Formulating the kinetic energy of an object - helical motion

  • #1
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Hi guys, I need your support to formulate the kinetic energy of an object:

- having mass m [Kg]
- rotating with angular velocity o [rad/sec] referred to an axis t [m] distant (and parallel) to the symmetry axis of the object
- moving along the direction of its symmetry axis with a costant velocity w [m/sec]

We can say that the motion describes an Helix.

Now, is it possible to write the kinetic energy making a sum of the rotating energy plus the translating energy, as Ek = 1/2*I*o^2 + 1/2*m*w^2 ? With I the inertia of the object calculated through the Huygens-Steiner Theorem for a parallel axis.
Please consider any object you like (sphere, cylinder..). It´s clear that the peripheral velocity v=o*t [m/sec] and w [m/sec] are ortogonal to each other, so what is the consequent formulation for the kinetic energy?

Thanks in advance!
Ennio
 
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  • #2
jfizzix
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In general, the kinetic energy of any object can always be broken up into two terms. The first is the kinetic energy of the center of mass, and the second is the kinetic energy of the object relative to its center of mass. The reason this is so is because the location of any point in an object can be written as the vector sum of the position of the center of mass of the object, and the position of this point, relative to the center of mass.

In this case, you would first find the velocity of the center of mass [itex]\vec{v}_{CM}[/itex], and use its magnitude square to get the kinetic energy (with the other factors).

[itex] KE_{CM} = \frac{1}{2} m |\vec{v}_{CM}|^{2}[/itex]

The kinetic energy of a rigid object relative to its center of mass is described with its angular velocity [itex]\vec{\omega}[/itex] relative to the center of mass, which requires knowing the moment of intertia tensor relative to the center of mass:

[itex] KE_{ relative}=\frac{1}{2} \vec{\omega}\cdot\mathbf{I}_{CM}\cdot\vec{\omega}[/itex]

Since the moment of inertia is a tensor (i.e., a matrix), we can take dot products on both the left and right side, where on the left is multiplication by a row vector, and on the right is multiplying by a column vector. Here, [itex]\vec{\omega}[/itex] is a vector pointing along the axis of rotation (in the direction given by the right hand rule), and with magnitude given by angular velocity in radians per second.
 
  • #3
18
2
Thanks for replying jfizzix!

Please see my Sketch below. Can we break up the kinetic energy to put it than together? Or it makes no sense.
KE = KEm + KErel
KE = 1/2*m*w^2 + (1/2*m*r^2 + m*r^2) (FYI the rotation axis was not meant the symmetry axis of the object)
Can it represent the total EK, or is it exactly as you have written in your previuous comment?

By the way sorry I do not yet know how to add ormulas
Thanks again

upload_2018-8-20_12-34-6.png





In general, the kinetic energy of any object can always be broken up into two terms. The first is the kinetic energy of the center of mass, and the second is the kinetic energy of the object relative to its center of mass. The reason this is so is because the location of any point in an object can be written as the vector sum of the position of the center of mass of the object, and the position of this point, relative to the center of mass.

In this case, you would first find the velocity of the center of mass [itex]\vec{v}_{CM}[/itex], and use its magnitude square to get the kinetic energy (with the other factors).

[itex] KE_{CM} = \frac{1}{2} m |\vec{v}_{CM}|^{2}[/itex]

The kinetic energy of a rigid object relative to its center of mass is described with its angular velocity [itex]\vec{\omega}[/itex] relative to the center of mass, which requires knowing the moment of intertia tensor relative to the center of mass:

[itex] KE_{ relative}=\frac{1}{2} \vec{\omega}\cdot\mathbf{I}_{CM}\cdot\vec{\omega}[/itex]

Since the moment of inertia is a tensor (i.e., a matrix), we can take dot products on both the left and right side, where on the left is multiplication by a row vector, and on the right is multiplying by a column vector. Here, [itex]\vec{\omega}[/itex] is a vector pointing along the axis of rotation (in the direction given by the right hand rule), and with magnitude given by angular velocity in radians per second.
 

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  • #4
A.T.
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Ennio said:
KE = 1/2*m*w^2 + (1/2*m*r^2 + m*r^2)

upload_2018-8-20_12-34-6-png.png
Doesn't look right to me. The parameters o and t don't even show up. The units of the second term are wrong. Is the object a solid cylinder, or just a cylindrical shell?
 

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  • #5
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It´s a solid cylinder and your are right:
KE = 1/2*m*w^2 + 1/2*(1/2*m*r^2 + m*r^2)*o^2
where I = 1/2*m*r^2 + m*r^2 is the inertia
Is the formulation correct? Can we sum up these two kinetic terms?

Reference https://www.physicsforums.com/threads/formulation-kinetic-energy-helix-motion.953647/


Doesn't look right to me. The parameters o and t don't even show up. The units of the second term are wrong. Is the object a solid cylinder, or just a cylindrical shell?
 
  • #6
A.T.
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KE = 1/2*m*w^2 + 1/2*(1/2*m*r^2 + m*r^2)*o^2
where I = 1/2*m*r^2 + m*r^2 is the inertia
I still see no t in there.
 
  • #7
18
2
I still see no t in there.
typing error --> Now KE = 1/2*m*w^2 + 1/2*(1/2*m*r^2 + m*t^2)*o^2 : )
Please is the formulation/concept correct ?Can we sum up these two kinetic Terms? A.T. thanks in adv.
 
  • #8
A.T.
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typing error --> Now KE = 1/2*m*w^2 + 1/2*(1/2*m*r^2 + m*t^2)*o^2 : )
Please is the formulation/concept correct ?Can we sum up these two kinetic Terms? A.T. thanks in adv.
Try it out. Dissolve the bracket, and combine the terms related to the two components of the COMs linear velocity. Compare with the formula given by @jfizzix .
 
  • #9
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2
Try it out. Dissolve the bracket, and combine the terms related to the two components of the COMs linear velocity. Compare with the formula given by @jfizzix .
Hi @A.T. , acc. to @jfizzix --> Kinetic terms separated: KEm = 1/2*m*(v^2+w^2) = 1/2*m*t^2*o^2 + 1/2*m*w^2 Joule where v=0*t m/sec
and KEr = 1/2 * 1/2*m*r^2 * o^2 = 1/4*m*r^2*o^2 Joule
It´s the same compared to my calculation, except the fact that I sum the terms!
KE = 1/2*m*w^2 + 1/2*(1/2*m*r^2 + m*t^2)*o^2 =
= 1/2*m*w^2 + 1/4*m*r^2*o^2 + 1/2*m*t^2*o^2

My last question is: makes sense to sum the kinetic Terms in order to write KEtot = KEm + KEr ?

Thanks again
 
  • #10
A.T.
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My last question is: makes sense to sum the kinetic Terms in order to write KEtot = KEm + KEr ?
Yes.
 

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