# Magnitude of Tension: Circular Motion [closed]

A tennis ball connected to a string is spun around in a vertical, circular path at a uniform speed. The ball has a mass m = 0.165 kg and moves at v = 4.76 m/s. The circular path has a radius of R = 0.95 m.

What is the magnitude of the tension in the string when the ball is at the bottom of the circle?

To solve the problem I did a summation of forces: $\Sigma$$F = F_T - F_G =$$\frac{mv^2}{R^2}$. When I add $F_G$ to the left hand side of the equation to solve for tension (using -9.8 for gravity) I get 2.32N. This is the incorrect answer. When I use positive 9.8 for gravity, I get 5.55N -- the correct answer.

I don't understand why I make gravity positive when solving this problem. Could someone please help me understand what is going on here?

## closed as off-topic by Gert, user36790, Daniel Griscom, Qmechanic♦Feb 1 '16 at 13:07

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Your error is putting $g = -9.8$ in your evaluation of $m\frac {v^2}{ r}+mg$ because you already have indicated the direction of the gravitational force with a negative sign in $\vec{F_{net}} = \vec{F_T} - \vec{F_{grav}}$

Put another way $\vec{F_{net}} = \vec{F_T} + \vec{F_{grav}} = \vec {F_T}+ m (-\vec g)$

If you have

$$\vec{F_{net}} = \vec{F_T} - \vec{F_{grav}}$$

and net force is zero in that instant, then

$$mg = \frac{mv^2}{r}$$

so

$$g = \frac{v^2}{r} = \vec{a_c}$$

You should be able to see here that changing the sign of $g$ will only change the sign of the centripetal acceleration $a_c$. I'm not sure how you are getting two different answers based on a sign change. Regardless, the magnitude of the $F_T$ vector will always be positive, since its components are squared using Pythagorean theorem.

• But the net force isn't 0 at that instant. Shouldn't there be a net acceleration and thus net force towards the center of the circle? – foobar34 Jan 31 '16 at 3:01
• Net force implies acceleration. If the ball is moving centripetally towards the center, then yes, there is a net force towards the center. Since the string is taut, the force due to gravity and the tension will be equal; in any other case, the ball would stop moving in a circular path (the string would break). This is implied by 'uniform speed' in the problem. – Zack Hutchens Jan 31 '16 at 3:04
• But the force of gravity is not equal to the force of tension in the string in this question. The solution I obtained is larger than just the force of gravity. – foobar34 Jan 31 '16 at 3:26
• Then perhaps I don't understand the question. Have I made a faulty assumption? The net force has to be zero because the line is taut. If force due to gravity was stronger, then the string would break, and if tension was stronger, we would have slack in the line and the ball moving in a non-circular path. Both contradict the situation. – Zack Hutchens Jan 31 '16 at 3:42