Centripetal force equals weight in horizontal circular motion?

Question

So imagine a tennis ball (or other object) attached to a string and spun in perfectly horizontal circular motion at a constant velocity. The two forces acting on the ball would be the force of tension (pulling the ball towards the center aka centripetal force) and the force of gravity (pulling the ball downwards aka weight).

In order for the ball to not succumb to gravity, its centripetal force must be greater, right?

So therefore, would the centripetal force have to be equivalent (or greater) than the force of gravity on the ball? Is the weight of the ball always going to be the minimal centripetal force required to keep the ball in motion?

On one level this seems almost intuitive (you want to balance the forces) but on the other I feel like gravity shouldn't even play a part in horizontal motion.

Any help in clearing up my befuddled brain is greatly appreciated.

Answer 1

You have to consider the two axes separately: lets call $x$ the horizontal plane, and $y$ the up and down (direction in which gravity acts). In the "free-body diagram" of the ball, there are also three forces to consider: 1) gravity, 2) centripetal, and 3) the tension in the string. The tension in the string needs to balance the combination of gravity and centripetal acceleration.

The centripetal acceleration itself, has nothing to do with gravity, it only has to do with the ball's circular motion. Instead of the ball spinning nearly-horizontally, you can imagine spinning it ever so slowly---and it remaining nearly vertical. Gravity hasn't changes, only the centripetal acceleration, and thus tension in the string---via force balance in the $x$ direction.

Because there is gravity in the $y$ direction, the string can never be fully horizontal, because some amount of the string tension must act in the $+y$ direction to counteract gravity. Thus, only the $y$ component of the string-tension is determined by gravity, and this occurs regardless of the centripetal force.

Answer 2

I will not be writing down everything, you should be able to write equation for FBD. now the tension in string is T and weight of ball is balanced by Tcos∅

so 1. if we also consider the string, the motion is conical 2. here, the gravity does plays a part (it determines the shape of cone) 3. weight equals Tcos∅, to your answer, Tension must be greater than weight of ball.

Answer 3

Your intuition is correct : this is a simple case of balancing forces.

The only 2 forces on the ball are the tension in the string and its weight. The resultant (=vector sum) of these 2 forces is the centripetal force. The centripetal force is not a 3rd force. It is just the name given to the amount of force which causes circular motion. In this situation it is the horizontal component of the tension in the string; the vertical component of tension is balanced by the weight of the ball.

There must be an unbalanced force here because the ball in circular motion is accelerating towards the centre of the circle.

There is no reason for the centripetal force to be greater or less than the weight of the ball. It is the tension in the string which must be greater than the weight of the ball.