- #1

erfz

## Homework Statement

There is a uniform rod of mass m = 1 kg and length L = 0.2 m fixed to the wall by an axis passing through its end.

A uniform ball of mass M = 0.1 kg and radius R = 2.85 cm is on the ground, below the axis of the rod, such that the rod's unfixed end is at the height of the ball's center-of-mass if the rod were to be at an angle of 0 degrees with the vertical.

The rod is set at angle 45 degrees to the left of the vertical and let go -- find the angular velocity of the rod before and after the collision, as well as the ball's translational and angular speeds after the collision. Assume the ball rolls without slipping.

I have written my approach below. I don't have the answer at hand, but I'm sure something is wrong here since the translational speed of the ball I get is 0.

## Homework Equations

##\vec{L} = \vec{r} \times m\vec{v}_{cm} + I_{cm}\vec{\omega}##

##L = I\omega##

## The Attempt at a Solution

I viewed this as an instantaneous collision and set the axis of rotation at the axis through the rod's end. So, the first step is finding the initial omega ##\omega_i## right before the collision using energy conservation about the rod's COM: $$m_{rod}g\frac{L}{2}(1-cos ~\theta) = 1/2 I_{rod}\omega_i^2$$ The moment of inertia of the rod about its end is ##1/3mL^2##. Plugging in the numbers, I get that ##\omega_i = 6.6 ~\text{rad/s}##

The angular momentum expression is: $$I_{rod}\omega_i = I_{rod}\omega_f + MLv_{ball,cm} + I_{ball,cm}\omega_{ball}$$ This second and third terms on the right come from the first equation I listed in "Relevant equations." I am defining counter-clockwise to be positive, so the translational contribution to the angular momentum of the ball (the second term on the right) should also be positive. However, the third term should be negative overall since the ball will be rotating clockwise. Since the ball isn't slipping, I can replace ##\omega_{ball} = -v_{ball,cm}/R##. So then: $$I_{rod}\omega_i = I_{rod}\omega_f + MLv_{ball,cm} - I_{ball,cm}v_{ball,cm}/R$$ I still need another equation to relate ##\omega_f## and ##v_{cm}##. I believe I can use linear momentum conservation for this: $$m_{rod}v_{rod,cm} = m_{rod}v'_{rod,cm} + Mv_{ball,cm}$$ The velocity of the rod's center-of-mass, I believe, should be such that ##v_{rod,cm} = \omega r##, where r is the distance from the axis to the center-of-mass of the rod (i.e. ##L/2##). So then this becomes $$m_{rod}\omega_i L/2 = m_{rod}\omega_f L/2 + Mv_{ball,cm}$$ Plugging in the values here, I get that relation that $$\omega_f = \omega_i - v_{ball,cm}$$ Now, if I substitute this into the angular momentum expression and solve, I get that ##v_{ball,cm}## is essentially 0, which intuitively cannot be correct. What is wrong with my approach?

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