A child sits down on one end of a horizontal seesaw of negligible weight, 2.14 m from the pivot point. No one balances on the other side. Find the instantaneous angular acceleration of the see-saw and the tangential linear acceleration of the child.
4 answers
I know that the angular acc is g/d or 9.8/2.14=4.58 rad/s^2
the tangential linear acceleration of the child. = g = 9.81m/s^2
instantaneous angular acceleration of the see-saw = a/r = 9.81/1.91 = 5.14rad/s^2
instantaneous angular acceleration of the see-saw = a/r = 9.81/1.91 = 5.14rad/s^2
where is the 1.91 coming from?
opps, switch that with the 2.14m
1 answer