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July 29, 2014

July 29, 2014

Posted by **johnny** on Wednesday, April 11, 2012 at 1:40pm.

At what angular speed does the door swing open immediately after the collision?

Calculate the energy of the bullet-door system and determine whether it is less than or equal to the kinetic energy of the bullet before the collision (what is the final and initial KE's)

- College physics -
**SocialistHero**, Friday, June 7, 2013 at 5:38pmI'll assume that 1.00 103 means 1x10^3

a)

Yes. Any object which has a velocity vector which does not pass directly through the chosen axis has angular momentum about that axis.

b)

No, mechanical energy is not conserved. Angular momentum is conserved, and any excess energy is dissipated within the door.

c)

W.before = W.after

Solve for both, then solve for final w. (note that W is angular momentum, w is angular velocity)

W.before = I.bullet * w.bullet + I.door * w.door

I.bullet = m.bullet * (.91 m)^2

w.bullet = V.bullet / (.91 m)

W.before = m.bullet * V.bullet * (.91 m) + 0

W.after = I.bullet * w.bullet + I.door + w.door

W.after = m.bullet * (.91 m)^2 * w.bullet + 1/3 * m.door * (1m)^2 * w.door

since w.bullet = w.door, we'll just call that w

W.after = w * (m.bullet * (.91 m)^2 + 1/3 * m.door * (1m)^2)

Set W.before = W.after and solve for w

m.bullet * V.bullet * (.91 m) + 0 = w * (m.bullet * (.91 m)^2 + 1/3 * m.door * (1m)^2)

.006 kg * 1000 m/s * .91 m = w * (.006 kg * (.91 m)^2 + 1/3 * 18.2 kg * (1 m)^2)

5.46 kgm^2/s = w * 6.0716 kgm^2

w = .8992 radians/second

d)

Before:

Ek = 1/2 * m.bullet * V.bullet^2

Ek = 1/2 * .006 kg * 1000000 m^2/s^2

Ek = 3000 joules

After:

Ek = I.bullet * w^2 + I.door * w^2

Ek = (.8992 rad/sec)^2 * (.006 kg * (.91 m)^2 + 1/3 * 18.2 kg * (1m)^2)

Ek = 4.909 joules

Note that kinetic energy after is significantly less than kinetic energy before.

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