Question

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A uniform stick 1.2 m long with a total mass of 300 g is pivoted at its center. A 4.5 g bullet is shot through the stick midway between the pivot and one end (seefigure). The bullet approaches at 250 m/s and leaves at 160 m/s. With what angular speed is the stick spinning after the collision?

The image attached contains the question in full as well as a diagram

Answer 1

Length of the stick is \(L=1.2 \mathrm{~m}\)

\(\mathrm{m}\) ass is \(M=0.3 \mathrm{cg}\)

mass of the bullet is \(m=4.5 \times 10^{-3} \mathrm{~kg}\)

initial veocity of the bullet \(v_{i}=250 \mathrm{~m} / \mathrm{s}\)

final velocity of the bullet \(v_{\mathrm{f}}=160 \mathrm{~m} / \mathrm{s}\)

position where the bullet struck is \(r=\frac{L}{4}\) change in angular momentum of the bull et is

$$ \begin{aligned} r m\left(v_{i}-v_{f}\right) &=\left(\frac{1.2}{4}\right)\left(4.5 \times 10^{-3} \mathrm{~kg}\right)(250-160) \\ &=0.1215 \mathrm{~kg} \mathrm{~m}^{2} / \mathrm{s} \end{aligned} $$

the mom ent of inertia of the stick

\(I=\frac{M L^{2}}{12}\)

\(=\frac{(0.3)(1.2)^{2}}{12}\)

\(=0.036 \mathrm{~kg} \mathrm{~m}^{2}\)

if \(\omega\) is the angular velocity then

\(r m\left(v_{i}-v_{f}\right)=I \omega\)

Thus

\(\begin{aligned} \omega &=\frac{r m\left(v_{i}-v_{f}\right)}{I} \\ &=3.375 \mathrm{rad} / \mathrm{s} \end{aligned}\)

Answer 2

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