Tension on a Vertical Circle- Conceptual Question

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SUMMARY

The discussion focuses on the tension in a string when a solid rubber ball, with a mass of 0.34 kg, is swung in a vertical circle with a radius of 0.5 m. At the leftmost point of the circle, where the string is horizontal, the tension is indeed present despite the downward velocity of the ball. The net force equation, F = mv²/R = FG + T, confirms that tension (T) is necessary to provide the centripetal force required for circular motion, even when the ball's velocity is directed downward.

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Homework Statement


GIVEN: mass = .34 kg
radius =.5 m

If a solid rubber ball attached to the end of a light string is swung at a constant speed in a vertical circle, what is the tension a quarter of the way around the circle (i.e. when the string attached to the ball is horizontal with the ground/ the ball is at the leftmost side of the circle) ?

My main question is a conceptual one; would there in fact be any tension exerted on the rubber ball? It seems at this point, since the velocity of the ball would be directly downwards, there would be no horizontal force (tension) exerted upon the ball to alter its course...

Is this correct? If so, how might I go about proving this?

Under normal circumstances (if the ball was at the top of the circle, for example) I would simply say that:
F (net force) = FG + T
F = mv2/R
mv2/R = FG + T
T = mv2/R - m*g

Thanks so much for any help!
 
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Well you are correct in saying that the tangential velocity at this point is directly downwards, however, have a think of where the acceleration is.

Hint; even if the magnitude of the velocity is not changing, the particle can still be accelerating.
 
oh of course! Thank you so much for your help :)
 

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