Rotating ball and string Rotational Kinetic Energy?

[Vontox7]

Homework Statement

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.

Homework Equations

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.

 

[tiny-tim]

Hi Vontox7!

… 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?

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 r²)ω² …

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

 

[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 !

 

[haruspex]

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

Strictly speaking, since the string is attached to the outside of the ball:
1/2 m(R+r)²ω² + 1/2 2/5 mr²ω²

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

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.

 

[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.

 

[tiny-tim]

Hi Vontox7!

Maybe I am not understanding what exactly is rotational kinetic energy.

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 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?

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.

 

[megankayler1]

I like basis physics