Why Is Potential Energy Calculated from the Bottom of the Circle?

[duplaimp]

Homework Statement

In the figure on the right, a 4.0 kg ball is attached to the end of a 1.6 m rope, which is fixed at O.
The ball is held at A, with the rope horizontal, and is given an initial downward velocity. The ball moves through three quarters of a circle and arrives at B, with the rope barely under tension.

Homework Equations

K1 + U1 = K2 + U2

The Attempt at a Solution

I tried with the following:

R = 1.6m
m = 4kg

K1 = \frac{1}{2}mv_{a}^{2}
U1 = mgR
K2 = \frac{1}{2}mv_{b}^{2}
U2 = mgR

But in a solution I found it was used U2 = mg2R

Why it was used 2R and not R? The distance to use in a circle is from the center to the particle or is from the bottom to the particle?
Because as A is in a distance R from center and B too, I tried with R for both but it seems that it needs to be done using a distance from bottom

 

[Maiq]

Why it was used 2R and not R? The distance to use in a circle is from the center to the particle or is from the bottom to the particle?

That U equation is for Potential energy. The equation is mass x g x distance. The distance you would use isn't too important. U1 and U2 just have to be measured from the same spot so that the change in potential energy would be mgR.

 

[duplaimp]

Ok, so from center A will be -R and B will be R?

 

[Maiq]

Sorry I guess i wasn't too clear. Its the height that you would be measuring so if you did use the center you would just need to look at how high each point is from the center.

 

[haruspex]

Why it was used 2R and not R?

If it were R, how fast would the ball be traveling at the top of the circle?

 

[PeterO]

Homework Statement

In the figure on the right, a 4.0 kg ball is attached to the end of a 1.6 m rope, which is fixed at O.
The ball is held at A, with the rope horizontal, and is given an initial downward velocity. The ball moves through three quarters of a circle and arrives at B, with the rope barely under tension.

Homework Equations

K1 + U1 = K2 + U2

The Attempt at a Solution

I tried with the following:

R = 1.6m
m = 4kg

K1 = \frac{1}{2}mv_{a}^{2}
U1 = mgR
K2 = \frac{1}{2}mv_{b}^{2}
U2 = mgR

But in a solution I found it was used U2 = mg2R

Why it was used 2R and not R? The distance to use in a circle is from the center to the particle or is from the bottom to the particle?
Because as A is in a distance R from center and B too, I tried with R for both but it seems that it needs to be done using a distance from bottom

I keep reading the problem, and don't find a question? Are you trying to calculate v_o perhaps?

 

[haruspex]

If it were R, how fast would the ball be traveling at the top of the circle?

Forget that - I misinterpreted your equations.
You have U = mgR at A. How could it also be mgR at B, given that B is height R above A?

 

[rude man]

Assuming the question is, 'what is the initial imparted velocity', use energy conservation. Note that the tension on the rope imparts force to the mass but since it's always orthogonal to the directin of motion, that force does no work.

So, simply, initial kinetic energy = change in potential energy.

 

[technician]

When the ball arrives at B the tension in the rope becomes zero. This does not necessarily mean that the ball has come to rest.
It is not 100% clear what the question is asking but, in my experience of questions of this sort, I would say the ball completes a circle.
If that is the case then the ball has KE at point B, as well as PE.

 

[haruspex]

When the ball arrives at B the tension in the rope becomes zero. This does not necessarily mean that the ball has come to rest.
It is not 100% clear what the question is asking but, in my experience of questions of this sort, I would say the ball completes a circle.
If that is the case then the ball has KE at point B, as well as PE.

I think you've made the same presumption I made in my first post on this thread, that the poster has made the usual mistake of forgetting about centripetal acceleration. But the question posed concerns the potential energy at B. duplaimp appears to have measured the PE at A relative to the bottom of the loop but that at B relative to A.

 

[technician]

Haruspex: I agree, some information missing. My inclination is that it would be a mistake to assume that the ball comes to rest at B.

 

[PeterO]

I think you've made the same presumption I made in my first post on this thread, that the poster has made the usual mistake of forgetting about centripetal acceleration. But the question posed concerns the potential energy at B. duplaimp appears to have measured the PE at A relative to the bottom of the loop but that at B relative to A.

This is the problem with the Original Post! There is no question posed !

 

[haruspex]

This is the problem with the Original Post! There is no question posed !

It doesn't say what question the poster was trying to answer, but it does pose this question from the poster:

U1 = mgR
U2 = mgR

But in a solution I found it was used U2 = mg2R

Why it was used 2R and not R?

Not wonderfully clear, I agree, but I think there's enough circumstantial evidence to interpret U1 and U2 as representing the PEs at A and B respectively. If so, the error is evident.

 

[PeterO]

It doesn't say what question the poster was trying to answer, but it does pose this question from the poster:

Not wonderfully clear, I agree, but I think there's enough circumstantial evidence to interpret U1 and U2 as representing the PEs at A and B respectively. If so, the error is evident.

Understood. Seems OP was unaware that the standard formula U = mgh had had the h value replaced by "the height above lowest point" value - R and 2R respectively.

 

[rude man]

Let
vi = initial velocity at A
vf = final velocity at B

EDIT:

Technician is right, the velocity at B is not zero. Matter of fact, it's v = vf = sqrt(gR). That result should be obvious: at the top, the centripetal force still has to be mv^2/R and that has to be equal to mg + T, but T(B) = 0.

So we have the complete picture: 1/2 mvi^2 = 1/2 mvf^2 + mgR. With vf known, vi is immediately determined.

There is no information missing.

 

[PeterO]

Let
vi = initial velocity at A
vf = final velocity at B

I think electrician is right, the velocity at B is not zero. Matter of fact, it's v = vf = sqrt(gR). That result should be obvious: at the top, the centripetal force still has to be mv^2/R and that has to be equal to mg + T, but T(B) = 0.

So we have the complete picture: 1/2 mvi^2 = 1/2 mvf^2 + mgR. With vf known, vi is immediately determined.

Good work, electrician.

It is not clear that OP had any difficulty coping with the centripetal acceleration etc - he just wondered why U was mgR in one position and mg2R in the other. Presumably he had forgotten that the R and 2R referred to the height above a reference point, not the distance from the centre.

 

[rude man]

It is not clear that OP had any difficulty coping with the centripetal acceleration etc - he just wondered why U was mgR in one position and mg2R in the other. Presumably he had forgotten that the R and 2R referred to the height above a reference point, not the distance from the centre.

So it's agreed there is no "missing information"?

The only possible questions are what are the initial & final velocities & I have answered both.

 

[PeterO]

So it's agreed there is no "missing information"?

The only possible questions are what are the initial & final velocities & I have answered both.

I eventually found the OP's question.

It is at the bottom

"But in a solution I found it was used U2 = mg2R

Why it was used 2R and not R? The distance to use in a circle is from the center to the particle or is from the bottom to the particle?
Because as A is in a distance R from center and B too, I tried with R for both but it seems that it needs to be done using a distance from bottom"

 

[rude man]

I eventually found the OP's question.

It is at the bottom

OK, so that's easy to find when you know the velocity at B.

 

[PeterO]

OK, so that's easy to find when you know the velocity at B.

You don't need to find the velocity anywhere - you just have to realize that the potential energy U is calculated using the vertical displacement above some reference point/level - and the the solution OP had seen that reference level was the bottom of the circle.
Point A is One Radius above the reference point, so the usual mgh had become mgR
Point B is two Radii above the reference point, so the usual mgh had become mg2R.