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If I have a disk which is pure rolling and it strikes with a ladder, so can I conserve angular momentum about point O? I think I can because normal reaction passes through O, so torque due to it will be zero. But using it, I am getting wrong answer. enter image description here

I have written following equations:

$L_i = L_f$

$mv(H-r) + \frac{MR^2}{2}\frac{v}{R} = (\frac{MR^2}{2}+MR^2) \frac{v'}{R}$

Please help me figure out the concept.

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  • $\begingroup$ Consider the limiting case of a high step, where the contact is at h=R and the wheel bounces back. Do you have the right physics in that limit? $\endgroup$
    – Bob Jacobsen
    Commented May 15, 2019 at 13:34
  • $\begingroup$ I think, the disk will lift off the ground a little, as on disk, a force in upward direction will be applied due to step, so as to maintain its pure rolling. $\endgroup$
    – Yash Mittal
    Commented May 15, 2019 at 16:56

2 Answers 2

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If I have a disk which is pure rolling and it strikes with a ladder, so can I conserve angular momentum about point O?

No you can't.

Conservation of momentum does not apply in the presence of a net external force. Gravity is the external force that provides an angular deceleration so initial momentum does not equal final momentum.

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Your assumption that angular momentum is conserved about $O$ is correct, since the only impulse that could give rise to an impulsive moment is the contact impulsive force at $O$. (To be slightly pedantic, it's inaccurate to call this a normal impulsive force, since a tangential component will also be present due to the no-slip condition)

I can spot a error in your angular moment balance which appears to be the culprit. Hint: $R > H$. Other than that, the physics is all good. :)

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  • $\begingroup$ Ok, Thanks. Is it correct to say that velocity which is tangent to Line Of Contact will be the only velocity after collision (thorough 2-D collision), as velocity along Line of Contact will be 0 after collision (perfectly inelastic) ? $\endgroup$
    – Yash Mittal
    Commented May 15, 2019 at 11:26
  • $\begingroup$ That's right, the velocity of the centre of mass must be perpendicular to the line of contact, due to the no-slip condition, the disc must pivot about the point of contact after the collision. Being able to enforce the no-slip condition, however, relies on the assumption that contact is maintained between the disc and the step after the collision! (If $H$ is large enough, it is possible the disc might bounce back!) $\endgroup$
    – Involute
    Commented May 15, 2019 at 11:37
  • $\begingroup$ But using this method, I am getting different answer. $\endgroup$
    – Yash Mittal
    Commented May 15, 2019 at 12:08
  • $\begingroup$ Ok, then in what way is your answer and their answer different? Are there extra details to the problem that you haven't considered? e.g. the disc is just a wire, a so the moment of inertia is $J = MR^2$ instead of $J = \frac{1}{2}MR^2$. Is pure rolling/no slip a valid assumption? etc. $\endgroup$
    – Involute
    Commented May 15, 2019 at 12:21
  • $\begingroup$ I mean, the answers I am getting from both methods discussed above are different, you can also check. But acc. to physics, there should be only one answer. So, which one is correct? $\endgroup$
    – Yash Mittal
    Commented May 15, 2019 at 16:57

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