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?

1 Answer

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