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[TheQueen'sOwn]

An Australian bushman hunts kangaroos with the following weapon, a heavy rock tied to one end of a light vine of length 2m. He holds the other end above his head, at a point 2m above ground level, and swings the rock in a horizontal circle. The cunning kangaroo has observed that the vine always breaks when the angle theta (measured between the vine and the vertical) reaches 60 degrees.

At what minimum distance from the hunter can the kangaroo stand with no danger of a direct hit?

So here's what I've done. First I broke the problem into two parts.
#1. Circular Motion.
#2. Projectile Motion.
Then I made a diagram (below) .



I can find the distance it travels after it is released with no problem. I just need to know it's initial velocity!

So basically, my question is....
Is the circular motion similar to the motion of a car on an Indy 500 track?
My reasoning behind that question is... I don't think this is similar, as a race car on the Indy 500 is travelling on a road that is constantly elevated at an angle... In this case... isn't it more of a 2-dimensional motion on a plane that is elevated at 30 degrees??

If you can't understand my insane ramblings above... could you please just try the question in any way you see fit?

This is killing me =(.
Thank you GAF... and in advance, probably Jinx =) :lol.

 

[TheQueen'sOwn]

Bump for the nightcrew =).

 

[Asbel]

It doesn't look like he intends to 'toss' the rock at the kangaroo to me. He wants to swing it like a sock full of quaters but the vine always breaks. Eh.. not pythagorean's thereom, just 2 x cos 30 because the length of the vine is 2. Sorry bout that.

 

[TheQueen'sOwn]

That would be sweet ... but I dunno if that would be the sole operation involved in solving a Grade 12 Academic Physics problem. My biggest problem right now is conceptualizing the whole thing. Honestly I hadn't thought of it that way.. so thank you.

However.. we just finished Projectile Motion and we're doing Circular Motion right now so I figured the question would be a combination of the two.

 

[Asbel]

I can understand that but you have no given velocities. I don't know how you can find the projectile distance.

 

[Dilbert]

Your diagram is wrong in a couple of important ways -- it's no wonder you're having trouble with the problem.

For one thing, the rope is dangling DOWNWARD from his hand. If you don't understand why, go tie something to the end of a string and try the experiment out yourself. If you hold the string above your head, the object will be dangling straight down. Now, slowly start swinging the object in a circle. The faster it swings, the greater the angle from vertical...and, the greater the tension in the rope has to be in order to maintain the circular motion. When the angle is 60 degrees, the rope snaps and the object flies off. There is no way for a rope to be going "up" the way you draw -- it's physically impossible.

We need to find the angular velocity (w) -- if you don't see why, read ahead to the next paragraph. When the rope breaks, we know the component of tension in the vertical dimension is equal to the weight of the rock due to equilibrium conditions...and the horizontal component of tension is related to angular velocity since that force is causing the rotational motion. Even though there are some things we don't know -- for instance, the mass of the rock -- the unknowns should cancel out if you set up the equations properly.

The second problem with the diagram is the top view, since the radius of the circle is not 2 meters. The rope length is actually the hypotenuse of a right triangle, but the radius of the circular motion is the x-component...in this case, R = (2 sin 60) meters. The linear velocity as it flies off is given by v = Rw, where R is the radius of circular motion and w is the angular velocity in radians.

Finally, you can solve the kinematic equation rather easily. When the rope breaks, the rock is at height 2 - (2 cos 60) meters. Calculate the time it takes to hit the ground by solving the 1-D kinematic equations in the y-dimension, then use that time and the velocity in the x-dimension (v from the paragraph above) to figure out the minimum safe range for the kangaroo.

Hope this helps -- I need to get back to work and can't work the actual equations.

 

[TheQueen'sOwn]

:lol :lol
That helps a lot!
I'll see what I can do with that tomorrow morning and I'll be sure to post the results.

Thanks guys! My birthday is monday and I'm trying to get all the homework out of the way early on .
G'night.

I can understand that but you have no given velocities. I don't know how you can find the projectile distance.

I had originally thought I could possibly use the equation.

v = sqrt( g * R * tantheta)

Which is used to find the exact speed for a car to maintain a set spot on a banked curve (without going up or down the bank). I would have used the length of the vine/string as R, theta would have been 30 degrees and gravity is..well.. gravity. I think this speed came out to be 3.36 m/s or about 12 km/h. Conceptually that seemed wrong to me and now I think I get it.. we'll see tomorrow.

 

[TheQueen'sOwn]

Sorry but I'm having trouble understanding this angular velocity and linear velocity stuff....
Here's my new (quickly made) diagram.



I know how to find the radius and the height.
Radius = 2sin60 = 1.732 m
Height = 2 - 2cos60 = 1m

I've found the angular velocity as being:

w^2 = g / (L)(cosx)

where w is the angular velocity in radians, g is the acceleration due to gravity, L is the length of the string/vine, and x is the measure of the angle between the vertical and the string/vine.

In this case,

w^2 = (9.8m/s^2) / (2m)(cos60)
w^2 = 9.8 radians/second
w = 3.13

Linear velocity was given to me... thanks Jinx... as v = wR

In this case,

v = (3.13)(1.732m)
v = 5.42 m/s



What does this linear velocity represent? How do I find the x- and y-component of this velocity? I just need one side of that stupid triangle above and I could finish this problem :lol *cries*.

 

[TheQueen'sOwn]

The velocity is perpendicular to the rope. It's at a 90 degree angle from the rope when it breaks.

Really?? Wow I really don't understand that =(.

Here's what I thought I might be done with:

And my beautiful concluding statement:
Therefore the kangaroo would have to be 2.99 m away from the hunter. Yeah I know.. seems like a small distance.... ah well :lol.. I give up.