Kinetic energy of a disk that turn around 2 axis

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  • #1
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I would like to know if kinetics energy of a disk with radius 'r', mass 'm' which turn around itself at Wb (relative rotation) and which turn around blue axis at distance 'd' is :

1/2md²Wa²+1/2mr²(Wa−Wb)² ???

With Wa > Wb

Look at image please.
 

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  • #2
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The prefactor for the second term looks wrong. That would correspond to a ring instead of a disk.
 
  • #3
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A disk is not a sum of rings ? What is the result for a disk ? I'm looking for a link but it's very difficult to find on Internet
 
  • #4
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A disk can be described as a sum of rings of different radius, but it is not a ring with radius r. Check its moment of inertia.
 
  • #5
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But you're sure my formula is for a ring ?
 
  • #6
34,989
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It is valid for a ring. It is valid for other shapes as well, but not for your disk. The prefactor is wrong.
 
  • #7
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Have you a link where I can find formulas for different shapes ?
 
  • #9
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Ok, but I don't find the formula: 1/2md²Wa²+1/2mr²(Wa−Wb)²

I would like to find the demonstration of this formula to be sure it is correct at least for a ring.
 
  • #10
jbriggs444
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The point was that Google can find the moment of inertia of a ring or of a disc. ##mr^2## is the moment of inertia of a ring of mass m and radius r.

A demonstration I find convincing is to imagine that one is applying the work neccessary to spin this arrangement up.

Start with the rod locked in place and spin up the disk to a rotation rate of ##\omega_a - \omega_b##. How much energy does that take? It should be ##\frac{I_b(\omega_a - \omega_b)^2}{2}##. Where ##I_b## is the moment of inertia of a disk or ring of radius r. (Which you can look up on Google)

Now unlock the rod and spin it up to a rotation rate of ##\omega_a##. Because the pivot on which the disk/ring is mounted is frictionless and mounted at the center of mass of the disk/ring, the energy required to do this is independent of the size, shape or rotation rate of the object on the end of the rod. It depends only on the disk/ring's mass. How much energy is this? It should be ##\frac{I_a\omega_a^2}{2}## where ##I_a## is the moment of inertia of a mass at a distance d. By definition, that's ##md^2##.

Having done this, the relative rotation rate of the disk or ring with respect to the rod will be ##-\omega_b## and we are in the correct final configuration. The energy in the configuration is the total of the energy that went into spinning it up.

There is no requirement that ##\omega_a \gt \omega_b##. The formula works regardless.
 
  • #11
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Ok, but I don't find the formula: 1/2md²Wa²+1/2mr²(Wa−Wb)²
There is also no formula for tons of other cases. All the discussion here was about a specific part of the formula. And there is a formula for the moment of inertia of a spinning disk. And somewhere on that page or linked pages you can also find the formula for the energy of a spinning disk. Put both together with the angular velocity you found and you'll see the second term of your formula has a wrong prefactor.
 

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