I am disgusted today

Scroll to see replies

This problem reminds me of polar coordinates problems. Here the only force is radial (tension in spring), and so the transverse component of the acceleration is zero, and the radial acceleration can be linked to the force by F = ma. Angular momentum conservation may also come in handy?

Reply 61

atsruser

Original post by A Slice of Pi

You're right. See my half-baked back-of-envelope solution.

Reply 62

TeeEm

Original post by atsruser

Set up a rectangular coord system with the spring aligned along the x-axis at

t=0

. Let the particle have polar coords

r, \theta

. Then the Lagrangian of the system is:

Unparseable latex formula:

L=T-V = \frac{1}{2}m(\dot{r}^2 + r^2\dot{\theta}^2)-\frac{1}{2}\lamda(r-l)^2

with associated Euler-Lagrange equations

\frac{d}{dt}(\frac{\partial L}{\partial \dot{q}}) - \frac{\partial L}{\partial q} = 0

for each

q \in \{ r,\theta \}

giving:

\frac{d}{dt}(m \dot{r}) -mr \dot{\theta}^2 + \lambda(r-l)=0 \Rightarrow m\ddot{r}= mr \dot{\theta}^2 +\lambda(l-r)

\frac{d}{dt}(m r^2 \dot{\theta}) = 0 \Rightarrow m r^2 \dot{\theta} = C

where the second eqn expresses conservation of AM.

So we have:

m\ddot{r}=C+\lambda(l-r) = K -\lambda r

with

K=C+\lambda l

Now write

\rho =\frac{\lambda r-K}{\lambda}

to give

m \ddot{\rho} = -\lambda \rho

which is SHM in

\rho

. I got bored at this point, so I'll let someone else finish it off/correct my errors.

I did mentioned earlier that it can be done with Lagrangian/Hamiltonian mechanics but it can be done with A level methods.
I am waiting for the "comoving_frame" to show his method ... as he made a rapid exit ...
(by the way I need no help ... I have already written a full solution as this is a question to go into my example books)

Reply 63

TeeEm

Original post by A Slice of Pi

correct route

Reply 64

A Slice of Pi

Original post by TeeEm

correct route

Is this the route you tried?

Reply 65

john2054

stop trying to cheat. if you can't do the work, you shouldn't be taking the exam!

Reply 66

username2176541

Original post by john2054

stop trying to cheat. if you can't do the work, you shouldn't be taking the exam!

What exam?

Reply 67

atsruser

Original post by TeeEm

I decided that it was time to introduce the youngsters to a bit of real mechanics.

Reply 68

TeeEm

Original post by A Slice of Pi

Is this the route you tried?

I tried ....?!

I made.
I solved in retrospect that is why I put some numbers after as my solution was all in variables

Reply 69

spotify95

Original post by Mathemagicien

Joined today, knows about TeeEm's booklets, very rude to TeeEm... clearly duplicate account of some troll

If it's a dupe of a troll then it needs reporting.

Reply 70

TeeEm

Original post by spotify95

If it's a dupe of a troll then it needs reporting.

It somebody with "some mathematical knowledge" which know the maths room well and my personal activity.
He has done nothing to be reported but I think I know who he is (usernamewise)

Reply 71

Krollo

Is there a way of doing this with polar coordinates? I might try, and inevitably fail, tonight

Posted from TSR Mobile

Reply 72

TeeEm

Original post by Krollo

Is there a way of doing this with polar coordinates? I might try, and inevitably fail, tonight

Posted from TSR Mobile

That is the way ...
I will post full solution before I retire tonight.

Reply 73

john2054

Original post by Xenon17

What exam?

or the coursework.

the fact is that this kid is clearly out of his depth, and essentially with no insight in to the subject matter, is asking us to answer the question for him.

this is cheating

Reply 74

username2176541

Original post by john2054

or the coursework.

the fact is that this kid is clearly out of his depth, and essentially with no insight in to the subject matter, is asking us to answer the question for him.

this is cheating

you think Teeems cheating?

Reply 75

TheBBQ

Original post by A Slice of Pi

Am I missing something on why we can completely omit weight?

Reply 76

john2054

I just don't like it when people come on here and say 'please tell me the answer to this question'. It is true this individual may well be genuine given the insight he has subsequently showed in to the subject matter. But i don't think that it hurts to question their credentials, when they come on here (tsr) doing this. he wouldn't be the first!!

Reply 77

Shumaya

Original post by john2054

shhh

Reply 78

TeeEm

Original post by john2054

or the coursework.

the fact is that this kid is clearly out of his depth, and essentially with no insight in to the subject matter, is asking us to answer the question for him.

this is cheating