At 8:22, You defined angular momentum of the door as Iw. But doesn't L = Iw hold only when the rigid body is rotating about the axis of symmetry? THe door is not rotating about the axis of symmetry. How does it hold for the door case?
in general it doesn't really matter where the pivot of rotation is, so long as you change the value of the rotational inertia to match it. for example, a rod rotated about its center of mass will have a lower value for its rotational inertia than a rod with a pivot point at its end (like the one in the video). this is because when we have the pivot point at the center of mass, more of the mass is concentrated near the center of rotation, basically meaning it takes less effort to rotate it compared to when the pivot is at the end, where more of the mass is further away from where its rotating. But for the equation L=Iw, so long as we substitute the correct value for I, then we can just use the equation as normal
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
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?
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!
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
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?
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.
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
thanks for making these videos man! i like your thorough explanations
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.
How did you determine direction of the pivot force on door?
At 8:22, You defined angular momentum of the door as Iw. But doesn't L = Iw hold only when the rigid body is rotating about the axis of symmetry? THe door is not rotating about the axis of symmetry. How does it hold for the door case?
in general it doesn't really matter where the pivot of rotation is, so long as you change the value of the rotational inertia to match it. for example, a rod rotated about its center of mass will have a lower value for its rotational inertia than a rod with a pivot point at its end (like the one in the video). this is because when we have the pivot point at the center of mass, more of the mass is concentrated near the center of rotation, basically meaning it takes less effort to rotate it compared to when the pivot is at the end, where more of the mass is further away from where its rotating. But for the equation L=Iw, so long as we substitute the correct value for I, then we can just use the equation as normal
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
ruclips.net/video/VGuiLI5CyQM/видео.html
@@PhysicsNinja Thank you
@Physics Ninja can I please send you questions and then you do videos solving them
Could you tell me, why can I not use conservation of energy to solve this problem where KEi=KEf+KErot?
Collision is not necessarily an elastic collision
in second situation ,if we find the cofficient of restitution of collision we are gettimg it negative why?
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
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?
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!
8:05 Why omega finals are equal ?
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
Anladım teşekkürler
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!
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 ❤
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
Hey what about mg which is the external force
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
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
great 💪💪💪
The momentum of the bullet before the collision is 2 not 20
Yes my bad, at least I got the leading 2 right
in first case e=.0005 and in secuond case e=.25