Free Rolling and Circular Motion

[Fu-Kwun Hwang]

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There are 3 translations, select to view translated versionPortuguês - Brazilian portuguese(by dmferreira: 2)English(by palanichamy: 0)English(by ramakrishna.m: 0)
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This applet shows the positon,velocity and accelerationvectors
of a point on the rim of a body rolling without slippingon a stationary surface.

1. The term freely rolling means that there is no slipping (v=rw)
at the point of contact with the ground--no skidding (v>rw) and no spinning in place (w>v/r),where v,r and w are the linear speed, radius and
the angular speed of the rolling body.
2. The center of the wheel has a linear speed (with respect to ground)
equal to that of any point on its rim (with respect to center).
3. For circular motion (in the reference frame
moving with the velocity of the center of the rolling body) :
The acceleration a of a point always points in a direction
opposite the position vector r,
and the velocity vector V is perpendicular to both of them.
4. Observe the velocity vectors of a point on the rim
of the rolling body.
The white vector represents the velocity of the center
with resepect to ground and
the red vector represents the velocity of the point
with respect to the center.
The sum of these two vectors is the cyan vector
representing velocity of the point relative to ground.
5. The green curve shows the trajectory of a point on the rim.

Registed user can get files related to this applet for offline access.
If java program did not show up, please download and install latest Java RUN TIME

There are 3 translations, select to view translated versionPortuguês - Brazilian portuguese(by dmferreira: 2)English(by palanichamy: 0)English(by ramakrishna.m: 0)
Higher number at the end means more translation been done.

or

[Stam]

Thnx very helpful but are there more like these. Where can i find?

[Fu-Kwun Hwang]

There are several hundreds physics related simulation at this web site.
If there is some simulation you would like to have and you can not find it.
Write down in detail what you want, and I will try to help!