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# Interesting mechanics question watch

1. "A firework is shot into the air (vertical) when at a point in its trajectory when it stops moving. It explodes spewing aprticles with speeds "u".

At a later stage show that it is a sphere.

Describe the motion and the sphere's radius."

Thats all I've been given, I'm guessing since we know the inital speed of expansion is "u" we could work out the dV/dr of the sphere? I'm not really sure.

I'm guessing it is not a sphere, more of an oblate sphereoid.
2. ignoring gravity, it will be a sphere, with its centre stationary, and dr/dt = u => r = ut

with gravity, it will still be a sphere, but its centre will drop in free fall, and its radius will still be r=ut
3. (Original post by elpaw)
ignoring gravity, it will be a sphere, with its centre stationary, and dr/dt = u => r = ut

with gravity, it will still be a sphere, but its centre will drop in free fall, and its radius will still be r=ut
I don't get your thinking process...
4. (Original post by Hexa)
I don't get your thinking process...
Maybe 'cos he's bordering on genius! And the rest of us... well, we're just mere mortals!
5. It wont be a sphere. Gravity will distort the shape of the would-be sphere.
6. Thats true.
7. (Original post by Ralfskini)
It wont be a sphere. Gravity will distort the shape of the would-be sphere.
Thats what I thought, but my teacher says you must prove otherwise!
8. I think it will be a sphere.

Consider a couple of particular cases - particles shooting out horizontally and particles shooting out vertically.

Horizontally
=======
Let one particle move left at u m/s
Let another particle move right at u m/s
Then the equator of your "sphere" will be expanding at,
2u m/s
====

Vertically
======
Let one particle move vertically upwards at u m/s. call this velocity u1.
Let another particle move vertically downwards at u m/s. call this velocity u2.

u2 = u + gt (downwards)
u1 = u - gt (upwards) = -u + gt (downwards)

Relative velocity between upward and downward moving particles is,

u2 - u1= (u + gt) - (-u + gt)
u2 - u1 = 2u
========

i.e. the north and south poles of your "sphere" are moving apart at 2u m/s - exactly the same as the equator.
Therefore, the "sphere" is expanding uniformly and is actually a spherical sphere, rather than an oblate sheroid.
And, of course, the centre of this expanding sphere is moving vertically downwards at 32 ft/s²
9. (Original post by Fermat)
I think it will be a sphere.

Consider a couple of particular cases - particles shooting out horizontally and particles shooting out vertically.

Horizontally
=======
Let one particle move left at u m/s
Let another particle move right at u m/s
Then the equator of your "sphere" will be expanding at,
2u m/s
====

Vertically
======
Let one particle move vertically upwards at u m/s. call this velocity u1.
Let another particle move vertically downwards at u m/s. call this velocity u2.

u2 = u + gt (downwards)
u1 = u - gt (upwards) = -u + gt (downwards)

Relative velocity between upward and downward moving particles is,

u2 - u1= (u + gt) - (-u + gt)
u2 - u1 = 2u
========

i.e. the north and south poles of your "sphere" are moving apart at 2u m/s - exactly the same as the equator.
Therefore, the "sphere" is expanding uniformly and is actually a spherical sphere, rather than an oblate sheroid.
And, of course, the centre of this expanding sphere is moving vertically downwards at 32 ft/s²

But havent you neglected gravity in your horizontal equations?
10. (Original post by Fermat)
the centre of this expanding sphere is moving vertically downwards at 32 ft/s²
What the...?
11. The sphere is subject to a uniform radial expansion of u m/s, and an acceleration of g m/s² in the vertically downwards direction.
12. (Original post by Ralfskini)
But havent you neglected gravity in your horizontal equations?
gravity doesn't affect the horizontal components, only the vertical ones.
The horizontal velocity will remain constant until the particle hits something.
13. (Original post by Fermat)
gravity doesn't affect the horizontal components, only the vertical ones.
The horizontal velocity will remain constant until the particle hits something.
Surely if you shoot an object horizontally, we see gravity having an affect? By pulling the particle dowanwards? And such as it would be with the sphere's equatorial objects as well?
14. A sphere is formed if the distance between where the rocket explodes and each particle is constant for each value of t.

d=sqrt(u^2t^2-ut^3gsinx+g^2t^4/4), which is clearly dependant on the angle x of projection from the horizontal of each particle, so the distance at time t relative to the origin is not constant.
15. (Original post by Fermat)
gravity doesn't affect the horizontal components, only the vertical ones.
The horizontal velocity will remain constant until the particle hits something.

I know. But what I mean is no particles will move horizontally because they are all under the influence of gravity so will follow a curved path of trajectory.
16. (Original post by Hexa)
Surely if you shoot an object horizontally, we see gravity having an affect? By pulling the particle dowanwards? And such as it would be with the sphere's equatorial objects as well?

thankyou, that's what I meant.
17. (Original post by Hexa)
Surely if you shoot an object horizontally, we see gravity having an affect? By pulling the particle dowanwards? And such as it would be with the sphere's equatorial objects as well?
But it wont affect the horizontal components only the vertical components, so the pieces moving horizontally will fall downwards with velocity
v = gt
but will continue in the horizontal direction with the same speed.
Since all the particles are affected the same by gravity, they will all move relative to centre which itself is falling and it wont distort the shape.
18. (Original post by Hexa)
Surely if you shoot an object horizontally, we see gravity having an affect? By pulling the particle dowanwards? And such as it would be with the sphere's equatorial objects as well?
Everything has a horizontal and vertical component.
gravity only acts in the vertical dirn. so only affects the vertical component. The resultant of a constant horizontal component and increasing vertical component results in a parabolic trajectory towards the earth. But the horizontal velocity is still constant.
19. he is saying that you split each velocity up into horizontal and vertical components.

if there is no gravity you get a sphere (this is obvious)

now if there is gravity each particle will have the same accelleration towards the ground, and the horizontal components wont be affected, so you can imagine it just like a ball under gravity, where the surface of the ball is the particles.
20. But the strange problem was is that my teacher asked us to *prove* that it *was* a sphere?

So who is right here?

And Fermat, I hope the 32 ft/s^2 was in jest. Because your brainpower leaves me dazzled.

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