- #1
klg.amit
- 12
- 0
Hi,
The following is a general question which doesn't have to do with any particular problem. (therefore I am not including the template).
I understand that when a circular object (e.g: a hoop) rolls on a *flat* surface without slipping with angular speed [tex]\omega[/tex], its contact point, and thereby its center, moves with speed [tex]\omega r[/tex].
My difficulty is to understand the situation such as a pure roll of a hoop on a cylinder. I would like to understand what is wrong about my following analysis of the situation:
Let the hoop have radius r and the cylinder radius R, then:
1. Let the hoop rotate through an angle [tex]\theta[/tex] then its contact point moved a distance [tex]r\theta[/tex].
2. Let at the same time the angle from the center of the hoop to the center of the cylinder vary by [tex]\varphi[/tex] then the same distance is [tex]R\varphi[/tex]
3. Therefore [tex]r\theta = R\varphi \Rightarrow r\omega=R\dot{\varphi} [/tex] by differentiating.
However my Professor says that since there is no slipping, the *center* moves with speed [tex]r\omega[/tex], not the contact point. But according to my analysis, the center should move somewhat faster, since it seems to me that it covers a longer arc than [tex]r\theta[/tex] which the contact point covered after a rotation through [tex]\theta[/tex].
A clarification: when I say the "contact point moved" I understand that it's changing all the time. I actually refer to the total distance covered by the "bottom" of the hoop.
Your kind help will be very much appreciated,
Amit
The following is a general question which doesn't have to do with any particular problem. (therefore I am not including the template).
I understand that when a circular object (e.g: a hoop) rolls on a *flat* surface without slipping with angular speed [tex]\omega[/tex], its contact point, and thereby its center, moves with speed [tex]\omega r[/tex].
My difficulty is to understand the situation such as a pure roll of a hoop on a cylinder. I would like to understand what is wrong about my following analysis of the situation:
Let the hoop have radius r and the cylinder radius R, then:
1. Let the hoop rotate through an angle [tex]\theta[/tex] then its contact point moved a distance [tex]r\theta[/tex].
2. Let at the same time the angle from the center of the hoop to the center of the cylinder vary by [tex]\varphi[/tex] then the same distance is [tex]R\varphi[/tex]
3. Therefore [tex]r\theta = R\varphi \Rightarrow r\omega=R\dot{\varphi} [/tex] by differentiating.
However my Professor says that since there is no slipping, the *center* moves with speed [tex]r\omega[/tex], not the contact point. But according to my analysis, the center should move somewhat faster, since it seems to me that it covers a longer arc than [tex]r\theta[/tex] which the contact point covered after a rotation through [tex]\theta[/tex].
A clarification: when I say the "contact point moved" I understand that it's changing all the time. I actually refer to the total distance covered by the "bottom" of the hoop.
Your kind help will be very much appreciated,
Amit