Rolling down an incline: why does friction not reduce the energy, and which shape reaches first?

6 answers & replies

• [+][–]Arjun Nair12thSample· 8d ago

  1. In pure rolling the point of contact is instantaneously at rest relative to the surface. Friction is static, acts at that point, and the point doesn't move, so the work done by friction is $f\cdot v_{contact}\,dt=0$. Friction does not dissipate energy here. What it does do is give a torque that makes the body spin, and it reduces the translational acceleration compared to a frictionless slide ($g\sin\theta$) by a fraction. Energy conservation holds: $mgh=\tfrac12mv^2(1+k^2/R^2)$.

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  • [+][–]Siddharth RaoDropperSample· 7d agoAuthor

    Contact point at rest means friction does zero work. This was the missing piece. Making a table of $k^2/R^2$ values and the race order as my formula sheet for this chapter. Thanks everyone.

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• [+][–]Kabir Joshi11thSample· 8d ago

  Friction on a rolling body does reduce the energy, that's why it eventually stops, so mechanical energy shouldn't be conserved on a rough incline.

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  • [+][–]Neel Chatterjee12thSample· 7d ago

    Two different things, Kabir. Rolling resistance and slipping (kinetic friction) do dissipate energy, but ideal pure rolling on a rough surface involves only static friction at a point that does not slide. That's the case NCERT uses. If the body slips, then yes kinetic friction heats the surfaces and the energy method needs a loss term.

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• [+][–]Tushar PawarDropperSample· 7d ago

  Race order: solid sphere > disc > hollow sphere > ring. Because $a=\dfrac{g\sin\theta}{1+k^2/R^2}$ and $k^2/R^2$ is: solid sphere $2/5$, disc $1/2$, hollow sphere $2/3$, ring $1$. Smaller value of $k^2/R^2$ gives larger $a$:

  • solid sphere $\tfrac57g\sin\theta$
  • disc $\tfrac23g\sin\theta$
  • hollow sphere $\tfrac35g\sin\theta$
  • ring $\tfrac12g\sin\theta$

  Mass and radius cancel completely. A big ring and a small ring with any masses reach together.

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• [+][–]Meera Patel11thSample· 7d ago

  3. For pure rolling the velocity of the centre is $v$. The bottom point (contact) has velocity $v-\omega R=0$. The top point has $v+\omega R=2v$. Any other point is in between. So the instantaneous axis of rotation is the contact line, and the top moves at twice the centre speed. Picture a rolling cycle wheel: the part touching the road is momentarily still, and the top races ahead of the cycle.

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