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Angular velocity after collision

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A solid uniform sphere of radius R and mass M rolls without slipping with angular velocity $\omega_0$
when it encounters a step of height $0.4 R$. Find the angular velocity immediately after inelastic impact with the rough step.

I tried to conserve momentum
As after collision the linear velocity is zero .

$\frac{3}{5}mR^2\omega_0 +(2/5)mR^2\omega_0 = (2/5)mR^2\omega_{final}$

By this I m getting the answer as $\frac{5}{2}\omega_0$

But the answer given as $\frac{5}{7}\omega_0$

asked Dec 22, 2016 in Physics Problems by koolman (4,286 points)
reshown May 16 by sammy gerbil
Please can you explain your calculation in more detail.

My thinking is that the collision with the step imparts a momentum impulse $\Delta P$ to the sphere. This impulse is sufficient to reduce the linear momentum of the sphere to zero, so $\Delta P=mv=mR\omega_0$.

At the same time this impulse $\Delta P$ creates an angular impulse which increases the angular momentum of the sphere. The angular impulse equals $\Delta P \times (R-h)$, which is also equal to (the tangential component of $\Delta P) \times R$.
See my edited question
I still do not understand your calculation. Please can you explain in words what you are doing, what you are thinking. And don't just say "conserving momentum"!

Also, your equation gives $\omega_{final}=\frac53\omega_0$ not $\frac52\omega_0$.
Just noticed that the step is "rough". There will be friction between the step and the cylinder. The cylinder will slip against the step; friction will exert a counter-clockwise torque (or angular impulse).
Now I have showed my calculation with explanation . Is it fine?
It looks ok to me, but there may be more things to consider. The question does not explain clearly what "inelastic impact" means and whether "rough step" means we have to account for friction. Are you sure that you have typed the whole question as it was given to you?
Sorry , I misread the answer . The answer given as $\frac{5}{7}\omega$$_0$
Might be the answer given as wrong . Are you sure I have done correctly
Hmm, yes $\frac57$ could be a misprint of $\frac52$. I think the calculation is correct, but whether the answer is correct depends on the assumptions.

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