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Re: Motion of unconstrained body
Posted:
May 9, 2000 7:26 AM


In article <FUAR4.1390$Eu2.34384@news2win.server.ntlworld.com>, "Michael Garner" <michael.garner.nospam@iname.com> writes: > [Mechanics Question] > > Four uniform rods, each of mass m and length 2l are joined rigidly together > to form a square frame ABCD of side 2l. The frame is placed with all four > sides at rest on a smooth horizontal table. An inextensible string has one > end attached to the corner A. A particle of mass 4m is tied to the other > end of the string. The particle, initially at A is projected with speed u > in the direction DA. Given that the speed of the particle immediately after > the string becomes taut is V, show that the initial angular speed of the > square frame about an axis through its centre of gravity, perpendicular to > the plane of the frame is w where w = (2V  u)/l. Show that V= 7u/11 and > that immediately after the string becomes taut the KE of the particle and > frame is 14mu^2/11. > > I've spent a _long_ time trying to figure out how to do this, but can't get > anywhere. So far, I've called the impulse acting J and said: > J= ch in lin mom of particle = 4m(uv) > > and moment of J = change in ang mom of sys = ch in ang mom of square frame + > momemnt of mom of particle > > so, lJ = (16/3)ml^2w + l*4mV > > so 4ml(uV) = (16/3)ml^2w + l*4mV > > 3(uV) = 4lw + 3V > > 4lw = 3(u2V) > > w = (3/4l)(u2V) > > But this isn't right, and I can't see how to do the next part. Could anyone > point out my error? Thanks in advance for any help.
I think you're assuming that the motion of the frame immediately after the impulse consists purely of rotation about its centre.
Robert Hill



