Not sure I agree with this. Can you clarify one point for me. After the collision the ball's speed relative to the origin is (3/4)v(0) However, after the collision, the center of mass of the rod is no longer at the origin, and the velocity of the ball relative to the center of mass of the rod is (3/4)v(0)- (1/4)v(0)=(1/2)v(0). It seems (to me) one should use this value in the conservation of angular momentum equation. Not (3/4)v(0). One could of course still use the original origin for the calculations, but the moment of inertia around that point is no longer (1/12)mL^2, but continually changing in time as it gets further away.
Shouldnt we use com of system instead of com of rod for such question as ball is also part of system so we should write angular momentum according to that.
Bro, why linear momentum is independent of de point of the collision?? I mean, why does the rod has the same CM velocity whatever the point of collision is?
I was struggling with this too, however, I think I found the solution: it is not true that the rod has the same CM velocity whatever the point of collision is. The best way to prove this is using the three laws of conservation of linear momentum, angular momentum and energy. Lets change the problem a bit and ignore friction. Lets say we are looking for the final velociy of the cm of the rod, and the ball depending where on the rod it strikes: We assume the bal and rod of length L both have mass M and that the ball has velocity v0: for linear momentum we find: mv0 = mv1 + mv2 with v1 the velocity of the ball after the collision, and mv2 the velocity of the CM. for angular momentum we find 0.5*m*L*v0 = 0.5*m*L*v1 + Iw where I = 1/12mL^12 as in the example for energy we have: 1/2 mv0^2 = 1/2 mv1^2 + 1/2mv2^2 + 1/2Iw^2 we have three equations to find three unknowns, v1,v2 and w. If you solve this you will find that collisions close to the center of the rod will result in higher linear velocity, while collisions close to the edge of the rod will result in high rotational motion with smaller linear velocity
@@arnoholtrop3168 Thank you! Well the problem is that we can't use conservation of energy because we have to also compute the loss, so how can we fix that
A rather late reply, but the solution is simple: add the heat loss to the right side of the energy balance equation. Of course we have 3 equations here, so we can have only a maximum of three unknowns in the problem in order to solve it. So of the 5 variables v0,v1,v2,w and heat loss, we need to know at least 2 to find the remaining variables, in any situation.
At th-cam.com/video/yu4YnzfIRFQ/w-d-xo.html you can't cancel the masses, as one is for the ball and the other is for the rod. Not the same. I was loving it up to this point. Nice explanation, but wrong cancellation. Saying it another way, m_{ball} is not the same as m_{rod}.
cant the problem be solved if we take the angular momentum conservation point as the point of hit of the ball with the ball instead of the center of mass.
This is making zero sense, so what if the ball hits the rod at its centre of mass? using the transitional equation I would get MV = (3/4)M + MVcm but with NO angular rotation (because I hit the centre of mass), so like depending on were I hit it I can create more energy lol.... something is wrong.
Thanks for making this. I am a little confused by your calculation of the angular velocity of the bar. You use the velocity of the ball after the collision (3v[0]/4) directly, but I was expecting you to need to subtract out the newly acquired velocity of the rod's center of mass (v[0]/4). Is this correct, or am I misunderstanding angular momentum? Thanks in advance for your time.
I went back through it and it is correct. It's similar to cons of linear momentum. You essentially are finding the change. You are just setting up the equation with final and initial. But you can't just subtract the velocities, you need to subtract the momenta.
5:15 took me to heaven. Nice video btw
😂😂
Not sure I agree with this. Can you clarify one point for me. After the collision the ball's speed relative to the origin is (3/4)v(0) However, after the collision, the center of mass of the rod is no longer at the origin, and the velocity of the ball relative to the center of mass of the rod is (3/4)v(0)- (1/4)v(0)=(1/2)v(0). It seems (to me) one should use this value in the conservation of angular momentum equation. Not (3/4)v(0).
One could of course still use the original origin for the calculations, but the moment of inertia around that point is no longer (1/12)mL^2, but continually changing in time as it gets further away.
That ahhhhhh at 5.15 🤣🤣
I needed that animation
Nice vid. Thanks.
what is the physics simulation software used?
Shouldnt we use com of system instead of com of rod for such question as ball is also part of system so we should write angular momentum according to that.
Its not an Entirely Elastic collision, so the com of the ball isn't a part of the system.
Bro, why linear momentum is independent of de point of the collision?? I mean, why does the rod has the same CM velocity whatever the point of collision is?
I was struggling with this too, however, I think I found the solution: it is not true that the rod has the same CM velocity whatever the point of collision is. The best way to prove this is using the three laws of conservation of linear momentum, angular momentum and energy. Lets change the problem a bit and ignore friction. Lets say we are looking for the final velociy of the cm of the rod, and the ball depending where on the rod it strikes: We assume the bal and rod of length L both have mass M and that the ball has velocity v0:
for linear momentum we find:
mv0 = mv1 + mv2
with v1 the velocity of the ball after the collision, and mv2 the velocity of the CM.
for angular momentum we find
0.5*m*L*v0 = 0.5*m*L*v1 + Iw where I = 1/12mL^12 as in the example
for energy we have:
1/2 mv0^2 = 1/2 mv1^2 + 1/2mv2^2 + 1/2Iw^2
we have three equations to find three unknowns, v1,v2 and w. If you solve this you will find that collisions close to the center of the rod will result in higher linear velocity, while collisions close to the edge of the rod will result in high rotational motion with smaller linear velocity
@@arnoholtrop3168 Thank you! Well the problem is that we can't use conservation of energy because we have to also compute the loss, so how can we fix that
A rather late reply, but the solution is simple: add the heat loss to the right side of the energy balance equation. Of course we have 3 equations here, so we can have only a maximum of three unknowns in the problem in order to solve it. So of the 5 variables v0,v1,v2,w and heat loss, we need to know at least 2 to find the remaining variables, in any situation.
Nice
At th-cam.com/video/yu4YnzfIRFQ/w-d-xo.html you can't cancel the masses, as one is for the ball and the other is for the rod. Not the same. I was loving it up to this point. Nice explanation, but wrong cancellation. Saying it another way, m_{ball} is not the same as m_{rod}.
Did you not read the question? It is given that mass of ball is m, and that of rod is m as well
cant the problem be solved if we take the angular momentum conservation point as the point of hit of the ball with the ball instead of the center of mass.
No we can't take it ,bcoz at the point of hit of ball with the ball is not fixed point after collision as rod is translating
This is making zero sense, so what if the ball hits the rod at its centre of mass? using the transitional equation I would get MV = (3/4)M + MVcm but with NO angular rotation (because I hit the centre of mass), so like depending on were I hit it I can create more energy lol.... something is wrong.
He did it wrong
Angular momentum must be conserved about the centre of mass of SYSTEM if the collision is not perfectly elastic
@@simachowdhury2093 Can you tell me how to do it right? Or give me resources?
Nothing is wrong, in that case the value of Vcm will be larger
@@simachowdhury2093we can conserve angular momentum about any point
Why does this dude scream when he cancels out the masses
Yeah, haha :D
Bro, physics is hard
Thanks for making this. I am a little confused by your calculation of the angular velocity of the bar. You use the velocity of the ball after the collision (3v[0]/4) directly, but I was expecting you to need to subtract out the newly acquired velocity of the rod's center of mass (v[0]/4). Is this correct, or am I misunderstanding angular momentum? Thanks in advance for your time.
I went back through it and it is correct. It's similar to cons of linear momentum. You essentially are finding the change. You are just setting up the equation with final and initial. But you can't just subtract the velocities, you need to subtract the momenta.
physicsfrac thanks!