wow, this is def the best physics video I have ever seen. In the case where the ball bounced back from the rod, wouldn't it have a negative momentum because the velocity is bouncing back in the negative direction? This would thus lead to the positive angular momentum gained by the rod as it moves in the positive correct, and angular momentum is still conserved. Is this correct? And just to make sure, angular momentum is conserved when there are no net external forces acting upon the system right? Thanks again and keep up the good content!
Mr Finn, I don't understand how the ball can have an angular momentum when it is only travelling in a straight direction. Wouldn't there just be a translation of linear to rotational momentum?
I understand your confusion. The concept is extremely counter intuitive. Forms of momentum do not convert to other forms of momentum like energies do per say. Its the sum of all the momentums. Our only real evidence of angular momentum in the ball is because to give another object angular momentum you must have had it before to give. That’s why I made the video. Just one of those ideas that hurts our brain
Another awesome video, Mr. Finn! Quick question - at 3:42, why do you draw r diagonally like that rather than just a horizontal line from the pivot point to where the ball hits on the rod? Is there an underlying concept that results in drawing it a certain way? Thank you!
@@themathandphysicstutor , I had a quick question. If r constantly changes, does the angular momentum of the object change since mass and linear speed stays the same?
@@lohitmurali2137 Angular momentum of a particle moving at a constant velocity remains the same. You will notice that the sine term in the cross product takes care of the fact that the radius vector's magnitude keeps changing. As the radius vector magnitude increases, the angle decreases, and the two effects cancel each other out. Consider a particle that starts at position r = , a distance D from the origin along the x-axis, and has a velocity in the y-direction, of v=. The radius vector at time t becomes r(t) = . Take the cross product L=r cross (m*v) and get: L = m * cross = m*D*v, in the z-direction. Because the 0 from the x-component of velocity multiplies with the V*t term in the position vector, you end up with an angular momentum that remains independent of t.
@@IanXMiller Yes. Also, if there were no pivot, the rod would both rotate and translate as a consequence of being impacted in this collision. The de-facto center of rotation will be its center of mass if the ball is no longer in contact with it. In the event that the ball sticks to it, the two will collectively rotate around the new center of mass according to conservation of angular momentum, and the center of mass will translate according to conservation of linear momentum.
Mr. Finn, why does the bullet have its own moment of inertia? When the bullet is in the rod, won't the mass of the rod increase by 0.02 kg? How does the bullet have its own moment of inertia?
Everything with mass has inertia. Everything that wants to spin has a moment of inertia. Everything moving in reference to a pivot point has angular momentum
The bullet has negligible moment of inertia around its own center of mass when its own physical size is negligible. But when you shift your reference point to a point off to the side of its path (a distance x), you use the parallel axis theorem that adds m*x^2 to its moment of inertia. It has a moment of inertia because it is is separated from the reference point, about which we opt to define the new moment of inertia.
I have a question... for the initial angular momentum (where the puck's momentum is all that counts), why did you substitute L for the "r" in the mv(t)r formula? Doesn't the r mean the radius of the puck when you are looking for the momentum of the puck?
if the ball has an initial angular momentum, but then it hits the rod at radius zero, this means that the rod has gained no angular momentum. So where did the angular momentum of the ball go?
I was referring to 3:10 in the video sorry for not being clear. Could you explain the angular momentum exchange for that diagram where the ball hits the pivot point for 3:10 in the video?
@@keshavshah752 At r=0 the ball doesn't have any angular moment because the angular moment of the ball initially is equal to m*v*r, with r being the distance at where the ball hits the rod and since that is zero the angular moment is zero.
Mr. Finn can you explain why at 11:40 for when you created the equation for when the ball bounced back, why did you make mV(tangential)L turn into m1/2V(tangential)L?
You can pick any reference point you want, as an origin for defining angular momentum. You could alternatively solve this problem from the point of view of the location of impact, in which case you'd have to solve for the pivot reaction impulse at the instant of the collision, and use that to calculate the angular impulse and subsequent change in angular momentum about the point of impact. However, it would be a lot more work to calculate the unknown reaction force at the pivot point. It is usually best to set up your reference points around the locations at which unknown reaction forces are applied to the system. That way, you don't need to solve for such reaction forces, and you can stick to the actions that are directly applied, instead of going through the extra step. In this case, by assigning your origin at the pivot point, you make it so there is no torque due to the pivot reaction forces.
Wow, this helped me understand that angular momentum is a lot like linear momentum. Thanks!
No problem! Thank you so much for watching!
This really helped, thank you! It cleared a lot up.
Great news! Awesome. Thanks for watching!
wow, this is def the best physics video I have ever seen. In the case where the ball bounced back from the rod, wouldn't it have a negative momentum because the velocity is bouncing back in the negative direction? This would thus lead to the positive angular momentum gained by the rod as it moves in the positive correct, and angular momentum is still conserved. Is this correct? And just to make sure, angular momentum is conserved when there are no net external forces acting upon the system right? Thanks again and keep up the good content!
Yes. Both of those points are correct
@@themathandphysicstutor you are the best
Mr Finn, I don't understand how the ball can have an angular momentum when it is only travelling in a straight direction. Wouldn't there just be a translation of linear to rotational momentum?
I understand your confusion. The concept is extremely counter intuitive. Forms of momentum do not convert to other forms of momentum like energies do per say. Its the sum of all the momentums. Our only real evidence of angular momentum in the ball is because to give another object angular momentum you must have had it before to give. That’s why I made the video. Just one of those ideas that hurts our brain
Another awesome video, Mr. Finn! Quick question - at 3:42, why do you draw r diagonally like that rather than just a horizontal line from the pivot point to where the ball hits on the rod? Is there an underlying concept that results in drawing it a certain way? Thank you!
I can draw R from wherever the ball is. It constantly changes as the ball approaches the rod
@@themathandphysicstutor Thank you so much! I think I understand what you mean!
@@themathandphysicstutor , I had a quick question. If r constantly changes, does the angular momentum of the object change since mass and linear speed stays the same?
@@lohitmurali2137 Angular momentum of a particle moving at a constant velocity remains the same. You will notice that the sine term in the cross product takes care of the fact that the radius vector's magnitude keeps changing. As the radius vector magnitude increases, the angle decreases, and the two effects cancel each other out.
Consider a particle that starts at position r = , a distance D from the origin along the x-axis, and has a velocity in the y-direction, of v=. The radius vector at time t becomes r(t) = . Take the cross product L=r cross (m*v) and get:
L = m * cross = m*D*v, in the z-direction.
Because the 0 from the x-component of velocity multiplies with the V*t term in the position vector, you end up with an angular momentum that remains independent of t.
Thanks a lot. But what if there is no pivot and the rod is just lying on horizontal?.In this case about what point will it rotate?
If there is no pivot or axis of rotation the rod spins around it’s center of mass
@@themathandphysicstutor Ohk. Thank you
So for no pivot then I would be 1/12ML^2 rather than 1/3ML^2 and the R for the ball would be L/2 rather than L?
@@IanXMiller Yes. Also, if there were no pivot, the rod would both rotate and translate as a consequence of being impacted in this collision. The de-facto center of rotation will be its center of mass if the ball is no longer in contact with it. In the event that the ball sticks to it, the two will collectively rotate around the new center of mass according to conservation of angular momentum, and the center of mass will translate according to conservation of linear momentum.
Mr. Finn, why does the bullet have its own moment of inertia? When the bullet is in the rod, won't the mass of the rod increase by 0.02 kg? How does the bullet have its own moment of inertia?
Everything with mass has inertia. Everything that wants to spin has a moment of inertia. Everything moving in reference to a pivot point has angular momentum
The bullet has negligible moment of inertia around its own center of mass when its own physical size is negligible. But when you shift your reference point to a point off to the side of its path (a distance x), you use the parallel axis theorem that adds m*x^2 to its moment of inertia. It has a moment of inertia because it is is separated from the reference point, about which we opt to define the new moment of inertia.
For example 1, could you also use conservation of kinetic energy to solve this problem?
You can use whatever way you want. Just know that every way won’t get you the right answer
I have a question...
for the initial angular momentum (where the puck's momentum is all that counts), why did you substitute L for the "r" in the mv(t)r formula? Doesn't the r mean the radius of the puck when you are looking for the momentum of the puck?
no it does not. R is the distance from the pivot point
if the ball has an initial angular momentum, but then it hits the rod at radius zero, this means that the rod has gained no angular momentum. So where did the angular momentum of the ball go?
It doesn't hit it at r=0. It hits the rod at r=L (The distance from the pivot point to the impact point)
I was referring to 3:10 in the video sorry for not being clear. Could you explain the angular momentum exchange for that diagram where the ball hits the pivot point for 3:10 in the video?
@@keshavshah752 At r=0 the ball doesn't have any angular moment because the angular moment of the ball initially is equal to m*v*r, with r being the distance at where the ball hits the rod and since that is zero the angular moment is zero.
If we are using just linear momentum conservation then ans would be different
why so??
Liner momentum is not conserved. Total momentum is
Mr. Finn can you explain why at 11:40 for when you created the equation for when the ball bounced back, why did you make mV(tangential)L turn into m1/2V(tangential)L?
Because 10 seconds before that I said that the ball was going to bounce back at 1/2V...
@@themathandphysicstutor if it bounced back, wouldn't it have a negative velocity?
At 12:30 isn't the fact that the pivot point is fixed provide an external force and hence a torque?
at the pivot point r=0 therefor torque is zero
You can pick any reference point you want, as an origin for defining angular momentum. You could alternatively solve this problem from the point of view of the location of impact, in which case you'd have to solve for the pivot reaction impulse at the instant of the collision, and use that to calculate the angular impulse and subsequent change in angular momentum about the point of impact. However, it would be a lot more work to calculate the unknown reaction force at the pivot point.
It is usually best to set up your reference points around the locations at which unknown reaction forces are applied to the system. That way, you don't need to solve for such reaction forces, and you can stick to the actions that are directly applied, instead of going through the extra step. In this case, by assigning your origin at the pivot point, you make it so there is no torque due to the pivot reaction forces.
So, how do we comment on the conservation of linear momentum?
Not sure what you mean.