Earn BTC/LTC by helping me with physics coursework (BSC 1st year)

[newuser20]

I am a first year who has problems with attendance, I also have some spare LTC

1 LTC reward for help with my coursework - of course I don't expect the answer but a point in the right direction and/or helpful resources to find the solution. This could even be an ongoing thing.

So here's the question: http://imgur.com/QnnztcU
The problem i have is deciding whether particle A has potential energy - and how it actually moves when falling ie where is the oscillation
My answer so far would be that
T(a) = 1/2mr2 theta(dot)2 + 1/2 m r(dot)2
And V, its potential is zero.
T(b) = 0 V(b) = -mg(L-r)
I hope this is the right as im not exactly 100% on the question asking for polar coords?
Part b seems alright - just using d/dt(dL/dtheta(dot) = etc
Part c wasnt a problem
But how do i approach part d?

There is also a second question

http://imgur.com/RniiZju
T = 1/2m r(dot)2 + 1/2 m r2 theta(dot)2
Its the same with both linear kinetic energy and rotational energy
V = mgr(sin(omega*t)
And again part b isnt an issue
But part C i am unsure how to start
Part D is just an initial value problem?
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[kuroman]

Arghh the moment I looked at your problem my head started to hurt, not because it hard just reminded me of some vicious problem I used to solve when I was in grad school almost a decade ago, also it is in English which is annoying for me since the terms aren't similar to my language

Just to make sure we are in a non relativity context right?

To help you with the first question, does A has potential energy ? of course it does, but it all depends from on the origin of your basis, if you consider the origin as being the surface on which A is has the potential energy of A = 0 (since potential energy is related to hight or to Z) hope it helps you.

T(a) correct
V(a)=0 if you consider the origin as above (if the movement is linear which is most likely the case here teta(dot) = 0 so you have one term check this out)
T(b)= 1/2mz(dot)^2 (=0 in polar coordinates) as for V(b) isn't simply = -mgr why l-r ? like I said earlier it depends on your Z axis origin so I'm not sure what you took here, but the argumentation is correct on it self

as for d if I understand correctly they are asking for the Trajectory equation, If they are asking for the use of r0 I suppose the work should be done in polar coordinate, the difference here is that teta = omega x t (as we have a circular movement here) omega related to frequency?

[newuser20]

Thanks for getting back to me! I'm sorry if the notation is different to what you are used to.

Non-rel, yes. I'm first year so no trick questions atm.

So lets talk more about the potential - A's potential is zero at what point  ? Where is it maximum  ? Is it mgr(1-cos(theta)) ?

As for T(b) L-r is the length of string left before r has reached the 'plughole' which i assume is the point of zero potential ?

And why is it not simply -mgr? I assume the z axis is straight down, the way gravity is acting. Is it a case of a tension in the string opposing motion  ?  What else would it be ?

Could you provide a link to the trajectory equation or describe it to me ?

What did you think of question 2?

[kuroman]

Thanks for getting back to me! I'm sorry if the notation is different to what you are used to.

Non-rel, yes. I'm first year so no trick questions atm.

So lets talk more about the potential - A's potential is zero at what point  ? Where is it maximum  ? Is it mgr(1-cos(theta)) ?

As for T(b) L-r is the length of string left before r has reached the 'plughole' which i assume is the point of zero potential ?

And why is it not simply -mgr? I assume the z axis is straight down, the way gravity is acting. Is it a case of a tension in the string opposing motion  ?  What else would it be ?

Could you provide a link to the trajectory equation or describe it to me ?

What did you think of question 2?

I didn't look at A since it is 4am here in Europe I look at it tomorrow

The potential energy for A is either = 0 or a Constant, depending on the origin you consider (+/- mgh (h=height) if you consider the origin of the Z axis the table on which A is on and if we assimilated A to be a single point (Center of gravity with a coordinate Z=0) then potential energy of A = 0 (which is the simplist to do)

I considered r0=0 that's why i've got T(b)= mgr since when a moves by r, b also moves by r, but what you are doing is correct aswell, you consider at t (time) that the h= (l-r) if a moved by r I prefer this,  just be careful of variation sens.

tomorow if I have time we can discuss the rest ^^ gn

[newuser20]

This is in for tomorrow so im going to bump up the reward to 1 LTC or BTC equivalent provided the sheet gets finished  

I get that V(b) is constant (zero) for q1

Is T(b) going to be zero ? Might it be equal to the radial velocity of A ?



How is R(o) = 0 ? Surely the start position is where r = L ? Or maybe just (1-(r-L)) or whatever

[newuser20]

Question 2:
T = 1/2 m r(dot)2 + 1/2 m r2 theta(dot)2 V = mgr(sin(theta)) maybe sin(omega*t)) im not 100%
L = T-V
eq of motion mr2 theta'' = -mgr*cos(theta)
From there how do you sub to get an eqution of motion ? Ansatz? I take it ive made a mistake and B should be similar to my answer from the previous part?

http://imgur.com/vEoYSUf

[newuser20]

what i have so far for q1:

T(a) = 1/2mr2 theta(dot)2 + 1/2 m r(dot)2
V(a) = 0
T(b) = 0 or maybe 1/2 mr(dot)^2
V(b) = -mg(L-r), (L-r)=z for simplicity

L = mgz + 1/2 mr^2 O'^2  O = theta for simplicity, the ' notation is dot (derivative wrt time)

Eq for motion:

d/dt (dl/dO') = dl/dO

mr^2O'' = 0 ? ?

From part C

C.P.F = T(b)

mrw^2 = mgr (w=omega)  -- this was given to me by a friend and im not sure where it come from but it follows through:-

*mr^2:

m^2 r^4 w^2 = m^2 r^3 g

(mr^2w)^2 =

Iw^2 =

P(O)^2 = m^2 r ^3 g, where P(O) is angular frequency sometimes noted as L

P(O)^2 / m^2 g = r^3

r = r_o = answer on sheet

From this I find it natural to assume  my answer to part b was incorrect

For part d)

I think you set r = r_o + c then sub this back into the equation of motion? Is this correct ?

[newuser20]

Giving multiple rewards out for each considerable additional 

[kuroman]

Just got home and I'll be taking a look at it in the next hour or so after grabing something to eat and reply to some other threads, but lets answer your first reply:

a-

T(b) is different than Zero T(b) = 1/2 m r(dot)2 ( because 1/2mr2 x theta(dot)2= 0 since Theta = constant  thus Theta(dot)= 0 since we have an accelerated linear movement (no rotation) correct me if I misread or missed something here) r0=0 is possible it all depends on the origin you take for your basis, but lets use your method (since it is how your teacher do things ) lets say r0=L at t=0 so H=0 (Z0) at an instant t, A would have moved by L-r(t), and B would have moved the same right; so Ec= - mgh = -mg(l-r) I think we agree at this point, just verify the signs

so L is the difference I trust you with that, (just for reference and a fun fact in all the countries I've been they don't use Lagrangien but Mechanical Energy Em, Em= Ec (Kinetic Energy) + Ep (Potentiel Energy)  with that you can calculate Work (dEm=dW(F) and dEm/dt = P (power) )

[newuser20]

Ah yes,

the Lagrange is just an exercise on how to use it im sure. We did a little on power and work last term.

Check this out and see what you think (there is 4 images)

http://imgur.com/VEsrMK3,BJYb514,zGdXDY9,mRvsdOc#0

[newuser20]

q 2 + 3

http://imgur.com/pXMZhbf,YmqMuqW,KfWFwC3,938TTNj#0

[kuroman]

Ah yes,

the Lagrange is just an exercise on how to use it im sure. We did a little on power and work last term.

Check this out and see what you think (there is 4 images)

http://imgur.com/VEsrMK3,BJYb514,zGdXDY9,mRvsdOc#0

correct that's the general expression except that I think Theta'=0 as you have a linear movement as I explained above (theta = constant), is there other informations not shown in the text?

[newuser20]

Ah yes there is and that bit i wasn't sure about


The last bit just says oscillations with frequencies
w = sqrt(3/2)* P(o)/m*(r_o)2

P(o) is the angular frequency, normally noted as L

[kuroman]

q 2 + 3

http://imgur.com/pXMZhbf,YmqMuqW,KfWFwC3,938TTNj#0

a= Correct

b= Euler Lagrange? correct aswell

For c, I don't understand what they mean by substitution
but anyways you have a second order linear differential equation (w= omega) -w²r+ r''=gsin(wt) [<=> (-1/2)r + (1/2w²)r''=(g/2w²)sin(wt) useless for really just if you want to show B and when you want to solve for the particular solution you can use this form]
differential equation of second order:
Ok let's first find the solution for r"-w²r = 0
The solutions of this m²-w² =0  <=> m= -/+w
thus solution r(t)= A1e(-wt)+A2e(-wt)

Now we need a particular solution r(t)= Bsin(wt)
we replace in -w²r+ r''=gsin(wt) =>  -w²Bsin(wt)) -w²B(sin(wt)=gsin(wt) => -2w²Bsin(wt)= gsin(wt) hence B= -g/2w² (I have a - more than the question weird, I must forgoten a  - somewhere I trust you with the revision, it's not hard
Then I supose the substitution is the combination of both to get the generale solution which is the one written

for question d, it's easy, just apply the limit conditions given and find A1 and A2

And that's answer the second exercise

[kuroman]

Ah yes there is and that bit i wasn't sure about


The last bit just says oscillations with frequencies
w = sqrt(3/2)* P(o)/m*(r_o)2

P(o) is the angular frequency, normally noted as L

Ok so Phi is our Theta right?, if I consider question C, the Theta is not a constant since there is an angular momentum and thus we should keep the equation as you wrote it but wierd, since if we leave the system on it own it will have a linear momevement