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?

The angular velocity of the door just after impact is equal to .

Given the following data:

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:

   ....equation 1.

For the door:

  ....equation 2.

For the bullet:

   ....equation 3.

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

  ...equation 4.

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

  ...equation 5.

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

Substituting the given parameters into the formula, we have;

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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

(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

(2)

where

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

is the angular velocity of the system after the collision

So (2) can be rewritten as

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

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