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I think I can calculate its rotational kinetic energy but at first we still need to find where center of mass is. Does anyone have these parameters?
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Rotational KE doesn't work here, the fist itself isn't spinning, it's just traveling in an arc
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I didn't get it. It's the reason to use rotational KE. Do you mean by spinning rotating around its center or what?(sorry if I misunderstood)
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Yeah, that's what I mean. The fist isn't spinning about it center
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The fist doesn’t need to move about its center for rot KE to be the more valid move. If it’s moving with circular motion, and the center of the imaginary circle that the moving object is drawing is stationary, then yeah use rot KE. Think like attaching a ball to a string and spinning it round and round, there you’d use rot KE.
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Correct but small nitpick. It is not more valid. Because if done correctly both should give the same result.
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No, using linear KE formula gives different result than rotational KE formula, for example rotating rod around its end will be mv²/8 for linear KE and mv²/6 for rotational.(v here is speed of its rotating end)
So yeah it's more valid
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If the object was rotating around its own axis then it would not be another way to calculate same thing. We would have to add it to the linear kinetic energy and find a total KE.
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No. I said it roates around its end like in Arc7Kuroi's example
To make this for finding total KE there should be both rotational and linear motion. But there is only rotational.
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I didn't say that it does either
There isn't any linear motion to calculate linear KE. Here total KE is just equal to rotational KE
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Here, Ron Brown's answer. Basically angular KE is when object rotates around its own axis. If it does not it is linear KE. That linear KE can be calculated as angular KE using 1/2 I w^2 but it is still linear KE.
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Not really. Here for total KE there is an example of rolling ball. Yeah, here It's rotating around its own axis but it doesn't mean "if it doesn't rotate around its own axis, then its linear KE, not rotational".
That formula you wrote isn't even for linear KE, according to your source too.
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I never said it was. It is clearly angular KE. What I said is that it is basically linear motion in a angulat plane. It is much easier to just calculate 1/2 m v^2. That gives the same result as 1/2 I w^2 but you don't have to calculate I this way.
And here from the source:
"Yes, an object (a car) traveling in a circle can be said to have rotational kinetic energy - but it is just the kinetic energy of motion as it goes around the circle.".
As you can see it is just regular KE but it can be written in a angular KE formula. But is can be calculated much easier with regular KE formula.
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Both of these are angular. Linear KE is the basic 1/2 m v^2. For angular KE you need to find I, the moment of inertia and w, angular velocity. Both of these depend on radius of rotation so of course angular KE would be different depending on the axis of rotation.
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Center of mass will have velocity of v/2, which means KE is mv^2/8
1/2 * (1/3 * m * L^2) * w^2 = 1/6 * m *v^2
As you can see this is a different result
I already said it's full length as he rotates around its end
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Why? Center of mass travels the same distance as rest of the fist. There is no need to cut it in half.
Where did you get I and w from? If you do not show that how can we see if it is correct? But from what I see you think that w and v are the same. They are not. V is m/s and w is radyan.
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No? Just end of rod travels maximal distance here, if we look at mass center then radius will be 2 times less, so does distance and speed(only angular velocity remains same) so we need to divide by 2
I is 1/3 * m * L^2 when rod rotates around its end
Since L * w = v( here as I said v is speed of its rotating end), then L^2 * w^2 = v^2
Also w is not rad, it's rad/sec
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What? Again where did that /2 come from. Center of the mass is on the arc fist travels. It still travels the same distance with rest of the fist.
It does not rotate around its end. It rotates around L distance. A completely different I is necessary.
Yeah I missed that.
Yes, after all it is velocity. I just forgot to add "/sec".
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Because distance to the axis is 2 times less than other end of it? It's same 180 degree arc but with 2x less radius, and so 2x less distance ( also here I'm talking about a rod-like object to say its mass center is in the middle) almost every part of fist has different speed value
L distance from what? It rotates around L/2 distance from mass center (again, here I'm talking about rod for simplification, just to show results aren't same)
For example I gave, no. But for the feat itself, yeah a different I is necessary.
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Does not matter. Center of mass itself is on the arc.
L as in arm lenght. But dies not matter.
It is different for the example too. In your example when you calculate linear KE you assume a mass going around a center in a circular path. When you calculate angular KE you assume mass is not going around a circular path, rather it goes around itself.
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But as I said none of that matters. KE of an object in circular motion is 1/2 m v^2.
Here another source.
Here another source.
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Rot. KE formula considers how the mass of the rod is distributed around the axis of rotation. It accounts for the fact that different parts of the rod are moving at different linear speeds, proportional to their distance from the axis of rotation.
Linear KE formula treats the rod as if all its mass were concentrated at its center of mass, moving at the same speed. It does not account for the rotational motion around the axis.
Like it could work if KE itselft was linear with speed, but since it's not, even if we use speed of center of mass for linear KE formula, those far from the middle cannot properly balance those closer from the middle (because their average KE won't be equal to KE of the middle). So taking full mass and speed of mass center won't give us sum of KE of all particles.
If you want a simply example, think of rod as particles with equal masses. If KE was linear with speed(let's say mv/2), average KE of particles 0.49 L and 0.51 L distances far from center of rotation will be equal to the middle one's. So do 0.48 and 0.52 and so on. (Of course 0.01 difference is just for example) then if we have N particles, and average speed of them all is speed of the middle particle (let's say v), we can say that sum of all their speed is v * N, if from formula we write 1/2 *m * v * N to find sum of all their KE(where m is mass of 1 particle) m * N will be total mass of the rod. As you see using linear KE worked but that won't be true as KE isn't linear with distance over time. In this case it should be linear with distance from axis.
Another one can be when you use average speed to find distance. If speed is increasing linearly, let's say from 0 to v at t time, its average speed will be v/2 at t/2 time so you can just find d = v/2 * t.(it's like that bc just like in example above, average speed of 0.25t and 0.75 t is speed at 0.5t and so on) However same thing won't be true when speed will increase with v=t^2 function as at t/2 time it won't be average speed and we can't find distance from that. What I want to say is just think of t axis as length of rod and v axis as KE from the graphic. In short If we look at all dots in the graph, dividing sum of all their speed by their number won't give us speed at t/2 time.
Sorry if it's complicated but I don't thing I can explain in another way at least for now
Linear KE formula treats the rod as if all its mass were concentrated at its center of mass, moving at the same speed. It does not account for the rotational motion around the axis.
Like it could work if KE itselft was linear with speed, but since it's not, even if we use speed of center of mass for linear KE formula, those far from the middle cannot properly balance those closer from the middle (because their average KE won't be equal to KE of the middle). So taking full mass and speed of mass center won't give us sum of KE of all particles.
If you want a simply example, think of rod as particles with equal masses. If KE was linear with speed(let's say mv/2), average KE of particles 0.49 L and 0.51 L distances far from center of rotation will be equal to the middle one's. So do 0.48 and 0.52 and so on. (Of course 0.01 difference is just for example) then if we have N particles, and average speed of them all is speed of the middle particle (let's say v), we can say that sum of all their speed is v * N, if from formula we write 1/2 *m * v * N to find sum of all their KE(where m is mass of 1 particle) m * N will be total mass of the rod. As you see using linear KE worked but that won't be true as KE isn't linear with distance over time. In this case it should be linear with distance from axis.
Another one can be when you use average speed to find distance. If speed is increasing linearly, let's say from 0 to v at t time, its average speed will be v/2 at t/2 time so you can just find d = v/2 * t.(it's like that bc just like in example above, average speed of 0.25t and 0.75 t is speed at 0.5t and so on) However same thing won't be true when speed will increase with v=t^2 function as at t/2 time it won't be average speed and we can't find distance from that. What I want to say is just think of t axis as length of rod and v axis as KE from the graphic. In short If we look at all dots in the graph, dividing sum of all their speed by their number won't give us speed at t/2 time.
Sorry if it's complicated but I don't thing I can explain in another way at least for now
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This is just a dot it doesn't have "different speeds"
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It still travels less distance
No I actually stated so many times around what it's rotating
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Look, I have shared sources saying they are the same thing. This is the third one.
Linear KE account for difference in speed with using center of mass. Thanks to that you do not have calculate moment of inertia.
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Probably it's a point object. I already showed a rotating rod example where linear and rotational KEs were different
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They are always calculated as point objects. Because center of mass is a point.
And your example is wrong. You calculate two different things.
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Whole object isn't a point though. You should look at point object with some mass to use linear KE
I just wanted to explain that we need to sum up KEs of all particles. But its mass center at the middle hasn't average KE among all of them as KE isn't increasing linearly as you get futher away. It should be their average to use total mass via multiplying KE by number of particles
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Of course it is not. Real objects occupy a volume in space. But if you use the center of mass you can calculate it as point object. That is the whole deal of center of mass. Sorry but do you actually understand what you are trying to say here? You are trying to say that center of mass can't be used to calculate KE because it is just a point when that is how it is always calculated. Center of mass accounts for all the mass and shape of the object.
Ok. Hope you sleep well.
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Ok, look at this. I will explain with 5 points for simplification.
Think of rod which consists of 5 points with equal mass(just remember that number of points won't change the reasoning here)
Now, x axis here is distance from center of rotation and y axis is KE. Center of mass is L/2 distance far from center.
Since KE ~ v^2 ~ R^2 we can say that if second point's KE is E, others will be 4E, 9E and 16E.(first one's is 0)
Now let's look at what happens if you treat center mass as whole object trying to calculate with linear KE formula:
Mass center's speed will be v, whole mass will be 5m, so KE of rod = 5m * v^2 / 2.
Since we took 4E for mv^2/2 here KE equals to 20E
But it's less than actual KE of rod as 0+E+4E+9E+16E = 30E
This happens because center of mass have average speed among all of the points of object, not average KE. That's why I was saying it could be true if KE was linear with speed, but it increases not linearly, making "point with average KE" being more away from center of rotation than mass center( d > L/2 ). But you could use linear KE formula for that point though.
Think of rod which consists of 5 points with equal mass(just remember that number of points won't change the reasoning here)
Now, x axis here is distance from center of rotation and y axis is KE. Center of mass is L/2 distance far from center.
Since KE ~ v^2 ~ R^2 we can say that if second point's KE is E, others will be 4E, 9E and 16E.(first one's is 0)
Now let's look at what happens if you treat center mass as whole object trying to calculate with linear KE formula:
Mass center's speed will be v, whole mass will be 5m, so KE of rod = 5m * v^2 / 2.
Since we took 4E for mv^2/2 here KE equals to 20E
But it's less than actual KE of rod as 0+E+4E+9E+16E = 30E
This happens because center of mass have average speed among all of the points of object, not average KE. That's why I was saying it could be true if KE was linear with speed, but it increases not linearly, making "point with average KE" being more away from center of rotation than mass center( d > L/2 ). But you could use linear KE formula for that point though.
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Not because "it's just a point", using linear KE on mass center won't give true result as center mass doesn't have average KE just to use overall mass with its speed(which is the same thing to multiply its KE by number of points)
Linear KE formula doesn't account for mass distribution. But rotational KE does
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Again, how are you saying something so wrong about something as basic as center of mass.
Here its definition: "The center of mass is the unique point at the center of a distribution of mass in space that has the property that the weighted position vectors relative to this point sum to zero.".
It is all about distribution. Its position changes depending on the shape of the object.
But this really turned into me trying to explain basic concepts.
Look, just google how to calculate kinetic energy in circular motion and look at the results. Every single one will use classic KE formula with center of mass.
I gave 3 sources and you are still using your faulty example as a reason to dismiss them. I really don't want to argue with you any longer.
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