91Ó°ÊÓ

Angular momentum (rarely, moment of momentum or rotational momentum) is the rotational analog of linear momentum. It is an important quantity in physics because it is a conserved quantity—the total angular momentum of a closed system remains constant.

Angular momentum is an important property of a rotating object and is expressed as the product of the moment of inertia and angular velocity.

Use conservation of angular momentum to find the device’s final angular momentum, and then solve for $\mathrm{Ó\neg }$

Initially, the device has no angular momentum and the bullet has a translational angular momentum ${\stackrel{→}{L}}_i.$ After the collision the bullet and the device have a combined angular momentum equal to ${\stackrel{→}{L}}_f.$

From the law of conservation of angular momentum, ${\stackrel{→}{L}}_i={\stackrel{→}{L}}_f$

Choose a lever-arm that simplifies the equations as much as possible. The initial angular momentum of the bullets is,

${\stackrel{→}{L}}_i=\stackrel{→}{r}×{\stackrel{→}{p}}_{bullet}$

Here,${\stackrel{→}{p}}_{bullet}$is initial momentum of the bullet.

The impact location is a good candidate since its coordinates are known, In fact, because the coordinates system is centered on the axle, the lever-arm vector will supply the negative of the coordinates of the impact location.

This gives a vector pointing from the location of impact to the center where the axis of rotation is. The lever-arm we have chosen is,

The particle is moving horizontally at $v=\text{550 m}·{\text{s}}^{\text{- 1}}$with a mass of $M=0.002\text{kg}$.

The momentum is simply,

Now take the cross product of and respectively. The cross product of two vectors is,

$\stackrel{→}{r}×\stackrel{→}{p}=\left[\begin{array}{c}{r}_y{p}_z-r_z{p}_y\\ r_z{p}_x-r_x{p}_z\\ r_x{p}_y-r_y{p}_x\end{array}\right]\begin{array}{}\end{array}\begin{array}{}\end{array}$

The first two rows are zero since there are no z-components in two vector and . The third row is,

$$\begin{gathered} r_x{p}_y-r_y{p}_x=\left(-0.155\text{m}\right)\left(0\text{kg}·\text{m}·{\text{s}}^{\text{- 1}}\right)-\left(-0.580\text{m}\right)\left(1.1\text{kg}·\text{m}·{\text{s}}^{\text{- 1}}\right) \\ =0.638\text{kg}·{\text{m}}^{\text{2}}·{\text{s}}^{\text{- 1}} \end{gathered}$$

The initial angular momentum is,

The final angular momentum of the system is equal to the initial angular momentum. The angular momentum of the devices is,

Angular velocity is equal to$\stackrel{→}{\mathrm{L}}/\mathrm{i}$ so now you need to find the moment of inertia of the device.

The moment of inertia of the device is the sum of the moments of inertia for the four balls, plus the embedded bullet. For a ball on a thin stick, the moment of inertia is,

So, the moments of inertia for the two sets of balls are, $I=M{R}^2$

$$\begin{gathered} I_1=2\left(\text{1.2 kg}\right){\left(\text{0.6 m}\right)}^2 \\ =0.86\text{4 kg}·{\text{m}}^{\text{2}} \end{gathered}$$

And

$$\begin{gathered} I_1=2\left(0.4\text{kg}\right){\left(0.2\text{m}\right)}^2 \\ =0.032\text{kg}·{\text{m}}^{\text{2}} \end{gathered}$$

The distance from the axis of rotation at which the bullet stuck is,

$$\begin{gathered} R=\sqrt{{\left(0.155\right)}^2+{\left(0.58\right)}^2}\text{m} \\ =0.600\text{m} \end{gathered}$$

So, the moment of inertia for the bullet is,

$$\begin{gathered} I_{bullet}=M{R}^2 \\ =\left(0.002\text{kg}\right){\left(0.600\text{m}\right)}^2 \\ =0.000\text{72 kg}·{\text{m}}^{\text{2}} \end{gathered}$$

So, the moment of inertia for the bullet is negligible.

So, if you ignore the contribution from the bullet, The total moment of inertia of the device is

Now, just divide the angular momentum of the device by its moment of inertia, and you will get the angular velocity. The angular velocity is,

Find the magnitude of the angular velocity as below.

$\begin{array}{rcl}\left|\stackrel{→}{\mathrm{Ó\neg }}\right|& =& \sqrt{{\left(0\right)}^2+{\left(0\right)}^2+{(0.712)}^2}\text{rad}·{\text{s}}^{\text{- 1}}\\ & =& 0.712\text{rad}·{\text{s}}^{\text{- 1}}\\ & & \end{array}$

Hence, the magnitude of the angular velocity is the angular speed is$0.712\text{rad}·{\text{s}}^{\text{- 1}}$ .