# Homework Help: Phy word Problem

1. Nov 11, 2011

### Miike012

I added two pictures.. One of the word problem and one of my diagram I made..

My work:

1.Wnet = Kf - Ki
-Ugf = Kf.... -mgy = mv^2/2 .... - gy = v^2/2

2. x^2 + y^2 = 1780^2
dy/dx = -x/y = velocity (v) at point (x,y)

3. gy = x^2/(2y^2)

4. gy = (1780^2 - y^2)/(2y^2)

.... anyways... I know this must be totally wrong..

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2. Nov 11, 2011

### I like Serena

Hi Miike012!

I have difficulty making sense of what you did.
What do the dots for instance mean in (1)?

And if x and y are coordinates in (2), how did you relate them to a force of 1780 N.
These things have different units and can not be compared....

1. Conservation of kinetic and potential energy
2. Centripetal acceleration

Can you write those in formula form?

Next is the decomposition of the forces into aligned forces.

3. Nov 11, 2011

### Miike012

1.Wnet = delta K
Ugi - Ugf = Kf - Ki
-Ugf = Kf

-gy = v^2/2

2.a = v^2/r

3. SumFy = Ty - w = (m)(ay)
4. SumFx = Tx = (m)(ax)

Are these correct equations?

4. Nov 11, 2011

### Miike012

"And if x and y are coordinates in (2), how did you relate them to a force of 1780 N."

I related 1780 to (x,y) by saying the end point of tension started at the origin, and because the motion is a circle I gave the following equation...
x^2 + y^2 = 1780^2... which relates it to x and y... would this not be correct????

5. Nov 11, 2011

### I like Serena

Right!

Can you write y in terms of theta?

Yes.
You need to relate this to your forces.

Yes, these are correct equations.

However, I suggest you try to find the forces aligned with the rope.
If you find the component of the weight aligned with the rope, you can find the resulting force along the rope.
The resulting force along the rope must match the centripetal acceleration.

6. Nov 11, 2011

### I like Serena

Can it be that you intended Tx^2 + Ty^2 = 1780^2?

x and y would be coordinates that have "meter" as the unit.
1780 is the critical tensional forces that has "newton" as the unit.

Tx and Ty would be the cartesian force components of the tensional force.
They also have "newton" as the unit.

7. Nov 11, 2011

### Miike012

3. SumFx = T - wx = (m)(ax)
4. SumFy = -wy = (m)(ay)

like that?

8. Nov 11, 2011

### I like Serena

Yes.

Can you relate those quantities to the angle theta?

9. Nov 11, 2011

### Miike012

While I relate the equations the theta.. I was wondering, how come in some situations in circular motion sometimes the SumFy or SumFx has zero acceleration??

10. Nov 11, 2011

### I like Serena

In circular motion the radial force is not zero, or it won't be circular motion.
The tangential force can be zero, and indeed is zero if we're talking about "uniform" circular motion.

11. Nov 11, 2011

### Miike012

Tan(theta) = wx/wy

cos(theta) = (...)/T

Last edited: Nov 11, 2011
12. Nov 11, 2011

### Miike012

Here was my reasoning

Tx should be T

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13. Nov 11, 2011

### I like Serena

Right.

But I'm getting confused with what x and y stand for exactly.
So I'll use "radial" and "tangential" to keep things clear.

What I was looking for is:

And in your earlier equation for the conservation of energy:
g (L sin(θ)) = v^2/2

These need to be combined with:

Can you work that out?

14. Nov 11, 2011

### Miike012

T- (wy)Tan(theta = (m)(ax)
-(wx)Cot(theta) = (m)(ay)

15. Nov 11, 2011

### Miike012

ok ill work on that

16. Nov 11, 2011

### Miike012

17. Nov 11, 2011

### I like Serena

How did you get that?
And what happened to the tensional force T?

18. Nov 11, 2011

### Miike012

im saying x and y because i labeled it with the following axis..

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19. Nov 11, 2011

### I like Serena

Ah, okay!

20. Nov 11, 2011

### Miike012

g (L sin(θ)) = v^2/2

Fradial = m acentripetal = m(v^2/r)

21. Nov 11, 2011

### Miike012

And im not sure what Fradial is
im guessing its the tension and wradial

22. Nov 11, 2011

### I like Serena

Yes.

Furthermore this force has to match the radial acceleration. So:

I you can rewrite this in only numbers and theta, you can solve for theta.

23. Nov 11, 2011

### I like Serena

Ah, okay.

Note that your r is the radius of the circular motion, which is equal to the length of the rope.
So r=L.

You should leave the sin(θ) in.

24. Nov 11, 2011

### Miike012

Thank you finally got it....
Sin(theta) = T/(m*g*2 + w)

25. Nov 11, 2011

### Miike012

How in the world did you piece everything together?? I was sitting here thinking I had to use the sum tan-gentle force equation....?