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Topic: Motion of unconstrained body
Replies: 2   Last Post: May 11, 2000 9:12 AM

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Robert Hill

Posts: 529
Registered: 12/8/04
Re: Motion of unconstrained body
Posted: May 9, 2000 7:26 AM
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In article <FUAR4.1390$>, "Michael Garner" <> 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(u-v)
> 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(u-V) = (16/3)ml^2w + l*4mV
> 3(u-V) = 4lw + 3V
> 4lw = 3(u-2V)
> w = (3/4l)(u-2V)
> 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

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