Jill of the Jungle swings on a vine 6.1 m long. What is the tension in the vine if Jill, whose mass is 61 kg,?

Jill of the Jungle swings on a vine 6.1 m long. What is the tension in the vine if Jill, whose mass is 61 kg, is moving at 2.9 m/s when the vine is vertical?


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Jill of the Jungle swings on a vine 6.1 m long. What is the tension in the vine if Jill, whose mass is 61 kg,?
Tension in the vine when it is vertical:


Tension = weight + centripetal force


T = mg + m(v^2 / r)


T = 61*g + 61(2.9 ^2) / 6.1


T = 610 + 84.1


T = 694.1 N
Reply:Note that Jill is moving in a circular path. Whenever something moves in a circle, it has a centripetal component of acceleration equal to this:





a_centripetal = v²/r





The "v" is 2.9m/s; and the "r" is 6.1m.





The direction of centripetal acceleration is always toward the center of the circle, which in this case is straight up. Since the centripetal acceleration is in a vertical direction, we can also say:





a_vertical = v²/r





Now we use "F=ma". The net vertical force on Jill must equal her mass times her vertical acceleration:





F_vertical = m(a_vertical)


F_vertical = mv²/r





Furthermore, there are only two vertical forces acting on her:


(1) the tension in the rope, T (upward)


(2) her weight, mg (downward)





Therefore, the net upward force on her is: (T−mg). So we can substitute that into the previous equation:





T−mg = mv²/r





They give you "m", "v" and "r"; and "g" is just 9.8m/s². Solve for T.