# Homework Help: Rotating ball and string Rotational Kinetic Energy?

1. Dec 8, 2013

### Vontox7

1. The problem statement, all variables and given/known data
Hello,
I am a bit confused on when rotational kinetic energy exists and when linear kinetic energy exists. For example when we spin a string with a ball attached to the end it has kinetic energy of 0.5mv^2 of the ball when we were solving for velocity of the ball at certain points in the vertical circle of the ball. But know that i learned rotational motion, why don't we consider rotational energy in that example instead of just mgh and 0.5mv^2? There was a question where there was a ball attached to a bar and the system of bar and ball attached to the bar are moving and we used just rotational kinetic energy to solve the question and didn't account for 0.5mv^2.

2. Relevant equations

3. The attempt at a solution
I was thinking maybe it has to do with something about the string maybe being negligible or something but still not sure. I would like some clarification on this matter please.

2. Dec 9, 2013

### tiny-tim

Hi Vontox7!
If the string has length R and the ball has radius r, then the KE is 1/2 mR2ω2 + 1/2 2/5 mr2ω2

= 1/2 m(R2 + 2/5 r22

r2 is usually so much smaller than R2 that the extra energy can be ignored.

3. Dec 9, 2013

### Vontox7

Okay so just simplifying KE = 1/2m(R^2+0)w^2 =1/2mR^2w^2 = 1/2mR^2(v^2/R^2) = 1/2mv^2 ?
Btw thank you very much for help it is greatly appreciated !

4. Dec 9, 2013

### haruspex

Strictly speaking, since the string is attached to the outside of the ball:
1/2 m(R+r)2ω2 + 1/2 2/5 mr2ω2
For the motion of the mass centre about the axis, you can handle it in either of two ways:
- as linear motion, mv2/2
- as rotational motion, mr2ω2/2
Since v = rω, these are the same.
If you also want to account for the rotation of the object about its mass centre then you add Iω2/2.

5. Dec 9, 2013

### Vontox7

The problem is I know doubt the method we used back in high school to solve questions of objects rotating in a circle like a pendulum problem and using energy to find the balls maximum height and so forth because back then we didn't account for rotation kinetic energy why? Maybe I am not understanding what exactly is rotational kinetic energy.

6. Dec 10, 2013

### tiny-tim

Hi Vontox7!
There's nothing special about rotational kinetic energy.

It's just the sum (strictly, integral) of the ordinary 1/2 mv2 kinetic energy of the individual parts.
The school pendulum solution has a simple L in it.

That L is √(R2 + 2/5 r2).

Since you can't measure L very precisely (and since the string or rod isn't totally negligible either), the school solution is correct, it just uses a slightly different L.