrac{{m{V^2}}}{2} + mgr(1 - \sin \theta ) = 0 + mgr$
$ \Rightarrow \dfrac{{m{V^2}}}{2} + mgr - mgr\sin \theta = mgr$
$ \Rightarrow \dfrac{{m{V^2}}}{2} - mgr\sin \theta = 0$
$ \Rightarrow m(\dfrac{{{V^2}}}{2} - gr\sin \theta ) = 0$
$ \Rightarrow gr\sin \theta = \dfrac{{{V^2}}}{2}$
$ \Rightarrow g\sin \theta = \dfrac{{{V^2}}}{{2r}}$
The normal reaction at that position on the hemisphere is given by the sum of centripetal force and the angle made by the weight ‘mg’.
The centripetal force is written as
${F_{centripetal}} = \dfrac{{m{V^2}}}{r}$
Therefore, it can be written that
$ \Rightarrow \dfrac{{m{V^2}}}{r} + mg\sin \theta $
Substituting the value of $g\sin \theta $ in the above equation, the normal reaction will be
$ \Rightarrow \dfrac{{m{V^2}}}{r} + \dfrac{{m{V^2}}}{{2r}} = \dfrac{{2m{V^2} + m{V^2}}}{{2r}}$
$ \Rightarrow \dfrac{{3m{V^2}}}{{2r}}$

Option D is the right answer.

Note: It is important to remember that the kinetic energy of a particle is directly proportional to the mass of the particle and the square of its velocity. If the speed of the particle is doubled then the kinetic energy of the body increases by four times. When potential energy is used by the particle, it changes into kinetic energy.