0
$\begingroup$

Say we have a cylinder rolling without slipping on a horizontal plane: enter image description here

Now, the instantaneous centre of rotation is P, so we can use $ v_A=\omega \ r_{AP}$ to find the linear velocity of the centre of mass, and calculate: $$\mathrm{KE}_{\mathrm{translational}} = 1/2 mv_A^2$$

However, my question is to do with computing the rotational KE. If P is the instantaneous centre and the cylinder is rotating about P, then why do we not use moment of inertia about P in the equation for rotational KE?

Why is it $\mathrm{KE}_{\mathrm{rotational}} =1/2I_A \omega^2$ and not $1/2 I_P \omega^2$?

The moment of inertia in rotational KE is defined to be the moment of inertia about the axis of rotation. Now I can see, of course, that the cylinder's axis of rotation over the course of its motion is A. However, isn't the bottom of the cylinder in contact with the ground always an axis of rotation too? What am I missing here?

Improve this question
$\endgroup$
3
  • $\begingroup$ Do you know the parallel axis theorem. $\endgroup$
    – mike stone
    May 15, 2022 at 17:57
  • $\begingroup$ Yes, but surely using that to find $I_P$ would give a different value to moment of inertia about the centre of mass? $\endgroup$
    – VJ123
    May 15, 2022 at 17:58
  • $\begingroup$ If you apply F = ma to a no slipping roll, the work done by the force corresponds to the sum of 2 terms. One is called translational and the other rotational kinetic energy. See physics.stackexchange.com/a/707449/195949 $\endgroup$
    – Claudio Saspinski
    May 15, 2022 at 18:37

1 Answer 1

Reset to default
2
$\begingroup$

The rotation/translation decomposition depends on the axis of rotation that you choose. If you consider the cylinder to be rotating about $A$ as it translates, $$\text{KE}_\text{translational}^A = \frac 1 2 m v^2$$ $$\text{KE}_\text{rotational}^A = \frac 1 2 I_A \omega^2.$$ If you choose $P$ as the axis of rotation, the instantaneous motion is purely rotational, i.e. there is no translation: $$\text{KE}_\text{translational}^P = 0$$ $$\text{KE}_\text{rotational}^P = \frac 1 2 I_P \omega^2.$$ In either case, the total kinetic energy is the same, as you can verify using the parallel axis theorem and the fact that $v=\omega r$.

Improve this answer
$\endgroup$
1
  • $\begingroup$ Ah I see - thank you so much for such a clear response! I realise where I was going wrong. $\endgroup$
    – VJ123
    May 15, 2022 at 18:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.