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When using the conservation of angular momentum in collisions like the one in the following image.

enter image description here

Where there is a hoop of radius ${R}$ that is resting on a horizontal frictionless surface and a point particle of mass ${m}$ moving at a constant speed ${v}$. Considering an axis trough the center of mass of the hoop. If I wanted to calculate the angular momentum before the collision, using the conservation of angular momentum, because there are no external forces acting on the system, is the next relation valid? $\vec{L_i}$=${I}$$\vec{w_i}$=$\vec{L_f}$=${I}$$\vec{w_f}$, where ${I}$ is the moment of inertia, $\vec{L_i}$ the initial angular momentum (before collision) and $\vec{w_i}$ the initial angular velocity, $\vec{L_f}$ the final angular momentum (after collision) and $\vec{w_f}$ the final angular velocity. I tried calculating the initial angular momentum and I got $\vec{L_i}$=${mvr/2}$, where I used the relation I wrote before and i'm not sure whether thats right or wrong.

Qmechanic
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santijrv
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1 Answers1

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Due to Newonw's 3rd law whatever impulse $J$ is applied to the hoop, an equal and opposite is applied to the particle. As a result, the total linear momentum of the system is conserved.

Angular momentum is also conserved as it represents the geometric offset by which the impulse is applied, by definition of $L = r \times p$. More in details, see the table below.

Particle Hoop Total
Initial Linear Momentum, $\boldsymbol{p}=m \boldsymbol{v}$ $m v$ 0 $ m v$
Initial Angular Momentum, $\boldsymbol{L} = {\rm I} \boldsymbol{\omega} + \boldsymbol{r} \times \boldsymbol{p}$ $r m v$ 0 $r m v$
Impulse Magnitude at contact point, $\boldsymbol{J}$ $-J $ $+ J$ 0
Final Linear Momentum $m v - J$ $J$ $ m v$
Final Angular Momentum $r m v -r\, J$ $ r J$ $r m v$

The final motion is determined by the final momentum expressions

Particle Hoop
Linear Momentum $m v^f = m v - J$ $m u^f = J$
Angular Momentum not defined $I \omega^f = r J$

where $v^f$ is the particle's final speed, $u^f$ is the hoop's final speed (of the center) and $\omega^f$ is the hoop's final rotational speed.

This gives the following solution(s)

$$ v^f = v - \frac{J}{m} $$

$$ u^f = \frac{J}{m} $$

$$ \omega^f = \frac{r J}{I}$$

If you have a specific collision constraint, such as plasticity, then use the motion above to determine the impulse. For example, $v^f = u^f + r\, \omega^f$ leads to

$$ J = \frac{v}{\tfrac{1}{m} + \tfrac{1}{m} + \tfrac{r^2}{I}} $$

which is consistent with the effective mass of the system.

John Alexiou
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