M7 Mechanics Question 1

Original post by zetamcfc

D3+ would start to get interesting actually :smile:

Omg triggered

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Reply 81

stochasym

12

outer surface....do you mean the external one? Are the available forces adequate for such a motion? I am missing sth.

Reply 82

zetamcfc

Original post by drandy76

Omg triggered

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C'mon you could do some proper graph theory and other stuff, it would be fun.

Reply 83

atsruser

11

Original post by TeeEm

stuck
@comoving_frame

Motion on a cone.jpg

OK, I'm struggling to finish this off at the moment but my approach was as follows:

Spoiler

(edited 8 years ago)

Reply 84

the bear

20

come on exam boards bring on M8 if you think you're hard enough

Reply 85

OP

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 (...)

Reply 86

drandy76

inverted cone outer surface.png4.8KB

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

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Reply 87

stochasym

12

Have a look, it is close enough but...

(edited 8 years ago)

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.

Reply 89

OP

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

Reply 90

A Slice of Pi

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?

Reply 91

OP

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

Reply 92

A Slice of Pi

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

Original post by depymak

Have a look, it is close enough but...

Your method is very good :smile:

. 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

Reply 94

OP

My solution
Motion on a cone.jpg

thanks, I misinterpreted the term "semi-vertical angle " !

Original post by A Slice of Pi

Your method is very good :smile:

. 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

Reply 96

atsruser

11

Original post by TeeEm

stuck
@comoving_frame

Motion on a cone.jpg

My solution:

We locate the particle on the cone of semi-vertical angle

\alpha

via cylindrical polar coords

r, \theta, z

, with

z

increasing downwards. The initial angular momentum of the particle about the z-axis is

J =maU

and the motion of the particle is subject to the constraint:

f(r,z)=r-z\tan \alpha = 0

Its KE,

T

, and PE,

V

, are given by:

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

V = -mgz

taking

V=0

to be at the vertex of the cone, and

V

decreasing downwards.

The Euler-Lagrange equations for a particle with generalised coords

q_i

subject to a holonomic constraint

f(q_1, q_2, \cdots, q_n, t)=0

are:

\displaystyle \frac{d}{dt}\big(\frac{\partial L}{\partial \dot{q_i}}\big)-\frac{\partial L}{\partial q_i} -\lambda \frac{\partial f}{\partial q_i} =0

where

L=T-V

and where we have

q_i \in {r,\theta,z}

. Also, the force of constraint conjugate to

q_i

is:

Q_{q_i} = \lambda \frac{\partial f}{\partial q_i}

and these forces act in the direction of increasing

q_i

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

(1)

m\ddot{r} -mr\dot{\theta}^2-\lambda =0

(2)

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

where

J

is the conserved angular momentum.

(3)

m\ddot{z} -mg+\lambda \tan \alpha =0

Now using (2) to eliminate

\dot{\theta}

from (1) gives:

m\ddot{r} - \frac{J^2}{mr^3} -\lambda =0 \Rightarrow \lambda = m\ddot{r} - \frac{J^2}{mr^3}

(4)

and using the equation of constraint to eliminate

z

from (3) gives:

m\ddot{r} \tan \alpha -mg+\lambda \tan \alpha =0 \Rightarrow m\ddot{r} = \tan \alpha (mg -\lambda \tan \alpha)

(5)

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

\lambda = \cos^2 \alpha (mg \tan \alpha - \frac{J^2}{mr^3}) = m\cos^2 \alpha (g \tan \alpha - \frac{a^2U^2}{r^3})

So the constraint force conjugate to

r

is:

Q_r = \lambda \frac{\partial f}{\partial r} = \lambda

But this is the constraint force in the direction of increasing

r

, and is thus the horizontal component of the normal reaction

R

so we have:

Q_r = R \cos\alpha \Rightarrow R = \frac{Q_r}{\cos \alpha} = m\cos \alpha (g \tan \alpha - \frac{a^2U^2}{r^3})

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.

Reply 97

stochasym

12

nice!

Original post by TeeEm

My solution
Motion on a cone.jpg

Reply 98

OP