What is the initial angular momentum of the ball, in joule seconds, right before the collision relative to the pivot point of the rod? Li

Let's go through each part of the problem.

Part (a) - Initial Angular Momentum of the Ball ( $L_i$ )

To find the initial angular momentum ($L_i$) of the ball relative to the pivot point just before the collision, we use:

$L_i = r \cdot m \cdot v \cdot \sin(\theta)$

where:

• $r$ is the perpendicular distance from the pivot to the point where the ball strikes the rod,
• $m = 0.0382 \, \text{kg}$ (mass of the putty ball),
• $v = 5.59 \, \text{m/s}$ (speed of the ball),
• $\theta = 48.2^\circ$.

The distance $r$ is given by $D = \frac{2}{3}L$: $r = \frac{2}{3} \times 0.84 \, \text{m} = 0.56 \, \text{m}$

Now we can plug in the values: $L_i = (0.56 \, \text{m}) \times (0.0382 \, \text{kg}) \times (5.59 \, \text{m/s}) \times \sin(48.2^\circ)$

Calculate $\sin(48.2^\circ)$, then substitute to find $L_i$.

Part (b) - Moment of Inertia of the Rod-Ball System ( $I_f$ )

The total moment of inertia $I_f$ with respect to the hinge includes:

1. The moment of inertia of the rod about the hinge, $I_{\text{rod}} = \frac{1}{3} M L^2$,
2. The moment of inertia of the putty ball at a distance $r$ from the pivot, $I_{\text{ball}} = m r^2$.

Moment of Inertia of the Rod

Given $M = 3.41 \, \text{kg}$ and $L = 0.84 \, \text{m}$: $I_{\text{rod}} = \frac{1}{3} \cdot 3.41 \, \text{kg} \cdot (0.84 \, \text{m})^2$

Moment of Inertia of the Ball

Given $m = 0.0382 \, \text{kg}$ and $r = 0.56 \, \text{m}$: $I_{\text{ball}} = 0.0382 \, \text{kg} \cdot (0.56 \, \text{m})^2$

Then, the total moment of inertia is: $I_f = I_{\text{rod}} + I_{\text{ball}}$

Part (c) - Angular Speed of the System After Collision ( $\omega$ )

Using the principle of conservation of angular momentum: $L_i = I_f \cdot \omega$

Solving for $\omega$: $\omega = \frac{L_i}{I_f}$

Let me know if you'd like the calculated values, further breakdown, or have any questions.

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Further Related Questions

1. How would the initial angular momentum change if the angle $\theta$ increased?
2. What impact does the mass of the rod have on the total moment of inertia?
3. How would the angular speed change if the collision was not inelastic?
4. How does changing the length of the rod affect the overall system's moment of inertia?
5. What would be the effect on $\omega$ if the speed of the ball increased?

Tip

In rotational dynamics, always remember to consider the perpendicular distance for calculating torques and angular momenta relative to a pivot.