# Tension of a string rotating a ball in a circular motion?

A ball of mass 5kg is attached to a string of length 120 cm and rotating vertically at a speed of 10 cm/s. what is the tension of the string when the ball is farthest to the right from the center? (neglect both the string's mass and air resistance)

I tried applying Newton's second law that says $$\Sigma force= ma => T-mg = ma$$ but that doesn't give the right answer and i don't know why. can somebody please help me, thanks

• What is the centripetal force needed to keep the ball in a spherical orbit? "Furthest to the right" means gravity is at right angles and doesn't come into the equation. – Floris Feb 25 '16 at 5:02
• so do you mean that tension is just ma? i.e. $T = ma$ – GamefanA Feb 25 '16 at 5:30
• The tension and the centripetal force will be equal and opposite, since they are the only horizontal forces. See if this helps! – Robbie Feb 25 '16 at 5:52
• How can they be equal and opposite if the centripetal force is pulling towards the center and the tension is also pulling towards the center? – GamefanA Feb 25 '16 at 6:35

The centripetal force is not a "separate" force. I think it's best not to think of centripetal forces, but just centripetal acceleration. An object with circular motion means that net sum of all the forces acting on the object results in circular motion... meaning the net acceleration towards the center of the circle is $\dfrac{v^2}{r}$

In your situation there are two forces acting on the ball. The tension in the rope and gravity. (there's no extra centripetal force).

$\Sigma F_{towards center} = m_{ball}a_{towardscenter} => T = m_{ball}\dfrac{v^2}{r}$

So gravity does not play a role here because gravity acts downward, and the direction towards the center of the circle is to the left.

Suppose the ball was at an angle of 45 degrees to the right of the upward direction. Then you'd have to consider the tension in the rope and the component of gravity acting towards the center. Specifically you'd get $T+m_{ball}gcos(45) = m_{ball}\dfrac{v^2}{r}$

But anyway, for your question $T = m_{ball}\dfrac{v^2}{r}$

• Good and very clear answer. Just one thing. in this forumula you provided $T+m_{ball}gcos(45) = m_{ball}\dfrac{v^2}{r}$ how can you take the Y component of gravity (I assume that's what the $gcos(45)$ is)? i thought that gravity is always pulling down – GamefanA Feb 25 '16 at 8:13
• Yes, gravity is downwards. But I'm taking my y-axis towards the center, and my x-axis tangent to the circle. Remember in that situation the ball is 45 degrees to the right of the top. So the direction towards the center is down and to the left. So I split up gravity into two perpendicular components. towards center of the circle (which is down and to the left at 45 degrees). And the component tangent to the circle (which is down and to the right). Only the component towards the center contributes to the centripetal acceleration. – Ameet Sharma Feb 25 '16 at 8:24
• I see. so if we instead took the Y component of the $T$ (i.e. $Tcos(45)$) add gravity to it as a whole, and then use Pythagorean theorem to add both the new Y component of the tension and the old, unchanged X component (i.e. $\sqrt{(Tcos(45)+m_{ball}g)^2 + (Tsin(45))^2}$ would that give an equivalent answer? – GamefanA Feb 25 '16 at 8:35
• Assuming your taking down as the y-axis, that would give the magnitude of the net force acting on the ball. We're only interested in the net force acting towards the center. So we choose axes appropriately. in the 45 degree case with the y and x axes diagonal, the net force towards the center is T+mgcos(45) (this component equals mv^2/r). The net force tangent to the circle is mgsin(45). So using the pythagorean theorem we get your result of $\sqrt{(T+mgcos(45))^2+(mgsin45)^2}$. The difference between what you did is I think you took the y-axis down, and I took it as towards the center. – Ameet Sharma Feb 25 '16 at 8:48

The motion of a mass tied to a string in a vertical circle includes following mechanical concepts.

It must satisfy

(i) availability of centripetal force to remain in a circular path

(ii) satisfy conservation of energy If we take a situation that the ball just reaches the topmost position with velocity equal to zero then the the tension in the string will be such that it is just taut.

Therefore the gravitational pull must be providing a force equal to magnitude of centripetal force (m.v^2)/r so one can get the value of speed and the total energy is known.

In other cases also where the body can reach the top and can cover it with some finite velocity the total energy conservation can be applied with due consideration of change in potential energy of the body.

I will advise you to take the topmost velocity to be say v(3) the bottom point as V(1) and at horizintal midway V(2) and relate the energies K.E. +P.E. at the three points to be equal. you have a info that at top point the only force is mg acting downward providing the centripetal force. that will facilitate with the value of V(3). Then you can calculate v(1) and then naturally V(2) can be easily computed.

The Tension in the string at the horizontal point where the speed of the ball is v(2) T= m(v(2))^2/r as the mg force is perpendicular to the string and not contributing to the tension.

• And another answer where you did not include the source (Hyperphysics' "Motion in a Vertical Circle"); and the beginning of this answer is not a direct quote, but a very close paraphrase. You have to clearly mark citations, and you should not be writing answers in this copy-pasting manner. – ACuriousMind Mar 2 '16 at 17:50
• The comment has been taken into consideration and text has been edited – drvrm Mar 4 '16 at 7:35

## protected by ACuriousMind♦Jun 10 '17 at 12:10

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