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Dynamics

2

A particle moves on the surface of a smooth cone z=r which has its axis vertical. Initially, the particle is at height z=a and its velocity is horizontal, speed v, in the θ direction.
By using ideas of angular momentum and energy, deduce that

z + a²v²/(2z²) + gz = v²/2 + ga where z is z dot (differentiated wrt t)

and show that the particle remains between two heights, which can be determined.

Thanks for any help/guidance you can give me with this...am really struggling...

Cheers

Reply 1

mikeyT

2

lgs98jonee

A particle moves on the surface of a smooth cone z=r which has its axis vertical. Initially, the particle is at height z=a and its velocity is horizontal, speed v, in the θ direction.
By using ideas of angular momentum and energy, deduce that

z + a²v²/(2z²) + gz = v²/2 + ga where z is z dot (differentiated wrt t)

and show that the particle remains between two heights, which can be determined.

Thanks for any help/guidance you can give me with this...am really struggling...

Cheers

I recommend that you write this on some paper with the dots in the right places first before reading through it properly.

I'll use R to be the vector r differentiated wrt t and θ to be dθ/dt.

So we know that r²θ=constant.
And at t=0, we know z=r=a and rθ=av [transverse velocity]
So r²θ=av

Now use the energy equation:
Kinetic energy + Potential Energy = constant
Generally:
½mR² +mgh = constant (where m is the mass of the particle, h is the height of the particle above 0)

But in this situation:
R=v and h=a

Hence we have:
½mv² + mga =constant

Equate the two energy equations to get:
½mR² +mgh =½mv² + mga

But R²= r²+r²θ²+z²
Sub this into the above equation and start fiddling around and realising z=r and you will arrive at the answer.

I'll post the next part in a few minutes.

Reply 2

lgs98jonee

OP

2

m277

I recommend that you write this on some paper with the dots in the right places first before reading through it properly.

I'll use R to be the vector r differentiated wrt t and θ to be dθ/dt.

So we know that r²θ=constant.
And at t=0, we know z=r=a and rθ=av [transverse velocity]
So r²θ=av

So r²θ=av...do you mean a²v (not trying to be picky..just checking)

m277

But R²= r²+r²θ²+z²

Is r=R...should I know this equation? Or is it a sequnce in your working? Thanks alot for this btw....v helpful

Reply 3

mikeyT

2

lgs98jonee

So r²θ=av...do you mean a²v (not trying to be picky..just checking)

Is r=R...should I know this equation? Or is it a sequnce in your working? Thanks alot for this btw....v helpful

The first point is r²θ=av because rθ=v and r=a so r(rθ)=r²θ=av.

The second point. R is the vector so is has the three directional components.
--------------
Multiply the equation by 2z² to get rid of the ½ and z² on the denominator of some of the terms.

Then rearrange to get all the z derivative terms on one side.
So now you have a function for z in terms of z. Let this be f(z)
So if the particle moves up or down then z will change as well. Can you see why?

So differentiate f(z), and you will see f'(a)>0 for a condition and f'(a)<0 for another condition, which shows that when the particle moves from a it either falls or rises depending on the initial condition.

Reply 4

lgs98jonee

OP

2

Ok, i still cant get it

I take
½mR²+mgh=½mv²+mga
so ½R²+gh=½v²+ga
and then sub in for R²? to give the following?

½(r²+r²θ²+z²)+gh=½v²+ga
so ½r²+½zav+½z².....but this doesnt seem to be going in the right direction :frown:

Reply 5

lgs98jonee

OP

2

ok ignire this...too tired :P

Reply 6

lgs98jonee

OP

2

could you differentiate imm and get

2 z. z: - a²v²/z³+gz. where z.=z got and z: = z double dot

2z:-a²v²/z³+g=0

and then carry on from there?

Reply 7

mikeyT

2

lgs98jonee

could you differentiate imm and get

2 z. z: - a²v²/z³+gz. where z.=z got and z: = z double dot

2z:-a²v²/z³+g=0

and then carry on from there?

You don't want a z: term in here because that will tell you nothing about where the particle will go.

Reply 8

lgs98jonee

OP

2

Isnt f'(z)=z:? since f(z)=z.

If not...what was I supposed to differentiate to?

Reply 9

mikeyT

2

Forget about what I said about f(z), its not really relevant to this question. I only wrote it because I thought it had something to do with it. (sorry, I've only had 2 hours of sleep today)

z + a²v²/(2z²) + gz = v²/2 + ga
Multiply the above by 2z²
Get: (using z. as dz/dt)

2z²z.² +a²v² +2gz³ = v²z² +2gaz²

Rearranging to get the z. term on its own:
2z²z.²= z²v² + 2gaz² -2gz³ -a²v²
2z²z.²=(z-a)(zv²+av²-2gz²)
The LHS is positive.
So if z<a then z-a<0 so zv²+av²-2gz²<0 also, hence you can solve this quadratic to find two values of z that the particle will remain between.

If z>a then zv²+av²-2gz²>0 and similarly you can find two values of z again.

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