Question

A simple pendulum sits on a merry go round moving with `4m/s` and having a radius of `2m`. Calculate the time period of the pendulum?

Answer 1

Period, I get 0.395 s.

Explanation:

The formula for the period of a pendulum is

`T = 2*pi*sqrt(L/g)`

To solve this problem we need to interpret g as the vector sum of the normal value of g and the centripetal acceleration due to that 4 m/s rotation. The reason we need to do that is that the "restoring force" is increased by the centripetal force.

The centripetal acceleration is given by

`a_c = v^2/r = (4 m/s)^2/(2 m) = 8 m/s^2`

The centripetal acceleration is perpendicular to straight down (the direction of normal g). Therefore the sum of the 2 accelerations is found using Pythagoras:

`a_"sum" = sqrt((9.8 m/s^2)^2 + (8 m/s^2)^2) = 12.65 m/s^2`

So, for this situation, we use `12.65 m/s^2` as the value of g in the formula for the period of a pendulum.

`T = 2*pi*sqrt(L/g) = 2*pi*sqrt((0.05 m)/(12.65 m/s^2)) = 0.395 s`

Hmmm, I know you said the answer was 0.30 s. I am continuing the request to "double check my answer".

I hope this helps,

Steve