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Posted by on Monday, October 31, 2011 at 5:31pm.

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.

  • Physics - , Monday, October 31, 2011 at 5:34pm

    I know that the angular acc is g/d or 9.8/2.14=4.58 rad/s^2

  • Physics - , Monday, October 31, 2011 at 5:54pm

    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

  • Physics - , Monday, October 31, 2011 at 5:57pm

    where is the 1.91 coming from?

  • Physics - , Monday, October 31, 2011 at 5:59pm

    opps, switch that with the 2.14m

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