How Does a Bullet Impact the Kinetic Energy of a Rotating Disk?

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The discussion centers on calculating the kinetic energy of a rotating disk after a bullet impacts it. A bullet with a mass of 5 g and an initial velocity of 100 m/s penetrates a stationary disk, which has a mass of 2 kg and a radius of 20 cm. The bullet exits the disk at a reduced speed of 80 m/s, and the impact raises questions about angular momentum and torque. Participants emphasize that the disk's rotation is influenced by the bullet's angular momentum before and after the collision. The conversation highlights the need to analyze forces acting on the disk to understand its rotational motion.
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Homework Statement



A bullet with 5 g of mass and a velocity of 100m/s goes through a disk that was rested initially the disk is solid and as a radius r=20cm. It's 2cm thick and as a 2kg mass that can spin with no friction in an axis that goes through the center. The bullet goes trough the disk perpendiculary to the axis and in between its thickness at a distance d=10cm of its center with a final speed of 80m/s.

Calculate the kinetic energy of rotation of the disk
 
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EBBAzores said:

Homework Statement



A bullet with 5 g of mass and a velocity of 100m/s goes through a disk that was rested initially the disk is solid and as a radius r=20cm. It's 2cm thick and as a 2kg mass that can spin with no friction in an axis that goes through the center. The bullet goes trough the disk perpendiculary to the axis and in between its thickness at a distance d=10cm of its center with a final speed of 80m/s.

Calculate the kinetic energy of rotation of the disk
What have you done so far? Hint: what can you say about the angular momentum (ie of the bullet before the collision and the bullet+disk after) relative to the disk axis?

AM
 
The angular momentum will variate it self so by saying that, there will be a torque due to the variation and with the torque we can calculate de work of the bullet and by that the kinetic energy of rotation
 
Why should the disk rotate at all? It appears that the only force on the is parallel to the axis.
 
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