Optimized Solving a Physics Word Problem - Can You Help?

[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..

Can someone please help me answer this problem?

 

[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...

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?

 

[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?

 

[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?

 

[I like Serena]

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.

 

[I like Serena]

"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.

 

[Miike012]

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

like that?

 

[I like Serena]

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

like that?

Yes.

Can you relate those quantities to the angle theta?

 

[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??

 

[I like Serena]

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.

 

[Miike012]

Tan(theta) = wx/wy

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

 

[Miike012]

Here was my reasoning

Tx should be T

 

[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:
w_radial = 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:
F_radial = m a_centripetal

Can you work that out?

 

[Miike012]

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

 

[Miike012]

ok ill work on that

 

[Miike012]

F(radial) = m(2*g*L*(wradial/w))/r

 

[I like Serena]

F(radial) = m(2*g*L*(wradial/w))/r

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

 

[Miike012]

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

 

[I like Serena]

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

Ah, okay!

 

[Miike012]

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)

 

[Miike012]

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

 

[I like Serena]

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

Yes.

The total radial force F_radial = T - w_radial.

Furthermore this force has to match the radial acceleration. So:
F_radial = m a_centripetal

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

 

[I like Serena]

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 F_radial = T - w_radial.

 

[Miike012]

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

 

[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...?

 

[I like Serena]

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

Good!

But there's still something to be done...
What is the relation between m*g and w?

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

Well, I guess there is a method to it.

First list all your data and relevant equations.

Then consider what the critical condition is.
In this case that the total radial force exceeds the critical threshold.
This will give you a formula.

And then use your data and relevant equations to find the unknown that is asked for.

 

[Miike012]

w= mg

the answer I got was 41 deg

 

[Miike012]

thank you for your help!

 

[I like Serena]

You're welcome!