Problem 6: In the figure below a 0.001 kg bullet with initial speed v₁ = 1800 m/sec is fired into a 0.5 kg block attached to the end of a 0.60 m rod of mass 0.50 kg. Point A is the pivot point for the rod and the bullet stays inside the block after the collision. What is the angular speed w of the system after the collision? The moment of inertia of the rod about point A is I = 1/3 M ² and the moment of a particle is I = mr². Treat the block as a particle and also treat the bullet as a particle. Answer: w = 4.5 rad/sec Bullet A Rod Block

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Chapter1: Units, Trigonometry. And Vectors
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Problem 6:
In the figure below a 0.001 kg bullet with initial speed v₁ = 1800 m/sec is fired into a 0.5 kg
block attached to the end of a 0.60 m rod of mass 0.50 kg. Point A is the pivot point for the rod
and the bullet stays inside the block after the collision. What is the angular speed w of the
system after the collision? The moment of inertia of the rod about point A is I = 1/3 M 1² and
the moment of a particle is I = mr². Treat the block as a particle and also treat the bullet as a
particle. Answer: w = 4.5 rad/sec
Bullet
A
Rod
Block
Transcribed Image Text:Problem 6: In the figure below a 0.001 kg bullet with initial speed v₁ = 1800 m/sec is fired into a 0.5 kg block attached to the end of a 0.60 m rod of mass 0.50 kg. Point A is the pivot point for the rod and the bullet stays inside the block after the collision. What is the angular speed w of the system after the collision? The moment of inertia of the rod about point A is I = 1/3 M 1² and the moment of a particle is I = mr². Treat the block as a particle and also treat the bullet as a particle. Answer: w = 4.5 rad/sec Bullet A Rod Block
KE = mv², KE = 1w², Ug = mgh, U₁ = ¹kx², E = KE + Ug + Us, E₁ = Ef
L = mrv, 1 =1w
Transcribed Image Text:KE = mv², KE = 1w², Ug = mgh, U₁ = ¹kx², E = KE + Ug + Us, E₁ = Ef L = mrv, 1 =1w
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