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Loren Booda

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Loren Booda

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arcnets

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I'm sure you're thinking of something else, so please be more specific.

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Loren Booda

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ahrkron

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What do you mean with that?

Why "potentially"?

You need to give more context. Where would the limit come from? Are you thinking relativity? How should friction be considered?

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russ_watters

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Loren Booda

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russ_watters

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Forward speed and rotational speed of a thrown ball are completely independent of each other.

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arcnets

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Loren,

are you maybe thinking of a ball being accelerated by just one force that acts tangentially upon its perimeter, so moving & (at the same time) spinning it? Without taking into account air friction or gravity? I think, this may be easily answered.

Could this help, if we use 'sphere' instead of 'cylinder':

https://www.physicsforums.com/showthread.php?s=&threadid=4385

are you maybe thinking of a ball being accelerated by just one force that acts tangentially upon its perimeter, so moving & (at the same time) spinning it? Without taking into account air friction or gravity? I think, this may be easily answered.

Could this help, if we use 'sphere' instead of 'cylinder':

https://www.physicsforums.com/showthread.php?s=&threadid=4385

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Loren Booda

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russ_watters

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There is a theoretical minimum spin:velocity ratio, but not a theoretical maximum for this case.

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arcnets

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Probably not. Because when spinning the ball by a single force, you will always have a tangential component of the accelerating force, thus making the ball move.Originally posted by Loren Booda

First, is it possible to spin an otherwise freely "thrown" ball (released at one point on its surface) without causing it to move linearly,

Remember Apollo days: Spaceships had always 2 jets located at opposite positions wrt. the center of mass (and in opposite directions), which were fired simultaneously to make the ship rotate. Yes you could rotate the ship with just one jet, but that would cause linear acceleration, too. I think they knew why they had to install 2 jets, even though that was more expensive.

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Loren Booda

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Can you prove the latter assertion?There is a theoretical minimum spin:velocity ratio, but not a theoretical maximum for this case.

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russ_watters

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Yeah, but frankly it would take me more effort than I feel like right now. If you've already looked at the equations though, you can easily figure it out for yourself. I'll get you started:Originally posted by Loren Booda

russ_watters Can you prove the latter assertion?

The lower the moment of inertia for a given mass, the faster something will spin, right? Thats kinda the definition of moment of inertia. Moment of inertia depends on mass distribution and there is no theoretical limit on how you distribute the mass. You can never quite get to zero moment of inertia, but you can get as close as you are willing to take the effort to get. Which means infinite rotational rate. And the linear speed for a given mass and radius will always be the same (assuming you mean IMPULSE, not just force).

So its a limit problem - derive or find the equation relating moment of inertia to rotational rate and take the limit as I -> 0.

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arcnets

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Yes sure, but in the initial post, Loren said 'ball of constant density'.Originally posted by russ_watters

How much spin and how much forward motion you get can still vary infinitely

I think this can be done if we treat it as an elastic collision problem.

Let's assume that a particle of mass M and velocity V is fired away tangentially from the ball's surface (much like in my Apollo example, for the 1-jet case). Assume that M << m where m is the ball's mass.

Initial energy: E

Initial momentum: p

Final energy: E

Final momentum: p

For a full sphere, I = 2/5 m r

so

E

If I am not mistaken, there are 2 unknowns: v, w.

And 2 Equations:

E

p

So it should be possible, right?

- #15

Loren Booda

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- #16

arcnets

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Loren,

I think I found out how to do this. Let's consider the same process (particle fired away tangentially). I think it should be treated as*inelastic*. Linear and angular momenta are conserved, we don't care about energy.

Thus...

p_{i} = 0

p_{f} = MV - mv

J_{i} = 0

J_{f} = MVr - Iw

Let's assume I = Cmr^{2} with some constant C.

Thus...

I. MV - mv = 0

II. MVr - Cmrw = 0

I. V = mv/M (into II)

II. mvr - Cmr^{2}w = 0 (m cancels. Good!)

II. v = Crw, or

<drumroll> rw/v = 1/C.

That is the desired ratio, isn't it?

Now, for a full sphere, C = 2/5, so rw/v = 2.5

The maximum C value is C = 1 (for a ring). So the minumum ratio is 1.

There is no minimum C value, so there is no maximum for the ratio. As russ_watters already said.

OK?

I think I found out how to do this. Let's consider the same process (particle fired away tangentially). I think it should be treated as

Thus...

p

p

J

J

Let's assume I = Cmr

Thus...

I. MV - mv = 0

II. MVr - Cmrw = 0

I. V = mv/M (into II)

II. mvr - Cmr

II. v = Crw, or

<drumroll> rw/v = 1/C.

That is the desired ratio, isn't it?

Now, for a full sphere, C = 2/5, so rw/v = 2.5

The maximum C value is C = 1 (for a ring). So the minumum ratio is 1.

There is no minimum C value, so there is no maximum for the ratio. As russ_watters already said.

OK?

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russ_watters

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Oops. In that case, there is only ONE ratio I think.Originally posted by arcnets

Yes sure, but in the initial post, Loren said 'ball of constant density'.

- #18

Loren Booda

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1. Is there no minimum C (or moment of inertia) value in general? Perhaps all I[>=]mr

2. Aren't we talking here about the constant density sphere with C=2/5?

3. With fixed r

- #19

arcnets

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2. I think, yes. A ball of constant density, as you said.

3. I holds when the accelerating force acts tangentially. You could of course hit the ball more 'head-on', thus causing less spin and more linear motion.

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