You are Here: Home >< Maths

# M7 Mechanics Question 1 Watch

1. (Original post by zetamcfc)
D3+ would start to get interesting actually
Omg triggered

Posted from TSR Mobile
2. outer surface....do you mean the external one? Are the available forces adequate for such a motion? I am missing sth.
3. (Original post by drandy76)
Omg triggered

Posted from TSR Mobile
C'mon you could do some proper graph theory and other stuff, it would be fun.
4. OK, I'm struggling to finish this off at the moment but my approach was as follows:

Spoiler:
Show

1. When the particle has fallen to a position where the cone radius is , we have the horizontal velocity by conservation of AM.

2. Now resolving perpendicular to the surface of the cone gives

where is the instantaneous radius of curvature of the cone along the plane defined by the normal and the direction of i.e. the horizontal. But this plane is an ellipse, and the particle is at the vertex, so we need the radius of curvature of this ellipse at the vertex. That can be shown to be where are the semi-major, minor axes.

Now I've found in terms of (I think) but is giving me huge grief, so at the moment, I'm stymied with this approach. Note that it will agree with the stated result if , which I thought I'd shown last night, but I'd made a mistake.
5. come on exam boards bring on M8 if you think you're hard enough

6. Taking the origin at the vertex and z increasing downwards

there are 4 equations
1. transversly conservation of angular momentum
2. radially (usual polar acceleration = Rcos(...)
3. r = ztan (...) which can be differentiated and used where needed
4. mz (double dot) = mg - Rsin (...)
7. (Original post by zetamcfc)
C'mon you could do some proper graph theory and other stuff, it would be fun.
Noooo, I barely tolerated all the definitions and simplex iterations last year, no more

Posted from TSR Mobile
8. Have a look, it is close enough but...
Attached Images
9. not right.pdf (848.7 KB, 68 views)
10. (Original post by TeeEm)

mz (double dot) = mg - Rsin (...)
This is where I'm having problems. I've drawn a diagram of the situation as described in the question, with the particle on the outer surface of the inverted cone. In my diagram, mz (double dot) is not mg - Rsin (...) but rather
-mg - Rsin (...) (i.e. both are minus and not plus and minus, if we take the downward to be positive then both terms will be positive). Can someone clear this up? I currently have the correct answer except the two terms have the same sign and not different signs.
Attached Images

11. (Original post by A Slice of Pi)
This is where I'm having problems. I've drawn a diagram of the situation as described in the question, with the particle on the outer surface of the inverted cone. In my diagram, mz (double dot) is not mg - Rsin (...) but rather
-mg - Rsin (...) (i.e. both are minus and not plus and minus, if we take the downward to be positive then both terms will be positive). Can someone clear this up? I currently have the correct answer except the two terms have the same sign and not different signs.
the cone is the other way round
12. (Original post by TeeEm)
the cone is the other way round
That would make more sense, but in the question wasn't it called an 'inverted' cone?
13. (Original post by A Slice of Pi)
That would make more sense, but in the question wasn't it called an 'inverted' cone?
I am teaching at present to have a discussion but the cone in that orientation is inverted
14. (Original post by TeeEm)
I am teaching at present to have a discussion but the cone in that orientation is inverted
Well, I think it's definitely debatable. I'd maybe include a diagram in questions like this
Attached Images
15. solution2.pdf (191.2 KB, 68 views)
16. (Original post by depymak)
Have a look, it is close enough but...
Your method is very good . I think its actually the angle 'alpha' in the diagram that is incorrectly labelled, which is a shame because the rest of it is perfectly correct
17. My solution
Attached Images

18. thanks, I misinterpreted the term "semi-vertical angle " !
(Original post by A Slice of Pi)
Your method is very good . I think its actually the angle 'alpha' in the diagram that is incorrectly labelled, which is a shame because the rest of it is perfectly correct
19. My solution:

We locate the particle on the cone of semi-vertical angle via cylindrical polar coords , with increasing downwards. The initial angular momentum of the particle about the z-axis is and the motion of the particle is subject to the constraint:

Its KE, , and PE, , are given by:

taking to be at the vertex of the cone, and decreasing downwards.

The Euler-Lagrange equations for a particle with generalised coords subject to a holonomic constraint are:

where and where we have . Also, the force of constraint conjugate to is:

and these forces act in the direction of increasing

So for this problem, the E-L equations give (details omitted):

(1)

(2) where is the conserved angular momentum.

(3)

Now using (2) to eliminate from (1) gives:

(4)

and using the equation of constraint to eliminate from (3) gives:

(5)

So (4) and (5) give us, after some algebra:

So the constraint force conjugate to is:

But this is the constraint force in the direction of increasing , and is thus the horizontal component of the normal reaction so we have:

as required.

Note that my other approach would have been much quicker if it wasn't for the fact that I couldn't find the radius of curvature of the ellipse in question. So while the workings above represent another triumph for M. Lagrange, I suspect that my other method was more Newtonian in character, and probably could have been completed easily by Mr Newton, who would have known all about ellipses.
20. nice!
(Original post by TeeEm)
My solution
21. (Original post by atsruser)
....
Lagrangian considerations make it at least M9 ....
22. (Original post by TeeEm)
Lagrangian considerations make it at least M9 ....
Well, I couldn't finish my quick 'n dirty method, and other people had done it the obvious way, so I had little choice but to bring out the big guns. Maybe hides the physical insight though, compared to getting down and dirty with the forces.

TSR Support Team

We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.

This forum is supported by:
Updated: March 9, 2016
Today on TSR

### 'Entry requirements are a form of elitism'

What do you think?

### I fancy Donald Trump...

Discussions on TSR

• Latest
• ## 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
Useful resources

### Maths Forum posting guidelines

Not sure where to post? Read the updated guidelines here

### How to use LaTex

Writing equations the easy way

### Study habits of A* students

Top tips from students who have already aced their exams

## Groups associated with this forum:

View associated groups
Discussions on TSR

• Latest
• ## 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