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For example, a solid disk rolling down a hill would include both rotational and linear kinetic energy. For the rotational kinetic energy ($\frac{1}{2}I\omega^2$) the angular velocity becomes $v/r$ but what kind of velocity is it? I always assumed it was tangential. In the case of the disk rolling down the hill, would the $v_{tan}$ be the same as the $v_{cm}$?

If so, why? Seems kind of intuitive but I don't know how to prove it.

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  • $\begingroup$ Is the hill profile straight or curved? $\endgroup$
    – Qmechanic
    Apr 21, 2016 at 9:21

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Rolling of a circular body, on a flat surface, and without sliding results in $v_\text{tan}=v_\text{cm}$ where $v_\text{tan}=ωr$ is the tangential speed of any point on the rim of the body in the center-of-mass frame of reference.

This is understood by studying the motion in the c.m. frame: there, the flat surface has velocity $v_\text{cm}$ (backwards). The no-sliding condition implies it to be equal to the rim tangential speed.

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  • $\begingroup$ The answer is wrong. The velocity of the center of mass is two times less than the tangential velocity. physics.stackexchange.com/a/649185/306037 $\endgroup$ Jul 2, 2021 at 17:04
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    $\begingroup$ Note that L.Levrel defines the tangential velocity as measured relative to the center of mass. $\endgroup$
    – R.W. Bird
    Jul 2, 2021 at 18:57
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The velocity of the center of mass is two times less than the tangential velocity.

Think about rolling a book over a wheel (like a roll of tape). The book moves forward with the speed v, and imparts the same tangential velocity to the roll of tape.

If the tangential velocity was the same as the velocity of the center of mass, then the book would never roll over the tape, they'd keep rolling together forever. In the real world, however, the book will eventually roll over the piece of tape and fall off, because their centers of mass aren't moving at the same velocity.

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View this answer for a more detailed explanation.

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  • $\begingroup$ Note that you are talking about the velocity at the top of the wheel measured relative to the ground. $\endgroup$
    – R.W. Bird
    Jul 2, 2021 at 18:59
  • $\begingroup$ Read more carefully the statement, including the conditions, before pronouncing it wrong. Or right. $\endgroup$
    – nasu
    Jul 2, 2021 at 19:22
  • $\begingroup$ My answer says exactly the same thing as John Rennie's answer which you're referring to. Please read what I call $v_{\tan}$, and it's accordingly to the standard use of "tangential" which indicates the "θ" component in polar coordinates (natural coordinates in the c.m. frame where there's only rotation). On the other hand your use of the term seems non standard (any reference to back it up?). $\endgroup$
    – L. Levrel
    Jul 3, 2021 at 22:04
  • $\begingroup$ I guess I don't fully understand his answer or the initial question. I've removed the part about him being wrong. $\endgroup$ Jul 4, 2021 at 9:13

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