Bullet fired at a Hinged door (rotational velocity problem)

In summary, the problem involves a bullet of mass 0.005kg and velocity 1000m/s colliding with a door of mass 17.2kg and width 1m. The bullet embeds itself 9.8cm from the side opposite the hinges, causing the door to swing open. The door has the same moment of inertia as a rod with axis at one end. The question asks for the angular speed of the door immediately after the collision and whether the energy of the system is less than or equal to the initial kinetic energy of the bullet. The moment of inertia is calculated as (1/3)(17.205kg)(1.0m-0.098m)^2 = 4.66
  • #1
Number47
5
0

Homework Statement


A 0.005-kg bullet traveling horizontally with a speed of 1.00 103 m/s enters an 17.2-kg door, imbedding itself 9.8 cm from the side opposite the hinges as in the figure below. The 1.00-m-wide door is free to swing on its hinges.

relevant img: http://www.webassign.net/sercp8/p8-56.gif

At what angular speed does the door swing open immediately after the collision? (The door has the same moment of inertia as a rod with axis at one end.)

Calculate the energy of the door-bullet system and determine whether it is less than or equal to the kinetic energy of the bullet before the collision.


Homework Equations


moment of inertia=(1/3)mL^2

KErotational= (1/2)*I*Wf^2


The Attempt at a Solution



I calculated the moment of inertia for the door and bullet system I=(1/3)(17.205kg)(1.0m-.098m)^2=4.66

Then i set the KE of the bullet equal to the KE of the door bullet system equal to each other and solved for Wf

Wf=sqrt((.005kg*(1000m/s)^2)/4.66) and got the answer to be 32.76rad/s which sounds way to big to be true. This is due just before midnight tonight, so thanks in advance for any help
 
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  • #2
Number47 said:
Then i set the KE of the bullet equal to the KE of the door bullet system equal to each other and solved for Wf
KE is not conserved--it's a perfectly inelastic collision. (If it were conserved, the second question would be rather trivial.)

What is conserved?
 
  • #3
The KE is not conserved since this collision is inelastic. Linear momentum is also not conserved due to external (tension) forces keeping the door on its hinge.

Angular momentum; however, should be conserved as there are no external torques. The tension forces are radial in direction.
 
  • #4
so how should i set this up then? as 2 rotational velocities with different moments of inertia?
 
  • #5
Doc Al said:
KE is not conserved--it's a perfectly inelastic collision. (If it were conserved, the second question would be rather trivial.)

What is conserved?

momentum would be conserved then. so do i set it up as initial linear momentum equals final rotational momentum?
 
  • #6
You can't have a linear momentum equaling an angular momentum, they are different phenomena which are measured in different units.

You want initial angular momentum equal to final angular momentum.
 

1. How does the rotational velocity of a bullet affect its impact on a hinged door?

The rotational velocity of a bullet refers to the speed at which it spins as it travels through the air. This velocity can greatly impact the bullet's trajectory and its impact on a hinged door. A higher rotational velocity can cause the bullet to penetrate deeper into the door, while a lower velocity may cause it to ricochet or bounce off the surface.

2. What factors can influence the rotational velocity of a fired bullet?

There are several factors that can influence the rotational velocity of a fired bullet, including the type of firearm used, the caliber and weight of the bullet, and the distance it travels. Additionally, external factors such as air resistance and wind speed can also affect the rotational velocity.

3. Is there a formula for calculating the rotational velocity of a bullet?

Yes, the formula for calculating the rotational velocity of a bullet is v = ωr, where v is the linear velocity of the bullet, ω is the angular velocity (measured in radians per second), and r is the distance from the center of rotation to the point on the bullet's surface.

4. How does the angle at which the bullet hits the door affect the rotational velocity?

The angle at which a bullet hits a hinged door can greatly impact its rotational velocity. If the bullet strikes the door at a 90-degree angle, it will have the greatest rotational velocity. However, if it hits at a shallower angle, the rotational velocity will be reduced as some of the bullet's linear velocity is converted into translational motion.

5. Can the rotational velocity of a bullet be controlled by adjusting the weight or shape of the bullet?

Yes, the rotational velocity of a bullet can be controlled to some extent by adjusting its weight or shape. A heavier bullet will typically have a lower rotational velocity, while a lighter bullet may spin faster. Additionally, bullet shape can also impact the rotational velocity, with more streamlined bullets typically having a higher velocity than those with a more irregular shape.

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