Gravitational potential energy of ball

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SUMMARY

The discussion centers on calculating the gravitational potential energy (Ug) of a 7.26 kg ball suspended from a 2.5 m rope, pulled back to a 45-degree angle with the vertical. The correct formula to determine the height (h) is h = 2.5 m * (1 - cos(45)), which results in a gravitational potential energy of 52 J. The conversation highlights the importance of understanding the reference point for potential energy calculations and the role of trigonometry in determining vertical displacement.

PREREQUISITES
  • Understanding of gravitational potential energy (Ug = mgh)
  • Basic knowledge of trigonometry, specifically cosine functions
  • Familiarity with the concept of reference points in physics
  • Ability to interpret and draw geometric diagrams
NEXT STEPS
  • Study the application of trigonometric functions in physics problems
  • Learn about different reference points for potential energy calculations
  • Explore the concept of energy conservation in pendulum motion
  • Investigate the implications of changing reference points on potential energy values
USEFUL FOR

Students studying physics, particularly those focusing on mechanics, educators teaching energy concepts, and anyone interested in understanding gravitational potential energy calculations.

pebbles
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Homework Statement


A 7.26 kg ball hangs from end of 2.5 m rope. The ball is pulled back until rope makes 45 degree angle with vertical. What is the gravitation potential energy?


Homework Equations



ug=mgh

The Attempt at a Solution


I thought I would just multiply mass times gravity times height, but the answer is 52 J.
What do I do with the angle I'm given?
 
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What are the maximum and minimum vertical heights of the ball?
 
0 m and 2.5 m...?

I provided the entire problem.
 
In the back of the book it says to find height do
h=2.5m * (1-cos(45))
Then plug that into Ug=mgh.
Why 1-cos(45)?
 
pebbles said:
0 m and 2.5 m...?

I provided the entire problem.

No. The 2.5 m corresponds to the total height of the rope. The ball is not lifted so that it's level with the top of the rope- it's pulled along like a pendulum such that the rope is at 45 degrees to vertical.

Edit: Think of swinging your arm such that it's 45 degrees to vertical. How far has your hand moved vertically?
 
Ok, I think I understand. It's going to be hard to explain, but in other words, when the ball is in the air and has that 45 degree angle, it's going to "lose" height; it's going to be "shorter" than hanging straight down. So that's why 1 is being subtracted from cos(45).
 
I'm not sure I understand your explanation. At 45 degrees the ball is higher than completely vertical- so I'm not sure why you claim it loses height.

Try answering my question. If you swing your arm by 45 degrees from vertical how far has your hand moved vertically (upwards)? (You could take the length of your arm to be e.g. 1.1 meters.)
 
i don't know. I'm flustered and confused.
 
OK, calm down!

This problem involves a little trig. It's best to draw a diagram.

Draw a vertical line of length, e.g. 2.5m between points A and B (with A at the top, B at the bottom.)

Next- draw a line between points A and C which is of the same length, but at 45 degrees to the AB line.

Tell me once you've done this.
 
  • #10
pebbles said:

Homework Statement


A 7.26 kg ball hangs from end of 2.5 m rope. The ball is pulled back until rope makes 45 degree angle with vertical. What is the gravitation potential energy?


Homework Equations



ug=mgh

The Attempt at a Solution


I thought I would just multiply mass times gravity times height, but the answer is 52 J.
What do I do with the angle I'm given?

Just a comment: to be exact, the question does not make sense because the gravitational potential energy depends on the position of the origin so all answers are possible depending on where you set your origin. Did they mean to calculate the change of potential energy as the ball is pulled back?
 
  • #11
kdv: The gravitational potential energy refers to the potential energy that the ball can convert into kinetic energy without cutting the rope. If we allow the rope to be cut- then yes we would have to specify at what height we take as the zero.
 
  • #12
christianjb said:
kdv: The gravitational potential energy refers to the potential energy that the ball can convert into kinetic energy without cutting the rope. If we allow the rope to be cut- then yes we would have to specify at what height we take as the zero.

But it is still arbitrary even if the rope is not cut. One may define the gravitational potential energy to be zero at th elowest point of the swing or at the ceiling or at any other point. The value mgy never appears isolated in the equations, it's only the combination mg(y_i-y_f) (the work done by gravity) that is relevant. So if the question is worded exactly the way given in the OP and without specifying what origin must be used, then the question is actually meaningless. My bet is that they meant to ask what is the change of potential energy as the ball is pulled from its initial position to its final position. That would make sense as a question because a difference of potential energy is the same no matter where the origin is chosen.
 
  • #13
kdv said:
But it is still arbitrary even if the rope is not cut. One may define the gravitational potential energy to be zero at th elowest point of the swing or at the ceiling or at any other point. The value mgy never appears isolated in the equations, it's only the combination mg(y_i-y_f) (the work done by gravity) that is relevant. So if the question is worded exactly the way given in the OP and without specifying what origin must be used, then the question is actually meaningless. My bet is that they meant to ask what is the change of potential energy as the ball is pulled from its initial position to its final position. That would make sense as a question because a difference of potential energy is the same no matter where the origin is chosen.

This is nitpicking and isn't helping the thread. The potential in this case is clearly the amount that can be completely converted to kinetic energy at the vertical. We're all aware that energy has to be defined with respect to a reference point- but it's absolutely clear in this case what the reference point is! If you want to debate this further then message me.
 
  • #14
christianjb said:
This is nitpicking and isn't helping the thread. The potential in this case is clearly the amount that can be completely converted to kinetic energy at the vertical. We're all aware that energy has to be defined with respect to a reference point- but it's absolutely clear in this case what the reference point is! If you want to debate this further then message me.

Sorry. Ten years of teaching college physics has forced me to become careful about making sure things are clear in the students mind before they do a calculation. It did not seem nitpicking to me since it is the very definition of potential energy that it may be defined with respect to any origin. I thought explaining this to the OP would give him/her a better understanding of the concepts (and that tehre is not a unique answer here) but if it's nitpicking, I apologize.
 
  • #15
kdv said:
Sorry. Ten years of teaching college physics has forced me to become careful about making sure things are clear in the students mind before they do a calculation. It did not seem nitpicking to me since it is the very definition of potential energy that it may be defined with respect to any origin. I thought explaining this to the OP would give him/her a better understanding of the concepts (and that tehre is not a unique answer here) but if it's nitpicking, I apologize.

Sorry if I was a little harsh.
 
  • #16
christian-as you explained is how i drew my diagram in the first place, so it's done.
 
  • #17
OK, good. Now for the evil next step.

Draw a line horizontally from C so that it intersects the AB line. Tell me when you're done.
 
  • #18
Your diagram should look like a triangular flag on a post. Does it?
 
  • #19
forgot to mention, i drew the horizontal line previously also.

now, i'll apply trig-->
2.5(cos45)=1.7678 and that's the height from A to the horizontal line that intersects AB
 
  • #20
OK. label 'D' the intersection of the horizontal with AB

So AD=2.5 cos(45)

But what is the vertical height the ball has been raised equal to on the diagram?
 
  • #21
...2.5 m...?
 
  • #22
No. The ball started off at B and is now at C. You can think of the total displacement from B to C as a vertical movement of y meters followed by a horizontal movement of x meters. What are x and y on your diagram?
 
  • #23
y=D, x=C...? yikes!
 
  • #24
I think you've got it- but the length of a line between (e.g.) E and F is called EF.

so y=??
x=??
where ?? stands for 2 letters.
 
  • #25
oh ok-->
y=BD
x=DC
 
  • #26
OK, so you've worked out that AD=2.5cos(45), and we know that

AB=2.5

But the vertical change in the ball is equal to BD.

So can you work out the length BD, now that you know both AB and AD?
 

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