# Kinetic Energy ratio

1. Nov 30, 2004

### americanangel

Alright, I have a test coming up next week and I was reading through my phsyics book and the text mentioned the ratio between the translation kinetic energy and the rotational kinetic energy. The only time it is mentioned is in one of the worked example in the text as a question "Find the ratio between the translational and rotational kinetic energy?"

What does that ratio actually mean? Is the sum of the translational and the rotational kinetic energy the work?

Thanks

2. Nov 30, 2004

### e(ho0n3

I suggest you google around if you don't know what the word 'ratio' means.

3. Nov 30, 2004

### Justin Lazear

The ratio of a to b is $\frac{a}{b}$.

--J

4. Nov 30, 2004

### americanangel

Either both of you do not understand the question or you enjoy making yourselves look stupid. I completely understand what a ratio is, you honestly don't get passed fourth grade math if you don't understand what a ratio is...My question was not what a ratio is, but rather what the ratio of the two represent. Apparently, no one else has read this post, since I did have anything question weaved in my original post. Honestly, if neither of you could be of an quality help, you should not be posting.

5. Nov 30, 2004

### Justin Lazear

The ratio of the translational and angular kinetic energies is

$$\frac{\mbox{translational KE}}{\mbox{angular KE}}$$

The work is defined in your book. It's doubtlessly also explained in your book. If you could read, you'd understand what it was.

Insulting people is fun, isn't it? So productive, too.

--J

6. Nov 30, 2004

### Silimay

The equation for translational kinetic energy is 1/2*m*velocity squared.
The equation for rotational kinetic energy, if I remember correctly, is 1/2*I*ω^2, where I is the moment of inertia and ω is the rotational speed. I hope that helps some. To get the total kinetic energy, just sum the two together. I think the ratio probably depends on the specific problem (what kind of objet is rolling, its mass, etc.).

7. Dec 1, 2004

### americanangel

Let me clarify, I didn't mean to insult anyone on the board I was only replying to the insults that I recieved and yes I do know how to read and do math. To make you feel a little better, I'm a double major in math and biochemsitry. I was asking you guys a question about the concept not the math, so no I did not appreciate the insult.

Why I asked the question was because the concept was not presented in the book. I had a feeling it was work, but there was no information in regards to work relating to rotational and translational energy in the text. The book explained the momemtum and energy concept but the only thing it said about work was that it was represented the change in kinetic energy, nothing about the ratio between the two forms of kinetic energy. I was asking the question to help clarify the information since my professor did not have any office hours this week. I thought that this board could have been of assistance

I thank you for the information posted within the insults, but apparently, this board is not run in a professional manner. Instead, you all enjoy making fun of those who actually need help. I say that not on the insults I recieved, but because of the overall tone of the board. The only time that you guys actually post some help is when you feel like it. I understand that we are all of the country and in same cases from different parts of the wall, but there should be a small degree of conduct and respect presented to each person in every post.

8. Dec 1, 2004

### Justin Lazear

The ratio of the translational and angular kinetic energies has no physical meaning other than that it is the ratio of translational and kinetic energies. It is most certainly not work, especially considering that the ratio is unitless, whereas work carries the units of energy.

The sum of the translational and angular KE is the total kinetic energy. In particular physical situations, it may have an equal magnitude as the work, but it is generally not.

--J

9. Dec 1, 2004

### americanangel

Thank you!!!

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