# Optimized Solving a Physics Word Problem - Can You Help?

• Miike012
Fradial = m(2*g*L*(wradial/w))/rFradial = m(2*g*L*(wradial/w))/rFradial = 1200 N.In summary, Miike012 added two pictures. One of the word problem and one of my diagram. Conservation of kinetic and potential energy, Centripetal acceleration, and decomposition of forces into aligned forces are all explained. The equation for the tension in the rope is also given.f

#### 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..

#### Attachments

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.

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?

"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?

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.
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.

"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.

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

like that?

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

like that?

Yes.

Can you relate those quantities to the angle theta?

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??

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.

Tan(theta) = wx/wy

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

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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:

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?

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

ok ill work on that

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

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

Ah, okay!

g (L sin(θ)) = v^2/2

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

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

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

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.

g (L sin(θ)) = v^2/2

Fradial = m acentripetal = m(v^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.

You should leave the sin(θ) in.

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

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.

w= mg

the answer I got was 41 deg