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We know that,in rolling without slipping,the point of the disc of radius $r$ which is in contact with ground,has two velocities acting on it. One is the linear velocity of center of mass $v_{\mathrm{CM}}$. The other is the tangential velocity $\omega r$. The two act in the opposite direction and the condition for pure rolling is $v_{\mathrm{CM}}=\omega r$. But since this makes the bottom point at rest, then what causes that point to rotate again with the disc? Is it inertia or something like that which makes that bottom point in motion again? Which phenomenon is responsible for this?

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  • $\begingroup$ I think that the external force (or torque ) at the center of mass causes that point to rotate again with the disc $\endgroup$
    – Eli
    Aug 8 at 19:10

1 Answer 1

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But since this makes the bottom point at rest,then what causes that point to rotate again with the disc?

Even though it is instantaneously at rest there are still forces acting on it. There is it’s weight, the contact force from the road, and the internal forces from the rest of the wheel. These forces combine to result in the observed acceleration.

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  • $\begingroup$ What do you mean by internal forces from the rest of the wheel and why not mention friction between the wheel and ground? $\endgroup$ Aug 6 at 22:25
  • $\begingroup$ @Not_Einstein friction is part of the contact force that I did mention. Inside the wheel there are forces of each part of the wheel pushing and pulling on its neighboring pieces. These forces occur between different parts of the wheel and hence are internal to the wheel $\endgroup$
    – Dale
    Aug 6 at 22:47
  • $\begingroup$ But don't those internal forces cancel out and hence have no effect on the motion of the wheel? $\endgroup$ Aug 7 at 1:06
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    $\begingroup$ @Not_Einstein the question wasn’t about the motion of the wheel as a whole, it was about the motion of one point on the wheel. The internal forces do not cancel out on a single point on the wheel. They are internal to the wheel but external to the point $\endgroup$
    – Dale
    Aug 7 at 2:29
  • $\begingroup$ @Dale the contact force is in this case constraint force ? $\endgroup$
    – Eli
    Aug 8 at 17:06

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