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.

 

[Nickie]

Hi there! Great job on finding the correct answer to this problem. Let me explain the reasoning behind adding the 2 in the equation.

First, let's look at the formula for kinetic energy: KE = 1/2(m)(v^2). This formula is used when an object is moving in a straight line with no rotation. In this case, the hula hoop is rolling on the ground, which means it is both translating (moving in a straight line) and rotating. This means that we need to take into account the rotational kinetic energy as well.

The formula for rotational kinetic energy is KE = 1/2(I)(ω^2), where I is the moment of inertia and ω is the angular velocity. In this case, the hula hoop has a moment of inertia of mr^2 (as you correctly substituted in the equation). But, instead of using ω (angular velocity), we need to use v (linear velocity) since we are solving for the linear speed of the hula hoop.

So, the correct equation to use in this case is: KE = 1/2(mr^2)(v^2/r^2) + 1/2(m)(v^2). The first part of the equation represents the rotational kinetic energy and the second part represents the translational kinetic energy. When we simplify this equation, we get: KE = 1/2(m)(v^2) + 1/2(m)(v^2), which can be written as KE = 2(1/2(m)(v^2)). This is where the 2 comes in.

To answer your question about when to use the formula for KE + Translational Energy, you would use this formula when an object is both translating and rotating. This is the case for the hula hoop in this problem. If an object is only translating, you would use the formula KE = 1/2(m)(v^2). If an object is only rotating, you would use the formula KE = 1/2(I)(ω^2).

I hope this helps clarify the reasoning behind adding the 2 in the equation and when to use the formula for KE + Translational Energy. Keep up the good work! :)