Will a ball on a string spin around its own axis when released?

1. Jan 26, 2013

leviterande

Hello! I couldnt find any answer to a wondering I have:

On one end of a light string, attach a heavy ball, and on the other end of the string hold it with your hand and whirl around the ball-on-string. You reach a high rotational speed and finally the string breaks at the string-ball attachment point. I understand that the ball will now fly off in a straight line.

But will the ball also spin around its own axis while flying in a straight line?

2. Jan 26, 2013

Staff: Mentor

No. Easiest way to see this is to consider the torque on the ball from the string at the moment of release.

(This assumes a perfectly spherical ball, a smooth and instantaneous release, and no little tail of string hanging off the side of the ball after release. Change any of these assumptions and the problem gets way more complicated)

3. Jan 26, 2013

mrspeedybob

Why should the string exert a torque on the ball?

4. Jan 26, 2013

leviterande

Thanxm that is is a little strange, I am a bit confused, would the ball spin if the string had been cut at where your hand holds the string instead?

btw all videos of hammer throws show that the ball spining while flying out in a straight line

5. Jan 26, 2013

Staff: Mentor

It doesn't - that's why the ball doesn't pick up any spin.

6. Jan 26, 2013

256bits

Yes the ball does spin after release by considering there is no torque acting at the time of release and point of release.

7. Jan 26, 2013

Staff: Mentor

Yes, because as the ball went flying off it would have this string tied to it, and the air resistance on the string would rotate the ball.

A hammer, or a ball? A hammer is about as far from an ideal spherically symmetric ball as I can imagine, and if the string is tied to the end of the handle it's working far from the center of mass of the hammer.

8. Jan 26, 2013

leviterande

hehe just what I expected opposite answers :D, I was 90% sure from reading peoples experiments and opinions that the ball indeed will spin upon release like this gif animation:

http://cosmoquest.org/forum/attachment.php?attachmentid=16619&d=1332906710

But Im not anymore sure, so I want to get a solid solution to this

Nugatory
hammer throws is an Olympic, its a heavy steel ball with a chain thrown, if air restance on the string causes the ball to rotate, then shouldnt the string trail behind? however the hammer throws show that steel ball rotates till it hits the ground

Last edited: Jan 26, 2013
9. Jan 26, 2013

mrspeedybob

Yes the ball will spin. I can think of 2 ways to explain why but the result is the same.

While the ball is attached to the string the same side always faces the string's anchor point, therefore the ball is spinning on its own axis at the same rate that it is orbiting the anchor point of the string. When it is released from the string there is no torque to cause it to change its angular momentum around its own axis.

The other way is to consider linear momentum of each side of the ball. For example, say you have a 10 cm ball at the end of a 1 m string orbiting at 1 rev / sec. The inside edge of the ball will be traveling at 6.28 m/s. The side of the ball opposite the string will be traveling a 10% larger circle so it will be traveling at 6.91 m/s. When the ball is released one side will be traveling faster then the other, but the ball as a whole is now traveling in a straight line, so it must be rotating.

10. Jan 26, 2013

leviterande

thanx , the second explanation sounds very logical, I heared very different answers from educated people but I thought however that the ball must somehow spin. On a danger to sound rude, how do I know that your answer is correct and final while Nugatory
got it wrong?

11. Jan 26, 2013

Staff: Mentor

You're right....thx...

No net torque from releasing the ball means that releasing it won't cause it to rotate about its axis, but also means that whatever rotation it had it will keep... and you're right, it's already rotating once per orbit.

12. Jan 26, 2013

mrspeedybob

That is not rude at all, in fact you have asked exactly the right question. The answer is to design and conduct an experimental test. Then determine if the experimental results match or disagree with the expected results. Finally, post your procedure and results for public review.

13. Jan 26, 2013

256bits

The ball possesses 2 angular momentum,
1, that of revolving around the central axis,
2, that of revolving around its own axis

Breaking the string converts #1 into translation.
The ball will continue to possess angular momentum #2 after release.

14. Jan 26, 2013

leviterande

Yeah I would love to do that but rotating a ball-on-a-string and then cutting the string exactly at the ball-string attachment while the whole thing whirls at good speed and then recording that in slow motion is kind of hard. so if I am not asking too much is there any papers, writings, simple diagrams or anything on the net you know that shows this exact phenomena(the spinning and not only the straight path)? everything I found on the net showed only that the ball went straight and didnt discuss anything beyond that.

15. Jan 26, 2013

Staff: Mentor

Not rude at all. If we correctly understand the physics, we'll all come up with same prediction for how the ball will behave. So if we come up with different predictions... Then one of us must have missed something, and asking questions is how we find out what is was.

16. Feb 1, 2013

sophiecentaur

Isn't the answer - Yes, because it has been spinning at the same rate as the rotation round your hand (the fixing point has always been facing towards you). If the string is detached cleanly, all force on the ball will disappear so the ball must maintain its angular velocity for ever on its tangential, straight line path?
For this answer, you need to assume that the ball has settled down into this condition. In actual fact, as the system is 'spun-up' the ball will oscillate (rotationally), facing about a mean direction that is along the string. If there is no friction, then the ball will continue doing this (yoyo motion) as the string is spun up and then for ever - until the string is disconnected. Then its angular speed could be any value within the range of its angular speeds on the way round. The oscillation rate will be totally unrelated to the rate of swirling of the string, too. It will depend upon its MI, angular velocity of the string and its mass.

17. Feb 1, 2013

rcgldr

While a ball on a string experiment would be difficult, you can watch a video of a hammer throw and note that the ball and chain continue to rotate. Towards the end of this video is an example. Due to lack of depth perception, there's an optical illusion that can make it look like the chain is spinning the wrong direction.

Last edited by a moderator: Sep 25, 2014
18. Feb 2, 2013

Jupiter6

I don't think you need to cut anything to prove the point. How about taking a short length of light string and attaching it to a ball, then swinging the ball around and letting it go. The string will act as visual indicator of the balls rotation while the ball itself moves in a straight line.

19. Feb 2, 2013

stevendaryl

Staff Emeritus
Another way to see the answer is to use a stick instead of a ball. You have a stick, and connected to one end of the stick is a string. The string is being spun around. Then the velocity of the near end of the stick is $\omega r$ where $\omega$ is the angular velocity, and $r$ is the length of the string. The velocity of the far end is $\omega (r+L)$ where $L$ is the length of the stick. Now if you cut the string, then immediately afterward, the velocities are unchanged. So the far end of the stick is moving slightly faster than the near end. So the stick is rotating (or stretching, I guess).

20. Feb 2, 2013

sophiecentaur

People seem to be making a massive meal of this. If the ball is rotating whilst is is being swung round then (and how can it not be?) is it not totally obvious that it will continue to rotate when no forces are acting on it? What more is there to say?