Centripetal Acceleration of Rock Problem

[Rozy]

Hi Everyone! I've been working on this Physics problem for the last week and I still can not figure it out! I'm sure I'm missing some very basic concept, but I'm not sure how to proceed:

An Australian bushman hunts kangaroos with the following weapon - a heavy rock is tied to one end of a light vine of length 2 m, he holds the other end above his head, at a point 2 m above ground level, and swings the rock in a horizontal circle, about his body. The cunning kangaroo has observed that the vine always breaks when the angle, measured between the vertical and the string, is 60°. At what minimum distance from the hunter can the kangaroo stand with no danger of a direct hit?



I understand that this problem is made up of both centripetal acceleration concepts and projectile motion concepts. I can figure out the projectile motion using the formula d = v1(t) + 1/2(a)(t)^2 and sine/cosine/tan. Because the radius of the circle and the length of the vine form a triangle with the "extra" vertical displacement being the opposite side, I used the angles to determine that when the hypotenuse (length of vine) was 2, the radian was cos(30)(2) = 1.73 and the vertical displacement was sin(30)(2) = 1 so the total vertical displacement is 3 metres. This means that the time the projectile is in flight is 0.782 seconds and I just need the horizontal velocity to solve for the horizontal displacement.

Ok, so after that mess , I know that all I need is the horizontal velocity of the rock the instant that the vine breaks and it is traveling as a projectile. This is where I get confused. The only two forces at work on this system are the force of tension (both vertical and horizontal components) and the force of gravity.

Fgravity = mg
Ftension = ma = mv^2/r

I'm just not sure how to proceed to solve for v...I would really appreciate any suggestions. I don't mind doing the "work", but I just don't know where to start...Thank-you!

 

[Doc Al]

I understand that this problem is made up of both centripetal acceleration concepts and projectile motion concepts. I can figure out the projectile motion using the formula d = v1(t) + 1/2(a)(t)^2 and sine/cosine/tan. Because the radius of the circle and the length of the vine form a triangle with the "extra" vertical displacement being the opposite side, I used the angles to determine that when the hypotenuse (length of vine) was 2, the radian was cos(30)(2) = 1.73 and the vertical displacement was sin(30)(2) = 1 so the total vertical displacement is 3 metres. This means that the time the projectile is in flight is 0.782 seconds and I just need the horizontal velocity to solve for the horizontal displacement.

Sanity check: Does the vine angle up from where its held or does it angle down?

Ok, so after that mess , I know that all I need is the horizontal velocity of the rock the instant that the vine breaks and it is traveling as a projectile. This is where I get confused. The only two forces at work on this system are the force of tension (both vertical and horizontal components) and the force of gravity.

Fgravity = mg
Ftension = ma = mv^2/r

As you said yourself, the force of tension has two components, horizontal and vertical. Vertically, there's no acceleration, so the vertical forces add to zero. Horizontally there's centripetal acceleration; so set the horizontal force equal to ma.

You'll get two equations (horizontal and vertical). Solve them together and you'll find the speed at which the vine breaks.

 

[Rozy]

Thank-you!

Thanks you for the "Sanity Check"

Nothing like a good and humiliating assumption problem, eh?

Anyways, the problem was actually incredibly simple once I changed my point of view. Thanx!