Formulating the kinetic energy of an object - helical motion

In summary, the kinetic energy of any object can be broken down into two terms: the kinetic energy of the center of mass and the kinetic energy of the object relative to its center of mass. For a solid cylinder rotating with angular velocity o and moving along its symmetry axis with constant velocity w, the kinetic energy can be written as 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 moment of inertia.
  • #1
Ennio
26
2
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
 
Last edited by a moderator:
Physics news on Phys.org
  • #2
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
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


jfizzix said:
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.
 

Attachments

  • upload_2018-8-20_12-34-6.png
    upload_2018-8-20_12-34-6.png
    8.9 KB · Views: 652
  • #4
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?
 

Attachments

  • upload_2018-8-20_12-34-6-png.png
    upload_2018-8-20_12-34-6-png.png
    8.9 KB · Views: 583
  • #5
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/
A.T. said:
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
Ennio said:
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
A.T. said:
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
Ennio said:
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
A.T. said:
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
Ennio said:
My last question is: makes sense to sum the kinetic Terms in order to write KEtot = KEm + KEr ?
Yes.
 

What is kinetic energy?

Kinetic energy is the energy an object possesses due to its motion. It is the energy that an object has when it is moving.

What is helical motion?

Helical motion refers to the circular or spiral motion of an object moving along a curved path. It combines both rotational and translational motion.

How do you formulate the kinetic energy of an object in helical motion?

The kinetic energy of an object in helical motion is given by the formula KE = (1/2)mv2 + (1/2)Iω2, where m is the mass of the object, v is its linear velocity, I is its moment of inertia, and ω is its angular velocity.

What factors affect the kinetic energy of an object in helical motion?

The kinetic energy of an object in helical motion is affected by its mass, linear velocity, angular velocity, and moment of inertia. Additionally, factors such as air resistance and friction may also affect the kinetic energy.

Why is understanding the kinetic energy of an object in helical motion important?

Understanding the kinetic energy of an object in helical motion is important because it helps us predict the motion of objects and calculate the amount of energy needed to move them. This information is crucial in various fields such as physics, engineering, and sports.

Similar threads

Replies
16
Views
1K
Replies
5
Views
787
  • Mechanics
Replies
12
Views
1K
Replies
1
Views
407
Replies
2
Views
1K
Replies
40
Views
2K
Replies
5
Views
703
Replies
30
Views
2K
  • Mechanics
Replies
20
Views
860
Replies
5
Views
810
Back
Top