V=wr. Is v tangential or center of mass?

[cjavier]

So...I have two different homework problems. I'm not asking for answers to these problems! Just clarification. In problem one I am asked: A thin disk with diameter d rotates about an axis through its center with 0.260J of kinetic energy.What is the speed of a point on the rim? I use E = 1/2Iω2 for kinetic rolling energy.
THEN I am told by chegg.com, which provided the correct answer, to use v=ωr to find the tangential speed, or the speed of a point at the edge of the disk. Keep in mind, this disk is not moving transitionally, only spinning.

In problem two: I am given a problem that needs me to find the energy of a rolling sphere when it reaches the bottom of the hill. When I use Ef = KErolling + KEtransitional I receive this equation:

Ef = 1/2Iω2 + 1/2mvCM2.

THIS IS THE CONFUSING PART, CHEGG TELLS ME TO REPLACE vcenter/mass with ωr.

Why am I able to do this? From my first homework problem, I established that v is equal to the speed at the tip of the rotating object. By replacing v with ωr for my kinetic energy, I am now saying that the transitional energy is dependent on the speed at the tip of the sphere!


Dazed and Confused

 

[Curious3141]

So...I have two different homework problems. I'm not asking for answers to these problems! Just clarification. In problem one I am asked: A thin disk with diameter d rotates about an axis through its center with 0.260J of kinetic energy.What is the speed of a point on the rim? I use E = 1/2Iω2 for kinetic rolling energy.
THEN I am told by chegg.com, which provided the correct answer, to use v=ωr to find the tangential speed, or the speed of a point at the edge of the disk. Keep in mind, this disk is not moving transitionally, only spinning.

In problem two: I am given a problem that needs me to find the energy of a rolling sphere when it reaches the bottom of the hill. When I use Ef = KErolling + KEtransitional I receive this equation:

Ef = 1/2Iω2 + 1/2mvCM2.

THIS IS THE CONFUSING PART, CHEGG TELLS ME TO REPLACE vcenter/mass with ωr.

Why am I able to do this? From my first homework problem, I established that v is equal to the speed at the tip of the rotating object. By replacing v with ωr for my kinetic energy, I am now saying that the transitional energy is dependent on the speed at the tip of the sphere!Dazed and Confused

Don't double or multiple post. I've moved my reply to your thread in General Physics here. Hopefully a Mentor will delete that thread.

You're wondering why v = rω represents both the speed at the tip of the rotating object and the translational (that's the right term, note the spelling) speed of the centre of mass.

This is only true for an object that rolls without slipping. When this condition is met, exactly one point of the sphere (the bottom-most) is in instantaneous contact with the surface at anyone point in time.

Now consider what happens when a sphere undergoing rotational motion at a constant rate (constant angular velocity) has rolled forward a length equal to exactly one circumference. Answer the following questions:

1) How much time does this take? Hint: think in terms of ω. How many radians would the sphere have turned through?

2) What distance has the sphere traveled forward in this time? This is the same distance the centre of mass has moved (translationally). This should be obvious when you consider the sphere is a rigid body that's not undergoing any deformation (change in shape), so every point in it has to be moving translationally forward at the same speed.

3) Hence, what's $v_{CM}$?