Why do things rise as they spin?

[rabcarl]

Consider a ball tethered to a post by a string. The faster it spins, the higher it gets until the bAll is the same height as the place that the string is attached to the post. What force accounts for this?

 

[jaydnul]

Gravity and centrifugal force i would guess. The faster the ball moves, the more it wants to go off in that direction. As it slows down gravity will over come the centrifugal force, pulling it downwards.

 

[HallsofIvy]

It's not so much a matter of "the faster it spins the higher it goes" as "the additional force necessary to make it go faster also makes it go higher".

 

[rabcarl]

So how does pulling inward on a body negate the effect of gravity? In physics I learned that the horizontal and vertical components of a vector don't affect each other. Sorry if I made a mistake as I'm far from a physics expert.

 

[rabcarl]

Sorry if that last post isn't clear. I meant that if you horizontally pull something (as in centripetal force) it shouldn't move the object vertically.

 

[russ_watters]

The answer is implied in the question: it isn't horizontal. No matter how fast it goes, it will always have some downward angle, even if it is too small to see.

 

[rabcarl]

Ok, I think I understand now. the ball wants to keep pushing outward away from the centripetal force due to its inertia or centrifugal force, but it can't because of the string, so the centrifugal force is re-directed upward where it's free to move. Is this correct?

 

[ehild]

The ball is not pulled horizontally: it is pulled by the string at an angle with respect to the vertical. Two forces act on the ball: the tension of the string (T) and gravity (G). The ball moves along a horizontal circle (of radius r) if the resultant of these forces is horizontal. That horizontal resultant is the centripetal force. T

The tension in the string has horizontal and vertical components. The horizontal component provides the centripetal force:

Tsinθ=Fcp=mrω².

The vertical component acts against gravity and if it is greater than G the ball will rise. The ball can move along a horizontal circle with angular speed ω if the vertical component of the tension cancels gravity:

Tcosθ=G

The radius of the circle is r=Lsinθ.

Tsinθ=mLsinθ ω²->

T=mLω²

and

cosθ=G/T=G/(mLω²).

The faster the ball spins the greater ω; the greater the tension in the string. Also, increasing ω involves decreasing cos(θ), that is, increasing the angle the string makes with the vertical. If θ is greater the ball spins higher.

ehild

 

[rabcarl]

Thanks so much! =D That was a great explanation. I don't know about you guys but I think about these kinds of things ALL the time and they really bother me if I can't figure them out. That's why I love pf.