Phy word Problem

  • Thread starter Miike012
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  • #1
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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..

Can someone please help me answer this problem?
 

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  • #2
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Hi Miike012! :smile:

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....

Anyway, let's start with the relevant equations.
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.
Which forces are there to start with?
 
  • #3
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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
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"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
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1.Wnet = delta K
Ugi - Ugf = Kf - Ki
-Ugf = Kf

-gy = v^2/2
Right!

Can you write y in terms of theta?


2.a = v^2/r
Yes.
You need to relate this to your forces.

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

Are these correct equations?
Yes, these are correct equations.

However, I suggest you try to find the forces aligned with the rope.
The tension is already aligned.
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
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"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????
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
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3. SumFx = T - wx = (m)(ax)
4. SumFy = -wy = (m)(ay)

like that?
 
  • #8
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3. SumFx = T - wx = (m)(ax)
4. SumFy = -wy = (m)(ay)

like that?
Yes.

Can you relate those quantities to the angle theta?
 
  • #9
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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
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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??
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
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Tan(theta) = wx/wy

cos(theta) = (...)/T
 
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  • #12
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Here was my reasoning

Tx should be T
 

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  • #13
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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:
wradial = w sin(θ)

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

These need to be combined with:
Fradial = m acentripetal

Can you work that out?
 
  • #14
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T- (wy)Tan(theta = (m)(ax)
-(wx)Cot(theta) = (m)(ay)
 
  • #15
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ok ill work on that
 
  • #16
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F(radial) = m(2*g*L*(wradial/w))/r
 
  • #17
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F(radial) = m(2*g*L*(wradial/w))/r
How did you get that?
And what happened to the tensional force T?
 
  • #18
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im saying x and y because i labeled it with the following axis..
 

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  • #19
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im saying x and y because i labeled it with the following axis..
Ah, okay!
 
  • #20
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sin(θ) = wradial/w

g (L sin(θ)) = v^2/2
g(L*wradial/w) = v^2/2
v^2 = g(L*wradial/w)(2)

Fradial = m acentripetal = m(v^2/r)
Fradial = m (g(L*wradial/w)(2)/r)
 
  • #21
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And im not sure what Fradial is
im guessing its the tension and wradial
 
  • #22
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And im not sure what Fradial is
im guessing its the tension and wradial
Yes.

The total radial force Fradial = T - wradial.

Furthermore this force has to match the radial acceleration. So:
Fradial = m acentripetal

I you can rewrite this in only numbers and theta, you can solve for theta.
 
  • #23
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sin(θ) = wradial/w

g (L sin(θ)) = v^2/2
g(L*wradial/w) = v^2/2
v^2 = g(L*wradial/w)(2)

Fradial = m acentripetal = m(v^2/r)
Fradial = m (g(L*wradial/w)(2)/r)
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.

I'm afraid that eliminating theta does not help you forward though.
You should leave the sin(θ) in.

What you do need is that Fradial = T - wradial.
 
  • #24
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Thank you finally got it....
Sin(theta) = T/(m*g*2 + w)
 
  • #25
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How in the world did you piece everything together?? I was sitting here thinking I had to use the sum tan-gentle force equation....?
 

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