# Angular momentum conservation during collision

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.

I have written following equations:

$$L_i = L_f$$

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

• 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? Commented May 15, 2019 at 13:34
• 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. Commented May 15, 2019 at 16:56

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.

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. :)

• 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) ? Commented May 15, 2019 at 11:26
• 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!) Commented May 15, 2019 at 11:37
• But using this method, I am getting different answer. Commented May 15, 2019 at 12:08
• 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. Commented May 15, 2019 at 12:21
• 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? Commented May 15, 2019 at 16:57