Conservation of Angular Momentum: Bullet and Door Collision Problem
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- čas přidán 25. 03. 2022
- Physics Ninja looks at a collision problem of a bullet and a door. Two cases are looked at and in each the final angular velocity of the door is calculated.
Conservation of Angular momentum is used in both cases.
thanks for making these videos man! i like your thorough explanations
How did you determine direction of the pivot force on door?
dear sir after the collision if I take the center of mass of the combined system (which is also 0.5 m away from the pivot) and calculate final angular momentum of this combined com w.r.t pivot (as a point particle 0.5 m away from pivot)i.e. I_com w= (15.01)*0.5^2*w_f, then in this way w_f is not coming as expected, why? please help!
in second situation ,if we find the cofficient of restitution of collision we are gettimg it negative why?
when u considered energy, why did you only consider KE? wouldnt there be Gpe aswell since initially the bullet is in the air, and after the combined system is displaced?
Hey what about mg which is the external force
Shouldn't the angular momentum initial be 2, not 20?
yes, sorry about that. The final answer for omega is correct. I must have misread my hand written notes when making the video.
great 💪💪💪
Do you have a lesson similar to this, but without the pivot point?
That is to say, the bullet hits the door, and receives a certain amount of angular velocity AND a certain amount of linear velocity
Angular Momentum - Clay Puck and Stick Collision
czcams.com/video/VGuiLI5CyQM/video.html
@@PhysicsNinja Thank you
@Physics Ninja can I please send you questions and then you do videos solving them
hello, I have a momentum question, and i have no idea how to go about it. If I were to ask you would be able to help me solve it?
Reach out onlinephysicsninja@gmail.com
This really helped a lot
Thanks
can u tell me by gravitaional force is not doing external torque
I think you misinterpreted the figure
It is not a hinged rod but a door 🚪 as seen from top of the door(say from ceiling point of view)
Gravity is acting into the plane of the figure
And the torque due to gravity is being cancelled by hinge reaction
Hope you understand
@@shivamraj04 oh okay got it so if a force is parallel to axis of rotation then it does no torque
8:05 Why omega finals are equal ?
In the next video of the playlist, the rod is not pivoted. And in that question, when the clay sticks the rod, the center of mass was changing. What's the difference with this problem? Why didn't we consider the changing center of mass if there is any change?
In this problem there is a pivot so the rotation will be about this point. It’s an extra constraint on the problem.
@@PhysicsNinja Oh, now I see. Thank you so much for the fast reply sir. Appreciated!
But angular momentum is conserved for the moment just after the collision right bcoz after that there would be mg torque acting about the pivot
Yes
in first case e=.0005 and in secuond case e=.25
How pivot produce external force? Is it due to centrifugal force?
It prevents the top from moving to right after the collision.
@@PhysicsNinja aah yes, i got it, thanks
8:21 by 'That axis' do we mean the middle point -center of mass- of the door sir ?
no, he meant the pivot point because angular momentum is conserved at that axis. You can see that if axis of rotation passes through com of the door, its moment of inertia would be (1/12)MR^2
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what does initial angular momentum mean .. initially bullet is travelling in a straight line.🤔 I got confused sir..
That is a great question. It’s weird but objects moving in a straight line can have angular momentum depending on the choice of origin. Angular momentum is L=r X p, but r depends on the choice of origin. I’ll post a short video on this next week.
@@PhysicsNinja waiting for that video sir ❤
The momentum of the bullet before the collision is 2 not 20
Yes my bad, at least I got the leading 2 right