Question regarding Conservation of Energy

[xxdrossxx]

Hello, I'm currently in a high school physics course, and I'm having trouble with a question in a "Work and Energy" chapter, which has focused on conservation of energy. I'm hoping that someone here can help me out.

A grappling hook, attached to a 1.5-m rope, is whirled in a circle that lies in the vertical plane. The lowest point on this circle is at ground level. The hook is whirled at a constant rate of three revolutions per second. In the absence of air resistance, to what maximum height can the hook be cast?

Thanks so much for any help that you can give.

 

[Pengwuino]

As per the forum rules, you need to show some work on your own or what you think you might need to do.

 

[xxdrossxx]

Woops, I'm sorry about that. Okay.

In earlier problems that I've seen like this, they've involved an object spinning in a circle on the vertical plane...but in those cases, the object just barely makes it over the top of the circle, so that the kinetic energy at the top of the circle is equal to 0. In the case of those problems, I would set potential energy at the top equal to the kinetic energy at the bottom. In other words, mgh = 0.5mv^2 and therefore gh = 0.5v^2.

This problem is different though, I think. It's saying that the hook is whirled at a "constant rate of three revolutions per second", which is equal to a velocity of 28.274 m/s (three revolutions = 3 * pi * 3.0 m = 28.274). However, if there's a constant velocity, then the hook is obviously moving completely around the circle...so wouldn't the "max height that the hook can be cast" simply be 3 meters, or the height at the top of the circle?

I think that I'm misunderstanding the wording of this question.

Thanks for any help that you can give.

 

[johnw188]

Woops, I'm sorry about that. Okay.
In earlier problems that I've seen like this, they've involved an object spinning in a circle on the vertical plane...but in those cases, the object just barely makes it over the top of the circle, so that the kinetic energy at the top of the circle is equal to 0. In the case of those problems, I would set potential energy at the top equal to the kinetic energy at the bottom. In other words, mgh = 0.5mv^2 and therefore gh = 0.5v^2.
This problem is different though, I think. It's saying that the hook is whirled at a "constant rate of three revolutions per second", which is equal to a velocity of 28.274 m/s (three revolutions = 3 * pi * 3.0 m = 28.274). However, if there's a constant velocity, then the hook is obviously moving completely around the circle...so wouldn't the "max height that the hook can be cast" simply be 3 meters, or the height at the top of the circle?
I think that I'm misunderstanding the wording of this question.
Thanks for any help that you can give.

Yea, it seems as if you are misunderstanding it. Think of how you'd throw a grappling hook. You'd twirl it around in your hand, and then release it. This question is asking, if you twirl the hook at a rate of three revolutions per second, what is the maximum height that the hook can reach?
Combine that with the fact that if you release the rope, a_{rad} immediatly becomes zero, and the problem shouldn't give you any more difficulties.

 

[xxdrossxx]

Yea, it seems as if you are misunderstanding it. Think of how you'd throw a grappling hook. You'd twirl it around in your hand, and then release it. This question is asking, if you twirl the hook at a rate of three revolutions per second, what is the maximum height that the hook can reach?
Combine that with the fact that if you release the rope, a_{rad} immediatly becomes zero, and the problem shouldn't give you any more difficulties.

Oh, of course! It makes perfect sense to me now. So I'd just do the following:

0.5 (28.274 m/s)^2 = (9.8 m/s^2) * h
h = 40.786 meters

Does that look correct?

 

[johnw188]

Oh, of course! It makes perfect sense to me now. So I'd just do the following:
0.5 (28.274 m/s)^2 = (9.8 m/s^2) * h
h = 40.786 meters
Does that look correct?

Couple of tips in general for physics.

1) Write everything out symbolically until the very last step, it keeps things easier to read.

\frac{1}{2} m v_0^2 = mgh
h = \frac{\frac{1}{2} m v_0^2}{mg}

The one thing you forgot is that you aren't releasing from ground level; you're releasing from when the hook is horizontal to you, so that your tangential velocity is pointing straight up to get the maximum height. As you know the rope is 1.5 m long, you have to add 1.5 m to your final height calculation.

2) Be careful of your significant figures. I didn't see what they gave you in the problem, but I doubt it was 5 for all the variables.