Classical Mechanics-Inclined plane tricky problem


by squareroot
Tags: classical, mechanicsinclined, plane, tricky
haruspex
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#37
Dec18-12, 12:38 AM
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Quote Quote by squareroot View Post
and where woud I use this u?
All this has done is measure x from a different point on the plane. Or to put it another way, we found the equilibrium position, which was not x=0. It is u=0.
squareroot
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#38
Dec18-12, 09:27 AM
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well yea, but what do i do with this u? and why is the equillibrium position at u=0?
haruspex
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#39
Dec18-12, 03:16 PM
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Quote Quote by squareroot View Post
well yea, but what do i do with this u?
You don't have to anything with it. Reformulating in terms of this u allows you to recognise the equation as being an exact fit for SHM. That allows you to figure out the frequency and amplitude in the usual way. Your problem was that extra constant term in the acceleration; rewriting it with u got rid of that.
and why is the equillibrium position at u=0?
Because that is where the acceleration will be 0.
squareroot
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#40
Dec18-12, 11:30 PM
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got it! but where do i use this u in equations?that s what i can t figure out.
squareroot
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#41
Dec18-12, 11:38 PM
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and why if i write it u like that, why does it the resemble a smh?
squareroot
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#42
Dec19-12, 12:09 AM
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what i want to say is how do i find k? with this u?
haruspex
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#43
Dec19-12, 01:05 AM
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Measuring x from the top you had:
F=mgsin(β)-b*x*mgcos(β)
which looks like SHM except for the constant acceleration term, mgsinβ.
But we note that this makes F, and hence the acceleration, 0 when x = tan(β)/b. So that must be the equilibrium point.
If we measure the distance from x = tan(β)/b instead, i.e. distance = u = x - tan(β)/b, we get F = -b*u*mgcos(β)
Since u is just a constant different from x, the acceleration of x is the same as the acceleration of u, so this is now recognisably SHM, and you can use your standard knowledge about SHM in relation to frequency, amplitude... The only thing to remember is that u is not measuring distance from the top.
squareroot
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#44
Dec27-12, 06:47 AM
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Ok, so now i can say that k=-b*mgcos(β) so that period T=2π√(m/k), T=2π√(m/-bmgcos(β)) , T=2π√1/(-bgcos(β)) , and the frequency is 1/T.

But how do i get from this equations to those which help me find the actual distance and time that i need?
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#45
Dec27-12, 02:35 PM
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Quote Quote by squareroot View Post
Ok, so now i can say that k=-b*mgcos(β) so that period T=2π√(m/k), T=2π√(m/-bmgcos(β)) , T=2π√1/(-bgcos(β)) , and the frequency is 1/T.

But how do i get from this equations to those which help me find the actual distance and time that i need?
The object starts at rest and we want the time until it is at rest again. Whether we measure distance as x or u does not affect the speed, so it's the time between successive points of zero speed in the SHM equation. How does that relate to the period of oscillation?
Similarly, amplitude is half the difference between extremes of position, and for that it won't matter whether we use x or u, so it's the amplitude you see in the SHM equation.


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