Question

A 10 gg bullet traveling at 440 m/s strikes a 12 kg , 0.90-m-wide door at the edge opposite the hinge. The bullet embeds itself in the door, causing the door to swing open. Part A What is the angular velocity of the door just after impact?

Answer 1

Answer:

1.22 rad/s

Explanation:

When the bullet strikes the door, the angular momentum of the bullet-door must be conserved.

The angular momentum of the bullet before the collision can be written as

$L_1 = mur$ (1)

where

m = 10 g = 0.010 kg is the mass of the bullet

u = 440 m/s is the initial velocity of the bullet

r = 0.90 m is the distance of the bullet from the hinge

The total angular momentum of the system after the collision is given by the sum of the angular momentum of the door + the angular momentum of the bullet, so

$L_2 = I \omega + m \omega r^2$ (2)

where

$I=\frac{1}{3}Mr^2$ is the moment of inertia of the door, where

M = 12 kg is the mass of the rod

r = 0.90 m is the width

$\omega$ is the angular velocity of the system after the collision

So (2) can be rewritten as

$L_2=\frac{1}{3}Mr^2 \omega + m\omega r^2 = \omega r^2 (\frac{1}{3}M+m)$

Equating (1) and (2) and solving for $\omega$, we find the angular velocity:

$mur=\omega r^2 (\frac{1}{3}M+m)\\\omega = \frac{mu}{r (\frac{1}{3}M+m)}=\frac{(0.010)(440)}{(0.90)(\frac{1}{3}(12)+0.010)}=1.22 rad/s$

Answer 2

The angular velocity of the door just after impact is equal to $1.22\;rad/s$.

Given the following data:

• Mass of bullet = 10 grams to kg = 0.01 kg

• Velocity = 440 m/s

• Mass of door = 12 kg

• Radius = 0.90 meter

To determine the angular velocity of the door just after impact, we would use the law of conservation of angular momentum:

Applying the law of conservation of angular momentum:

$L_T =L_d + L_b$    ....equation 1.

For the door:

$L_d = \frac{1}{3} Mr^2 \omega$   ....equation 2.

For the bullet:

$L_b = mr^2 \omega$    ....equation 3.

Substituting eqn. 2 and 3 into eqn. 1, we have:

$L_T = \frac{1}{3} Mr^2 \omega + mr^2 \omega$   ...equation 4.

The angular momentum of the bullet before collision is given by:

$L_T = mvr$   ...equation 5.

Equating eqn. 5 and eqn. 4, we have:

$mvr = \frac{1}{3} Mr^2 \omega + mr^2 \omega\\\\\omega = \frac{mv}{r(\frac{M}{3} +m)}$

Substituting the given parameters into the formula, we have;

$\omega = \frac{0.01 \times 440}{0.9 \times (\frac{12}{3}\; + \;0.01)} \\\\\omega = \frac{4.4}{0.9 \times (4\; + \;0.01)}\\\\\omega = \frac{4.4}{0.9 \times 4.01}\\\\\omega = \frac{4.4}{3.609}\\\\\omega = 1.22 \;rad/s$

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