Physics, 02.04.2020 23:58 beausisugpula

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?

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Answer from: minie66

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 kgVelocity = 440 m/sMass of door = 12 kgRadius = 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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A 10 gg bullet traveling at 440 m/s strikes a 12 kg , 0.90-m-wide door at the edge opposite the hing...

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