Jackson Garner
2022-09-25
Answered

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)

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i2n1f4a

Answered 2022-09-26
Author has **6** answers

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 $\frac{{v}^{2}}{r}$

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

$\mathrm{\Sigma}{F}_{towardscenter}={m}_{ball}{a}_{towardscenter}=>T={m}_{ball}{\displaystyle \frac{{v}^{2}}{r}}$

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 $T+{m}_{ball}gcos(45)={m}_{ball}{\displaystyle \frac{{v}^{2}}{r}}$

But anyway, for your question $T={m}_{ball}{\displaystyle \frac{{v}^{2}}{r}}$

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

$\mathrm{\Sigma}{F}_{towardscenter}={m}_{ball}{a}_{towardscenter}=>T={m}_{ball}{\displaystyle \frac{{v}^{2}}{r}}$

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 $T+{m}_{ball}gcos(45)={m}_{ball}{\displaystyle \frac{{v}^{2}}{r}}$

But anyway, for your question $T={m}_{ball}{\displaystyle \frac{{v}^{2}}{r}}$

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