Tarzan swinging on a vine -- At what angle does the vine break?

[Vc91]

Homework Statement

Tarzan who weighs 688 N swings from a cliff at the end of a vine 18 m long. From the top of the cliff to the bottom of the swing he descends by 3.2 m. The vine will break if the force exceeds 700 N. Find the angle with the vertical at which the vine breaks.

Homework Equations

T-mgcos(x) = mv^2/r
Mgh initial = mgh final + 1/2mv^2 final

The Attempt at a Solution

I am having trouble figuring out what to put for initial and final heights. For initial height I got rcosx and final is (r-rcosx)? Therefore due to conservation of energy
Mgrcos(x)=(r-rcos(x))mg + 1/2 mv^2. Can someone please let me know if this is correct?

 

[J Hann]

Doesn't (r - r cos x) imply that the vine breaks at x = 90 deg.?
I think the problem assumes the initial angle from the vertical is 90 deg.

 

[Vc91]

Would it be rcosx-r then? I'm just not sure how to find the appropriate height since the problem says that the height from the top of the cliff to the bottom of the swing is 3.2 m but I know that you are not supposed to just simply plug in 3.2 m for h in the conservation of energy equation.

 

[gneill]

Doesn't (r - r cos x) imply that the vine breaks at x = 90 deg.?
I think the problem assumes the initial angle from the vertical is 90 deg.

From the start of the swing to the bottom of the arc is only 3.2 m . So no, the vine does not start at 90°. The starting angle will have to be found.

 

[J Hann]

Would it be rcosx-r then? I'm just not sure how to find the appropriate height since the problem says that the height from the top of the cliff to the bottom of the swing is 3.2 m but I know that you are not supposed to just simply plug in 3.2 m for h in the conservation of energy equation.

I missed the fact that the vine is 18 meters long so you could find the initial angle
using 18 meters for the length of the vine and the vertical drop from that
point is 3.2 m to the bottom of the swing.
(Then, of course, he really wouldn't drop 3.2 m if the vine breaks
before he reaches the vertical point of the swing)
The problem seems to be a little vague on that point, but I think
that somehow you need to find the initial angle because the
tension on the vine at any point depends on the angle from the vertical at that point.

 

[haruspex]

The starting angle will have to be found.

That is not necessary.

 

[haruspex]

For initial height I got rcosx and final is (r-rcosx)?

Did you draw a diagram?

What point are you taking as zero height?
If the lowest point of the swing, that would be at angle 0, yes? Check your general expression for height at angle x against that.

 

[J Hann]

I missed the fact that the vine is 18 meters long so you could find the initial angle
using 18 meters for the length of the vine and the vertical drop from that
point is 3.2 m to the bottom of the swing.
(Then, of course, he really wouldn't drop 3.2 m if the vine breaks
before he reaches the vertical point of the swing)
The problem seems to be a little vague on that point, but I think
that somehow you need to find the initial angle because the
tension on the vine at any point depends on the angle from the vertical at that point.

That is not necessary.

I get h / L = .0174 for the vertical distance fallen..
If that is the case, then what does 3.2 m have to do with the problem
unless that is used to determine the starting angle.

 

[haruspex]

what does 3.2 m have to do with the problem
unless that is used to determine the starting angle.

It is needed to solve the problem, and you could use it to find the start angle, but doing so is unnecessary extra work.