Angular Momentum Prob: Find Wfinal & Angle

[Idoubt]

Homework Statement

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w= ang velocity

L= ang momentum

I = moment of inertia

Homework Equations

r cross F = torque


moment of inertia of a horizontal square surface = 1/6 m a²

The Attempt at a Solution

The time taken should not depend on the rotation and should simply be a/v

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Assuming the angle above, the torque the man applies on the square should be independent of the rotation on the square and should only depend on the position of the man on the side of the square. Then from the geometry, the torque is

r*F = rFcos(theta) but r = (a/2)/ cos theta

then Torque = (a/2)F = (a/2)md²r/dt²

Integrating ang momentum L = [( ma/2) (dr/dt)]

Now the velocity of the man with respect to the platform, V = dr/dt - rw

so dr/dt = v+rw

sub in ang momentum eq, I get,

L = Iw = mav/2 + marw/2

at t = a/v, r is the diagonal = a/(2)^1/2

so L_final = 1/6ma²w_final=mav/2 + ma²w/2(2)^1/2

then w_final = V/((1/3)-1/(2)^1/2))a

this is as far as I got, I get this nagging feeling that I have over thought this and made unnecessary complications, and also I can't integrate this W_final to get the angle since its not a function of time, just the final ang velocity. also not sure how to get r as a function of time, pls tell me what I am doing ( or thinking ) wrong.

 

[Idoubt]

Hmm now it occurs to me that if he has applying a force on the platform, he can't possibly walk with a constant velocity, so there must be an initial force, but after that there can be no net force over time. so his ang momentum must be constant?

 

[Idoubt]

Thinking of this prob with constant angular momentum,

L = Iw = r*P = r*mv = rmvcos(theta)again, r = a/2cos(theta) then, Iw = 1/6ma²w = amv/2

then w = 3v/a,

integrating, theta = 3vt/a

at t = a/v, theta = 3 radians, I think it should be = -3 rad, with my sign convension,I'm not so sure about r_t,

My best guess is,

W_{man measured from space} = W_{man from platform} - W_platform

Here W_platform was the w that we solved earlier

then, V_t/r_t = V/r_t - W_p

since, r is the same in both frames.

so, V_t + r_tW_p = V

diff w.r.t time,

d²r/dt² + W_pdr/dt = 0

This is the S.H.M D.E, with soln,

r_t = Acos ( (W_p)^1/2t + phase angle)

here A has to be the max value of r, ie the diagonal of the square a/(2)^1/2So r_t = a/(2)^1/2cos ((Wp)^1/2t + phase angle)Can some1 tell me if I'm even close?

 

[Idoubt]

And my best guess is very very wrong that's not the S.H.M Diff eqn, if i solve that equation, i get an exponent in my soln and that can't be right, I'm sure r_t has to be a sin or cos function.