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I have 2 questions

  1. A small solid sphere with a translation velocity $v$ m/s is climbing up an inclined surface of height $h$. what is the minimum $v$ required to climb this height

This question is solved by applying conservation of mechanical energy on translational motion of ball

  1. A disc of mass $M$ and radius $R$ rolls up on an inclined plane to a height $h$. if velocity of disc is $v$, the maximum height that the disc can reach is?

This question is solved by applying conservation of mechanical energy on both rotational and translational motion of disc

My problem is, how can we decide to apply conservation of mechanical energy on one component of kinetic energy or both?

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    $\begingroup$ If rotation is involved and is changing then you must use both $\endgroup$
    – Steeven
    Commented Sep 23, 2021 at 8:37
  • $\begingroup$ thank you! altho, in the first question only translational energy is converted to potential but in the second question both transational and kinetic energies are converted to potential. i dont understand how. changing rotation happens when? pls help $\endgroup$ Commented Sep 23, 2021 at 8:48
  • $\begingroup$ If that's true then in question one the rotation doesn't change. Maybe the surface is slippery or something like that. Of the rotation did change, then some energy went to or from the rotational kinetic energy in which case it must be included in the calculations. In general, everything that changes ought to be included in the calculations. $\endgroup$
    – Steeven
    Commented Sep 23, 2021 at 9:05

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In the first question they say " a SMALL solid sphere". This, and the fact that the radius of the sphere is not given, are hints that you should neglect the rotational kinetic energy. The rotational KE depends on the moment of inertia which depends on the radius (size) of the rotating object. If the radius is small the rotational KE is (or could be) small compared with the translational KE. If this is a school book problem, as the radius is not given you cannot consider rotational KE anyway. So there is no point to even consider if you need or not to consider rotational KE.

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  • $\begingroup$ oHH that makes sense thank you so much! $\endgroup$ Commented Sep 24, 2021 at 1:56

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