Question

A particle of mass m oscillates along the horizontal diameter AB inside a smooth spherical shell of radius R. At any instant the kinetic energy of the particle is K. Then the force applied by particle on the shell at this instant is –

• A.
• B. $\frac { 2 \mathrm {~K} } { \mathrm { R } }$
• C.
• D.

Answer 1

Answer: (C)

Answer 2

Let velocity of particle at point P is v.

From conservation of mechanical energy

$\frac { 1 } { 2 }$mv² = K = mgh

Let N be the normal reaction between the particle and the shell at this instant. Then

N – mg sin q =

or N = mg+$\frac { 2 \mathrm {~K} } { \mathrm { R } }$=+$\frac { 2 \mathrm {~K} } { \mathrm { R } }$ (mgh = K)

\ N == force on shell