A simple pendulum of length l and having a bob of mass M is suspended in a car. The car is moving on a circular track of radius R with a uniform speed V. If the pendulum makes small oscillation in a radial direction about its equilibrium position, what will be its time period?

A simple pendulum of length l and having a bob of mass M is suspended

1.	A simple pendulum of length l and having a bob of mass M is suspended in a car. The car is moving on a circular track of radius R with a uniform speed V. If the pendulum makes small oscillation in a radial direction about its equilibrium position, what will be its time period?
Answer» Solution :The pendulum bob is subject to two distinct ACCELERATIONS: (i) the acceleration DUE to gravity, `vecg`, acting down-wards, (ii) the centripetal acceleration, `veca=v^(2)/2`, acting in a horizontal direction FIG. So, the effective, or relative, acceleration due to gravity on the pendulum bob is, `g^(vecr)=vecg-veca=vecg+(vec(-a))` Its magnitude, from Fig., is `g.=sqrt(g^(2)+V^(4)/R^(2))`. Then, the time period of the pendulum becomes, `T.=2pisqrt(L/(g.))=(2pisqrtl)/((g^(2)+V^(4)/R^(2))^(1//4))`