(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

A uniform square board ABCD had mass 5kg and sides of length 60cm. The board is hinged at A so that it can rotate freely in a vertical plane. A light string of length 1 metre is attached to B and to a point E, 80cm above A. The bards rests in equilibrium with AB horizontal and D vertically below A. It is now raised through a right angle so that B is vertically above A and AD is horizontal, and released from rest in this position. When the board returns to its original position the string becomes taught and the board is brought immediately to rest. Find the impulsive tension in the string when this occurs and the horizontal and vertical components of the impulse on the hinge.

3. The attempt at a solution

This is what I did for the first bit.

[itex]I_A = I_{AB} + I_{AD} = 2\times (\frac{4}{3} \times 5 \times 0.3^2) = \frac{6}{5}[/itex] (perpendicular axes rule)

Then by energy changes, angular velocity just before it stops is [itex]\omega = \sqrt{24.5}[/itex]

So the impulse is the change in angular momentum which is [itex]\frac{6}{5} \times \sqrt{24.5} = 5.9396...Ns[/itex] = the wrong answer.

The right answer is 17.5 Ns. I don't know what I did wrong.

Also I have no idea about the second part.

It would be really amazingly immensely helpful if anyone could point me in the direction of some angular momentum/impulse on the axis notes, because I am really not getting it.

**Physics Forums | Science Articles, Homework Help, Discussion**

Dismiss Notice

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Homework Help: Angular momentum - impulse on the axis

**Physics Forums | Science Articles, Homework Help, Discussion**