How does the kinetic energy of a swinging ball change with respect to height?

In summary, the conversation discusses the concept of conservation of energy in a system involving a ball on a string swinging vertically. The conversation also touches on the relationship between tangential and angular velocity, and the complexity of understanding energy in a rotating system. The original question involves two balls of different masses attached by a string and rotating vertically, and the conversation delves into potential and kinetic energy changes in this scenario.
  • #1
Rainbowbarf756

Homework Statement


Imagine a ball on a string that we swing vertically so that the hight changes. By conservation of energy the velocity of the ball must change right? Because at the highest point of the swing it will have maximum GPE but at the bottom, minimum right? Watching many videos has just left me more confused as they say tangential velocity is constant but angular velocity will change but the ball is not rotating itself, plus the angular velocity only changes with the radius, which is kept constant. Please help I am beyond confused. It's more of a personal project than homework but it's doing my head in
TL;DR: How does kinectic energy of a ball swinging in a full vertical circle change with respect to hight?

Homework Equations


E=1/2*m*v^2
E=1/2*I*Omega^2 (I know very little about this equation, not on my syllabus and can't really understand anything online about it)
Omega = 2*pi*f
f =m*v^2/r

The Attempt at a Solution


I found how gpe changes with respect to hight as (1-cos(theta))*radius of rotation or (1-cos(2pi*time/period))*radius of rotation I was thinking maybe I could use this with gpe(before) + ke(before) = gpe(after) + ke(after) but this results in a violation of the conservation of energy...
 
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  • #2
Rainbowbarf756 said:

Homework Statement


Imagine a ball on a string that we swing vertically so that the hight changes. By conservation of energy the velocity of the ball must change right? Because at the highest point of the swing it will have maximum GPE but at the bottom, minimum right?
Yes, assuming no energy is added to or subtracted from the system during the motion. Often textbooks do consider vertical circular motion to have constant speed. This would require input of energy while the ball is rising and ouput of energy while coming down.
Watching many videos has just left me more confused as they say tangential velocity is constant but angular velocity will change
That's strange. The tangential velocity will always change in circular motion due to change in direction of the velocity. (Velocity is a vector quantity.) But the speed might be constant. But then the angular velocity would also be constant. Can you give a link to one of these videos?
 
  • #4
I appreciate the help btw ^~^
 
  • #5
Rainbowbarf756 said:
https://www.khanacademy.org/science...m/torque-tutorial/v/rotational-kinetic-energy
- I get that the velocity changes direction but in the ke equation we use the magnitude of v right? I mean how would the kinectic energy vary at an angle of 0 compared to an angle of pi/2 if the magnitude is still constant?
I looked at this video, but it didn't have anything about a ball on a string traveling overhead in a circle.
 
  • #6
Yes I was watching it to help understand how kinectic energy changes with rotation, the overall question I'm trying to answer is actually more complex than stated above: If two balls of mass m1 and m2 and diameter d1 and d2 are attached by a string and the system rotates vertically how does gpe and ke change over time, I have worked out how hight changes with respect to time of the projectile, found how the CoM is defined using the diameters and masses of the balls the next step I think is to find how the gpe and ke changes in one of the rotations then apply that to the second mass and merge all my equations together somehow but I think my problem lies in the understanding in the energy in a rotating system.
 
  • #7
Rainbowbarf756 said:
Yes I was watching it to help understand how kinectic energy changes with rotation, the overall question I'm trying to answer is actually more complex than stated above: If two balls of mass m1 and m2 and diameter d1 and d2 are attached by a string and the system rotates vertically how does gpe and ke change over time, I have worked out how hight changes with respect to time of the projectile, found how the CoM is defined using the diameters and masses of the balls the next step I think is to find how the gpe and ke changes in one of the rotations then apply that to the second mass and merge all my equations together somehow but I think my problem lies in the understanding in the energy in a rotating system.
I am having trouble visualizing the problem you are describing. Can you please provide a diagram.
 
  • #8
The whiteboard is the original question the paper is my generalisation
IMG_20170927_202811.jpg
IMG_20170926_151137.jpg
 
  • #9
Rainbowbarf756 said:
The whiteboard is the original question the paper is my generalisationView attachment 211839 View attachment 211840
The whiteboard question is unclear. Is there anything holding the whole system up, against gravity, or is the pair of rotating balls in free fall?
 
  • #10
Imagine it rotates forward I guess but ultimately I will model it as a projectile with initial velocity u and at an angle theta
 
  • #11
For the time being I was thinking of it where it has gpe but is not affected by gravity does that make sense? So it doesn't fall.
 
  • #12
Rainbowbarf756 said:
For the time being I was thinking of it where it has gpe but is not affected by gravity does that make sense? So it doesn't fall.
No, that does not make sense. If there is gravity but the system is not falling there must be another force somewhere, and that will change things.
 
  • #13
In that case imagine it as a projectile hight changes with respect to time as u*sin(theta) -g*t
 
  • #14
Rainbowbarf756 said:
In that case imagine it as a projectile hight changes with respect to time as u*sin(theta) -g*t
You mean, the vertical velocity follows that equation.
Ok, so you can separate the motion of the mass centre of the system, a simple parabolic trajectory, from the motion of the two balls relative to that.
 
  • #15
Yeah but then I need to consider the gpe and ke change of the projectile motion (pretty easy) but the part I do not understand is how the gpe and ke would change for the individual ball motions and then how to put that with the projectile motion
 
  • #16
Rainbowbarf756 said:
Yeah but then I need to consider the gpe and ke change of the projectile motion (pretty easy) but the part I do not understand is how the gpe and ke would change for the individual ball motions and then how to put that with the projectile motion
A mere matter of addition. Having found the GPE of the mass centre as a function of time, and the relative GPEs of the balls as a function of time, add them together,
You can do the same with the velocities, adding them vectorially.
 
  • #17
So the kinetic energy of the ball will just be it's gpe plus the ke of the CoM?
 
  • #18
Rainbowbarf756 said:
So the kinetic energy of the ball will just be it's gpe plus the ke of the CoM?
No, that's not what I wrote.
The GPE of the system will be the GPE of the common mass centre. To find the GPE of the individual balls, apportion that in proportion to the masses, then adjust by their heights relative to the common mass centre.
To find the KE of a ball, the safest way is to find its velocity in the ground frame. To find that, find the velocity of the common mass centre then add (vectorially) its velocity relative to that.
 
Last edited:
  • #19
The tangential velocity of the ball relative to the CoM? So if CoM was going 10 and the ball had tangential velocity of 12 it's relative velocity would be 2?
 
  • #20
Rainbowbarf756 said:
The tangential velocity of the ball relative to the CoM? So if CoM was going 10 and the ball had tangential velocity of 12 it's relative velocity would be 2?
Velocity is a vector, it has a direction as well as a magnitude. You understand how to add vectors, yes?
 
  • #21
Yes okay so the velocity of the projectile is just ucos and usin but how about the ball how do I calculate that?
 
  • #22
Rainbowbarf756 said:
Yes okay so the velocity of the projectile is just ucos and usin but how about the ball how do I calculate that?
Can you calculate what that would be without gravity?
 

1. What is kinetic energy?

Kinetic energy is the energy an object possesses due to its motion. It is defined as the work needed to accelerate a body of a given mass from rest to its current velocity.

2. How is kinetic energy related to the height of a swinging ball?

The kinetic energy of a swinging ball is directly proportional to its height. This means that as the height of the ball increases, so does its kinetic energy. This is because at a higher height, the ball has more potential energy, which is converted into kinetic energy as it swings downwards.

3. Does the mass of the ball affect its kinetic energy when swinging?

Yes, the mass of the ball does affect its kinetic energy when swinging. The kinetic energy of a swinging ball is directly proportional to its mass. This means that a heavier ball will have more kinetic energy than a lighter ball when swinging at the same height.

4. How does the speed of the swinging ball change with respect to height?

The speed of a swinging ball changes with respect to height in a non-linear manner. As the ball swings downwards, its speed increases due to the conversion of potential energy into kinetic energy. However, as it swings back up, the speed decreases as the kinetic energy is converted back into potential energy.

5. Is there a maximum height at which the swinging ball will have the most kinetic energy?

Yes, there is a maximum height at which the swinging ball will have the most kinetic energy. This height is the point at which the ball has the most potential energy, and therefore will have the most kinetic energy when it swings downwards. However, as the ball swings back up, the kinetic energy will decrease again as the potential energy is converted back.

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