# Trouble getting started

1. Mar 3, 2006

### ussjt

One type of slingshot can be made from a length of rope and a leather pocket for holding the stone. The stone can be thrown by whirling it rapidly in a horizontal circle and releasing it at the right moment. Such a slingshot is used to throw a stone from the edge of a cliff, the point of release being 10.0 m above the base of the cliff. The stone lands on the ground below the cliff at a point X. The horizontal distance of point X from the base of the cliff (directly beneath the point of release) is thirty times the radius of the circle on which the stone is whirled. Determine the angular speed of the stone at the moment of release.

I having problems setting this up. Any help you can give would be great.

2. Mar 3, 2006

### Staff: Mentor

What are you able to figure out? Can you find the time it takes for the stone to hit the ground? Can you write an expression for the horizontal distance as a function of time? How does angular speed relate to linear speed?

3. Mar 3, 2006

### xman

Try using the kinematic equation
y=y_{0} +v_{0y}t-1/2*g*t^2...(1)
Note the velocity in the x direction is constant throughout flight, and use this plus the relationship given for x and the radius to relate the velocities. Finally, use the relationship between the tangential velocity at release and angular velocity. Should find everything works cancels nice and neat and you should obtain a numerical value with the given information in the problem.

4. Mar 3, 2006

### ussjt

ok I figured out the time.

I am a little confused by what you mean.

5. Mar 3, 2006

### Staff: Mentor

Good. Now continue with what I suggested. (The "trick" is to translate everything you know into mathematical form and see what you can figure out.)

6. Mar 3, 2006

### ussjt

I am having problems figuring out the horizontal distance. Once I have that I can use that to determine the radius and I have the time already.

7. Mar 3, 2006

### Staff: Mentor

You won't be able to figure out the distance in the same way that you figured out the time. Instead, come up with expressions for:
(1) horizontal distance as a function of time
(2) linear speed in terms of angular speed

If you combine those two expressions (using the information given) you'll be able to solve for the angular speed in terms of time (which you already know).

8. Mar 3, 2006

### ussjt

but when trying to get the horizontal distance...there are 3 unknowns. V_1 V_2 and X_2. SO how fo you get it as a function of time?

9. Mar 3, 2006

### tony873004

v1 and v2 are the same since gravity doesn't accelerate in the horizontal component. X2 is known. It's given in terms of radius, not an exact number. Can you write and expression that solves for horizontal distance? And yes, there will be an unknown in it, but that's ok since you're not actually going to solve that equation.

10. Mar 3, 2006

### ussjt

ok I understand that x2 = 30*r.

x2 = v1t + .5at^2
x2 = v1t
x2/t = v1
30r/t = v1

then use:

Vt = rw

I am doing this right?...I am not sure because I am tired right now.

11. Mar 3, 2006

### tony873004

So far so good. Now that you know x2, can you write an expression for velocity in terms of distance (which you're calling x2) and time?

12. Mar 3, 2006

### ussjt

wouldn't the velocity just be the change in distance over time?

30r/t

13. Mar 3, 2006

### tony873004

Yes! Only 1 more step. You've figured out that Vt = rw and Vh = 30r/t. Since Velocity tangent is the same as Velocity horizontal, this simplifies to:

V=rw
V=30r/t

Can you figure out what to do from here? (btw. you probably know this but that's not a w, it a $$\omega$$)

***Edited to fix my typo of V=30r. Changed to V=30r/t.

Last edited: Mar 3, 2006
14. Mar 3, 2006

### ussjt

so it is rw=30r.

where w= (-)/t ?

15. Mar 3, 2006

### tony873004

Yes, except I made a typo. I should have said:
V=rw
V=30r/t

Which would give you:
rw=30r/t

It looks like you've figured out that the r's cancel. Just plug in your value for t and that's it.

16. Mar 3, 2006

### ussjt

got it now...thanks a lot.