What linear speed must a 0.0500-kg hula hoop have if its total kinetic energy

[Lo.Lee.Ta.]

"What linear speed must a 0.0500-kg hula hoop have if its total kinetic energy..."

Hi, you guys! I found the answer, but I am a bit confused as to the method of solving this problem... Thanks for your help! :)

1. What linear speed must a 0.0500-kg hula hoop have if its total kinetic energy is to be 0.104 J? Assume the hoop rolls on the ground without slipping.2. I(hoop)= mr^2
KE= 1/2(I)(ω)^2
v= ωr
KE= 1/2(I)(ω^2) + 1/2(m)(v^2)

3. Okay, so this problem can be solved by KE= 1/2(I)(ω)^2

Replaced I with: mr^2

Replaced ω with: v/r

So then it's: KE= 1/2(mr^2)(v/r)^2

The r^2's cancel out, so then it's:

KE= 1/2(m)(v^2) BUT, the person on Yahoo Answers made it: KE = 1/2(2)(m)(v^2)

WHY did they add the 2 to the equation, thereby cancelling the 1/2?

It led to the right answer of 1.44m/s^2, but how do you know to do that?

*Also, the person on Yahoo Answers said you could find the answer also with this formula:

KE = 1/2(I)(ω)^2 + 1/2(m)(v)^2 ...WAIT A SEC, is this what the guy on Yahoo Answers did?

KE= 1/2(mr^2)(v^2/r^2) + 1/2(m)(v)^2 The r^2's cancel. Wouldn't that be the same thing as writing:

KE= 2(1/2(m)(v^2)) and THAT'S how the 2 comes in?

Is this right reasoning?
*Oh, and HOW do you know when to use the formula for KE + Translational Energy?
At first I made the mistake of using: KE= 1/2(m)(v^2)...

Thank you SO MUCH for your help! :D

 

[jedishrfu]

well the hoop is rolling on the ground so there a translational component to the KE and the v's for both components would be the same since there is no slippage.

Your initial KE eqn handles the rotational component only.

 

[Lo.Lee.Ta.]

Oh, okay. Thanks!
I think I know what you mean.

So since the hoop is rolling on the ground, kinetic energy is not only being spent on the rotation of the hoop, but also the moving of the hoop from one spot to another on a surface...?

But if there was a hoop somehow being rotated in the same spot in the air, we would not use the translational energy + rotational energy, right? We would only have rotational energy in that case?

And also- what about an actual person hoola-hooping? The hoop is in the air, but it is moving from one side the other. Would we say that is translational energy + rotational energy, just rotational energy, or just translational energy?

Thank you so much for helping! :D

 

[jedishrfu]

Oh, okay. Thanks!
I think I know what you mean.

So since the hoop is rolling on the ground, kinetic energy is not only being spent on the rotation of the hoop, but also the moving of the hoop from one spot to another on a surface...?

Yes.

But if there was a hoop somehow being rotated in the same spot in the air, we would not use the translational energy + rotational energy, right? We would only have rotational energy in that case?

Yes.

And also- what about an actual person hoola-hooping? The hoop is in the air, but it is moving from one side the other. Would we say that is translational energy + rotational energy, just rotational energy, or just translational energy?

I think rotational only although a different "I" would be needed since its rotation is not about the center of the hoop.