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Since this is actually a somewhat sophisticated problem, I decided to post the methodology here, in case someone else could use it.A 75kg student runs at 5m/s, grabs a rope supported 10m over his head, and swings out over a lake. He releases the rope when his velocity is 0. what is the angle when he realeases the rope? what is the tension in the rope just before he realeases it? and what is the max tension in the rope?

**The first part: solving the angle of release.**

This is just an exercise in energy conservation and trigonometry. When the student ascends, he is trading his kinetic energy for gravitational potential energy.

0.5 m v

^{2}= m g y

Solving, we have y = 0.5 * 75 * 5

^{2}/ 9.8 = 1.28 meters.

We can easily see that the kinetic energy held by a 75 kg mass moving at 5 m/s is the same magnitude of the same 75 kg mass's potential energy at a height of 1.28 meters.

To solve for the angle, remember that the rope is 10 meters long. Draw a circle, and remember that any point on that circle can be expressed at (r cos θ, r sin θ).

We're solving for the y-coordinate. If you assume the center of the circle is at (0, 0), the student grabs the rope at the point (0, -10). The angle at this point, with the rope pointing straight down, is (3/2) π. The y-coordinate at the top of the swing then is -8.27 meters.

So θ = arcsin(-8.27/10) = -1.05 radians. We can add 2π to this without changing its meaning, so let's call it 4.71 radians. The angle at the end of the swing is 4.71 radians. We already said the angle at the start of the swing was (3/2)π radians, or 5.22 radians. The difference is thus 5.22 - 4.71 radians = 0.51 radians = 29.3 degrees. The student swings up by an angle of 29.3 degrees.

**The second and third questions**

To solve for the tension in the rope, you have to know a bit about circular motion.

First, we know that to maintain circular motion, the body on the end of the rope has to be pulled in towards the center of the circle -- this is called the centripetal force, and is equal to

F

_{centripetal}= ma = m v

^{2}/ r

where a is, of course, the centripetal acceleration, and F = ma is just Newton's second law of motion.

The rope has to bear this tension, and be able to provide this force, to keep the student moving along his circular path.

In addition to this force caused by the student's inertia, there is an additional force due to his weight. This force is just F

_{weight}= m g cos θ, taking θ to be zero at the bottom of the swing and 29.3 degrees at the top of it.

So the total tension in the rope is the sum of these two

F = F

_{centripetal}+ F

_{weight}= m v

^{2}/ r + m g cos θ

We now have to solve v for all points along the swing, since it isn't constant. The student begins his swing with velocity 5 m/s, and ends it with velocity 0 m/s.

Call the student's original kinetic energy, before grabbing the rope, K = 0.5 m v

^{2}= 937.5 J.

The remaining kinetic energy at any point in the swing is just (K - mgy), so the velocity can be solved as

0.5 m v

^{2}= (K - mgy)

v

^{2}= (2/m) (K - mgy)

Plugging this into our expression for total force, we have

F = (2K - 2mgy)/r + mg cos θ

The tension when he let's go is easily solved: plug in y = 1.28 meters, θ = 29.3 degrees and you get F = 640 N.

The tension at the start of the swing is the greatest, since his entire weight is applied to it. At any other point in the swing, his weight is applied less.

At the start of the swing, plug in y = 0, θ = 0 and you get F = 920 N.

- Warren