Total Kinetic Energy of a Body

[suffian]

[SOLVED] Total Kinetic Energy of a Body

I was reading my physics text and it came up with the standard formula for the kinetic energy of a rigid body in motion: the sum of the translational and rotational kinetic energy. That got me wondering whether we could include the internal energy by simply taking the sum of the kinetic energy of each particle when viewed in a frame where the translational and rotational velocity appear to be zero. In other words the total kinetic energy of the body should be (where v_Hi is the velocity in addition to trans and rot):

$$K = \frac{1}{2}\,Mv_\text{cm}^{\;\; 2} + \frac{1}{2}\,I_\text{cm}\,\omega^2 + \sum\frac{1}{2}\, m_i v_\text{Hi}^{\;\; 2} = \frac{1}{2}\,Mv_\text{cm}^{\;\; 2} + \frac{1}{2}\,I_\text{cm}\,\omega^2 + H$$

So I took a hand at the calculations, but the best I could seem to come up with was the following:

$$K = \frac{1}{2}\,Mv_\text{cm}^{\;\; 2} + \frac{1}{2}\,I_\text{cm}\,\omega^2 + H + \omega \, \cdot \, \sum m_i \, (r_i \times v_\text{Hi})$$

In the last term the quantities are vectors. This would seem to imply that the total kinetic energy of a body is more or less than the sum of its translation, rotational, and internal energy (assuming the internal energy is as i have defined it). Have I made some sort of mistake or is there some better way of looking at it?

 

[eljose79]

Hello,

Thank you for bringing up this interesting question about the total kinetic energy of a body. I would like to offer some insights and clarification on this topic.

Firstly, let's define the terms used in the formula you provided. The translational kinetic energy (Kt) is the energy associated with the linear motion of an object, and it is given by Kt = 1/2 mv^2, where m is the mass of the object and v is its velocity. On the other hand, rotational kinetic energy (Kr) is the energy associated with the rotational motion of an object, and it is given by Kr = 1/2 Iω^2, where I is the moment of inertia and ω is the angular velocity of the object.

Now, let's consider the internal energy (Hi). This is the energy associated with the random motion of particles within the body, and it is typically taken into account when calculating the total energy of a system. However, it is important to note that the internal energy is not directly related to the motion of the body as a whole, but rather to the motion of the individual particles within the body. Therefore, it cannot be simply added to the translational and rotational kinetic energies to obtain the total kinetic energy of the body.

In your formula, you have considered the velocity of each particle (vHi) in a frame where the translational and rotational velocities are zero. However, this does not accurately represent the internal energy of the body. The internal energy of a body is a function of temperature, and it is related to the average kinetic energy of the particles, not their individual velocities. Therefore, your formula does not accurately represent the total kinetic energy of a body.

In conclusion, the correct formula for the total kinetic energy of a rigid body in motion is indeed the sum of the translational and rotational kinetic energies, as given by K = 1/2 Mvcm^2 + 1/2 Icmω^2. The internal energy cannot be directly added to this formula, as it is not related to the motion of the body as a whole. I hope this helps to clarify your doubts. Keep up the curiosity and good luck with your studies!