Hey there! Sign in to join this conversationNew here? Join for free
    • Thread Starter
    Offline

    18
    ReputationRep:
    Far easier than the first ....

    Name:  2.jpg
Views: 206
Size:  49.8 KB
    Offline

    11
    ReputationRep:
    (Original post by TeeEm)
    Far easier than the first ....

    Name:  2.jpg
Views: 206
Size:  49.8 KB
    Are you sure that this is right? I get an extra term of \frac{J^2}{m^2r^2} where J is the conserved AM.
    • Thread Starter
    Offline

    18
    ReputationRep:
    (Original post by atsruser)
    Are you sure that this is right? I get an extra term of \frac{J^2}{m^2r^2} where J is the conserved AM.
    I hope so but these days I can never be sure

    this is

    conservation of AM
    shape constraint
    energy = constant
    Offline

    11
    ReputationRep:
    (Original post by TeeEm)
    I hope so but these days I can never be sure

    this is

    conservation of AM
    shape constraint
    energy = constant
    This is a rush job, so may be errors.

    The initial AM J=maU.

    Use cylindrical polars. We have z=f(r). Taking z and V(z)=mgz increasing upwards, and V(0)=0, we have

    L = \frac{1}{2}m(\dot{r}^2 + r^2 \dot{\theta}^2 + \dot{z}^2)-mgz

    and E-L equations for (r,\theta,z):

    (1) \frac{d}{dt}(m\dot{r}) - mr\dot{\theta}^2=0 \Rightarrow \frac{1}{2}\frac{d}{dr}(\dot{r}^  2)- r\dot{\theta}^2=0

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

    (3) \frac{d}{dt}(m\dot{z}) + mg = 0 \Rightarrow \frac{m}{2}\frac{d}{dz}(\dot{z}^  2) + g = 0 \Rightarrow \dot{z}^2 + 2gz = C_1

    Now \dot{z} = \frac{d}{dt} f(r) = f'(r) \frac{dr}{dt} = \dot{r} f'(r) so:

    \dot{r}^2 (f'(r))^2 + 2gf(r) = C_1 (4)

    Substituting (2) into (1) gives:

    \frac{d}{dr}(\dot{r}^2)- \frac{2J^2}{m^2r^3}=0 \Rightarrow \dot{r}^2 + \frac{J^2}{m^2 r^2} =C_2 (5)

    and adding (4) and (5) gives:

    \dot{r}^2(1+[f'(r)]^2) + 2gf(r) +\frac{J^2}{m^2 r^2}  = K
    • Thread Starter
    Offline

    18
    ReputationRep:
    (Original post by atsruser)
    This is a rush job, so may be errors.

    The initial AM J=maU.

    Use cylindrical polars. We have z=f(r). Taking z and V(z)=mgz increasing upwards, and V(0)=0, we have

    L = \frac{1}{2}m(\dot{r}^2 + r^2 \dot{\theta}^2 + \dot{z}^2)-mgz

    and E-L equations for (r,\theta,z):

    (1) \frac{d}{dt}(m\dot{r}) - mr\dot{\theta}^2=0 \Rightarrow \frac{1}{2}\frac{d}{dr}(\dot{r}^  2)- r\dot{\theta}^2=0

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

    (3) \frac{d}{dt}(m\dot{z}) + mg = 0 \Rightarrow \frac{m}{2}\frac{d}{dz}(\dot{z}^  2) + g = 0 \Rightarrow \dot{z}^2 + 2gz = C_1

    Now \dot{z} = \frac{d}{dt} f(r) = f'(r) \frac{dr}{dt} = \dot{r} f'(r) so:

    \dot{r}^2 (f'(r))^2 + 2gf(r) = C_1 (4)

    Substituting (2) into (1) gives:

    \frac{d}{dr}(\dot{r}^2)- \frac{2J^2}{m^2r^3}=0 \Rightarrow \dot{r}^2 + \frac{J^2}{m^2 r^2} =C_2 (5)

    and adding (4) and (5) gives:

    \dot{r}^2(1+[f'(r)]^2) + 2gf(r) +\frac{J^2}{m^2 r^2}  = K
    shouldn't the r2 in the extra term be a2
    Offline

    11
    ReputationRep:
    (Original post by TeeEm)
    shouldn't the r2 in the extra term be a2
    Not as far as I can see. a,U have already been absorbed into J, so they won't appear again.
    • Thread Starter
    Offline

    18
    ReputationRep:
    (Original post by atsruser)
    Not as far as I can see. a,U have already been absorbed into J, so they won't appear again.
    I cannot see anything wrong with you ... maybe I got it wrong
    will check my solution before I post
    • Thread Starter
    Offline

    18
    ReputationRep:
    (Original post by atsruser)
    Not as far as I can see. a,U have already been absorbed into J, so they won't appear again.
    I found what is happening we are both correct ...
    Stupidity on my part ...
    the extra bit is a function or r, which I absorbed into the potential, rellabelling as f(r), forgetting I had already used it to describe the equation of the curve
    • Thread Starter
    Offline

    18
    ReputationRep:
    my solution
    Name:  2.jpg
Views: 115
Size:  36.7 KB
    Attached Images
     
 
 
 
  • See more of what you like on The Student Room

    You can personalise what you see on TSR. Tell us a little about yourself to get started.

  • Poll
    Brussels sprouts
    Useful resources

    Make your revision easier

    Maths

    Maths Forum posting guidelines

    Not sure where to post? Read the updated guidelines here

    Equations

    How to use LaTex

    Writing equations the easy way

    Student revising

    Study habits of A* students

    Top tips from students who have already aced their exams

    Study Planner

    Create your own Study Planner

    Never miss a deadline again

    Polling station sign

    Thinking about a maths degree?

    Chat with other maths applicants

    Can you help? Study help unanswered threads

    Groups associated with this forum:

    View associated groups
  • See more of what you like on The Student Room

    You can personalise what you see on TSR. Tell us a little about yourself to get started.

  • The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

    Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

    Quick reply
    Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.