A simple pendulum, suspended from the roof of a car, is oscillating. What will be the time period of the pendulum when the car moves in a circular path at a constant speed?

A simple pendulum, suspended from the roof of a car, is oscillating. W

1.	A simple pendulum, suspended from the roof of a car, is oscillating. What will be the time period of the pendulum when the car moves in a circular path at a constant speed?
Answer» Solution :Two forces act on the pendulum bob when the car moves in a circular path at constant speed: (1) weight of the bob mg VERTICALLY downwards, (2) CENTRIFUGAL force in the bob `momega^(2)R` horizontally. (m = mass of bob, `omega` = angular velocity of bob, r = radius of the circular path) `therefore` Resultant effective force on the bob, `F=sqrt(m^(2)g^(2)+m^(2)omega^(2)r^(2))=msqrt(g^(2)+omega^(4)r^(2))` Hence, acceleration of the bob `a=F/m=sqrt(g^(2)+omega^(4)r^(2))` `therefore` Time period `T=2pisqrt(L/a)=2pisqrt(L/((g^(2)+omega^(4)r^(2))^(1//2)))` Alternative Method: When the car along with the point of suspension of the pendulum moves on a circular path horizontally, a freely falling body (along the vertical line through the point of suspension) is subjected to centripetal acceleration horizontally and acceleration due to gravity vertically. The centripetal acceleration is `omega^(2)r" "(omega` = angular velocity of point of suspension, r = radius of the circular path). `therefore` Acceleration of the body falling freely with RESPECT to the point of suspension = `sqrt(g^(2)+omega^(4)r^(2))` Hence, time period of OSCILLATION of the pendulum `T=2pisqrt(L/((g^(2)+omega^(4)r^(2))^(1//2)))`