The BEST EVER illustration to solve these type of problems. Really helped a lot. THANK YOU PROFESSOR WALTER LEWIN...
@p.s.chandravanshi25367 жыл бұрын
Subscribed...
@benheideveld46175 жыл бұрын
Although professor Lewin unfortunately did not make the viewer aware of it, in the video you saw that the exact same impulse J imparts an amount of kinetic energy to the rod THAT DEPENDS ON d!!!! That is remarkable and counter intuitive, isn’t it! The same jolt J imparting more or less energy. It is like getting energy for nothing. How is this possible?! I will tell you. Put a ruler in front of you on a flat slick surface. Push it perpendicular to the direction of its greatest length with your finger at the center of the ruler. Notice and remember the resistance your finger receives from the push. Now push the ruler in the same direction, but close to the top of the ruler. Notice that the ruler seems to put up less of a resistance to your push, like its inertia is lower. This is in fact what is happening. Pushing at the middle of the ruler makes it recede because you make the center of mass start to move. Pushing at the top of the ruler not only makes the center of mass accelerate at exactly the the same rate as when you hit it in the middle, on top of that you start to make it rotate at a rate α=τ/Ι where Alfa is the angular acceleration, tau is the torque and I is the moment of inertia for rotation around the center of mass. This results in an additional acceleration a[rotational]=αL/2 where a[rotational] is the additional acceleration and L/2 is half the length of the ruler, the distance between the application point of the force and the middle of the ruler. So the total acceleration of the point of the rod where it is pushed by the force F for a short but finite time Δt caused is a[total] = a[translational] + a[rotational] = = F/M + τL/2I = F/M + FL²/4I = = F/M + FL² •12/4ML² = F/M + 3F/M = 4F/M This means pushing at the top of the ruler feels the same as pushing a 4 times less massive ruler at the middle! But how does this explain that the same jolt J imparts more kinetic energy to the ruler when it is more off-center? That is because the total transfer of kinetic energy is the total work W that is performed by the force F during the time Δt! That work is equal to W=F•s Where W is the amount of Kinetic energy transferred to the rod and s is the distance that the point of contact moves during the amount of time Δt that the force F is applied. We can compute the displacement s by applying s = at²/2. When the ruler or rod is hit in the middle we get a[total] = a[translational] s = at²/2 = (Δt)² • F/2M with a[total] = a[translational] + a[rotational] s = at²/2 = (Δt)² • 4F/2M s is four times greater when we hit the ruler with the same force during the same time imparting the same jolt J. But because s is four times greater, the amount of kinetic energy is four times greater, because the work W of the jolt is force times distance s, and although F is the same, a is four times greater during the same time interval Δt and because the displacement, though quadratic with time, is only linear with a, the work W is 4 times greater and therefore the kinetic energy transferred to the ruler is four times greater, given the same jolt J, because the displacement during the jolt is four times greater! Just cross checking this with the result from the video. Professor Lewin derived the formula KE = 0.5 + 24 d² at the top of the 0.5 m ruler d = 0.25 m KE = 0.5 + 24 • (0.25)² = 0.5 + 1.5 = 2.0 And that indeed is four times the value of KE the kinetic energy of 0.5 for d=0.
@lecturesbywalterlewin.they9259
5 жыл бұрын
this should NOT surprise you. An impulse changes the momentum by a FIXED amount. Take a satellite, fire its rocket for 5 sec (fixed change of momentum). Do that when the satellite before rocket fire has speed 1 m/sec then do it when the satellite before rocket fire has speed 100 m/s. Then calculate the KE increase in both cases!!!!
@benheideveld4617
5 жыл бұрын
Lectures by Walter Lewin. They will make you ♥ Physics. Thank you professor Lewin. Zeer vereerd met uw respons! We all know you in The Netherlands since your appearance on De Wereld Draait Door. It is indeed counter intuitive that a satellite driven by a rocket engine outputting constant force, when we neglect declining propellant mass for now, and therefore adding the same amount of momentum to the satellite every second, is increasing the kinetic energy of the satellite that is proportional to t² when it starts accelerating at t=0, and thereby the tempo of increase of kinetic energy is increasing proportional to (2t+1), which is increasing with time. This fact only becomes plausible to our intuition when we remember that decelerating the satellite with constant force of friction F will lead to an amount of work W=F•s where s is the braking distance. Because s increases with the square of velocity, we understand that kinetic energy must increase with the square of velocity. Adding kinetic energy to a moving object is a more efficient undertaking than when it is at rest. I found your example of hitting a ruler so enlightening because first of all it hits home that a force applied to the top of the rod accelerates the center of mass at the same rate as hitting it in the middle of the ruler. Second of all the hit at the top is not hitting a moving object, but like an object with reduced inertia, because there are two mechanisms at work that both are making the object recede from the jolt. I though that was an interesting point to explain in full depth. Vriendelijke groet uit Amsterdam!
@lecturesbywalterlewin.they9259
5 жыл бұрын
assume the impulse changes the speed of the satellite by 1 m/s. If the speed was 1 m/s it will now be 2 m/s, if the speed was 10 m/s it will now be 11 m/s 2^2 is 4 1^2=1. Thus the KE has increased by 3. 10^2=100 11^2=121 thus the KE has increased by 21.
@shyam6468
2 ай бұрын
@@benheideveld4617hey, what about the situation when we don't know if the impulse is same or not? Like assume a case in which a small mass m is hitting a rod at the end and then getting stuck to it?
@ptyptypty36 жыл бұрын
wow, I gather that the Velocity of the Center of Mass is always THE SAME regardless of where the Impulse strikes the Rod???????????.... that is, regardless of the value of "d" ?? that is amazing!!!
@benheideveld4617
5 жыл бұрын
Philip Y You are so right and it is counter intuitive. Also the force you apply to a rod always acts 100% on the center of mass, whatever d is, so how ever far away from the center. Although professor Lewin unfortunately did not make the viewer aware of it, in the video you saw that the exact same impulse J imparts an amount of kinetic energy to the rod THAT DEPENDS ON d!!!! That is remarkable and counter intuitive, isn’t it! The same jolt J imparting more or less energy. It is like getting energy for nothing. How is this possible?! I will tell you. Put a ruler in front of you on a flat slick surface. Push it perpendicular to the direction of its greatest length with your finger at the center of the ruler. Notice and remember the resistance your finger receives from the push. Now push the ruler in the same direction, but close to the top of the ruler. Notice that the ruler seems to put up less of a resistance to your push, like its inertia is lower. This is in fact what is happening. Pushing at the middle of the ruler makes it recede because you make the center of mass start to move. Pushing at the top of the ruler not only makes the center of mass accelerate at exactly the the same rate as when you hit it in the middle, on top of that you start to make it rotate at a rate α=τ/Ι where Alfa is the angular acceleration, tau is the torque and I is the moment of inertia for rotation around the center of mass. This results in an additional acceleration a[rotational]=αL/2 where a[rotational] is the additional acceleration and L/2 is half the length of the ruler, the distance between the application point of the force and the middle of the ruler. So the total acceleration of the point of the rod where it is pushed by the force F for a short but finite time Δt caused is a[total] = a[translational] + a[rotational] = = F/M + τL/2I = F/M + FL²/4I = = F/M + FL² •12/4ML² = F/M + 3F/M = 4F/M This means pushing at the top of the ruler feels the same as pushing a 4 times less massive ruler at the middle! But how does this explain that the same jolt J imparts more kinetic energy to the ruler when it is more off-center? That is because the total transfer of kinetic energy is the total work W that is performed by the force F during the time Δt! That work is equal to W=F•s Where W is the amount of Kinetic energy transferred to the rod and s is the distance that the point of contact moves during the amount of time Δt that the force F is applied. We can compute the displacement s by applying s = at²/2. When the ruler or rod is hit in the middle we get a[total] = a[translational] s = at²/2 = (Δt)² • F/2M with a[total] = a[translational] + a[rotational] s = at²/2 = (Δt)² • 4F/2M s is four times greater when we hit the ruler with the same force during the same time imparting the same jolt J. But because s is four times greater, the amount of kinetic energy is four times greater, because the work W of the jolt is force times distance s, and although F is the same, a is four times greater during the same time interval Δt and because the displacement, though quadratic with time, is only linear with a, the work W is 4 times greater and therefore the kinetic energy transferred to the ruler is four times greater, given the same jolt J, because the displacement during the jolt is four times greater! Just cross checking this with the result from the video. Professor Lewin derived the formula KE = 0.5 + 24 d² at the top of the 0.5 m ruler d = 0.25 m KE = 0.5 + 24 • (0.25)² = 0.5 + 1.5 = 2.0 And that indeed is four times the value of KE the kinetic energy of 0.5 for d=0.
@soniatiwari9986
3 жыл бұрын
@@benheideveld4617 bruh..
@virenderbhardwaj7570
Жыл бұрын
@@benheideveld4617 but isn't the work=F.s where s is the displacement of COM of the rigid body?
@Shyzah6 жыл бұрын
old school expo marker... havn't seen one of those in a while
@_John_Sean_Walker
6 жыл бұрын
Yes, and all black and whte video.
@heramb5754 жыл бұрын
If we consider taking angular momentum about the point passing through where the impulse hit the rod ,before collision 0 angular momentum and after collison there is angular momentum due to linear translation of COM (which would be Jd) ..but there was no torque acting along this line and at 19:02 you had mentioned that the angular momentum before and after is 0 and would be conserved.
@michaels333
Жыл бұрын
Here Professor is correct. The angular momentum about the point where the impulse is applied, after the Impulse is zero. I also found this nonintuitive and decided to integrate the angular momenta of tiny masses dm translating about the point where the impulse was applied, and after a page of math the whole thing cancels out to zero. It seems, while the cm of mass is moving forward relative to that point, the rotation of the bar as a whole (especially the parts far from the point of impact) are enough to cancel that out. For anyone curious the integral was L about the point = Integral from -l/2 to l/2 of [(v_cm + ωy)(y-d)*ρ*dy] where ρ is the mass density of the bar.
@user-bs6he4mr4v4 жыл бұрын
Force units are Kg*meter/(second^2) not Kg*meter/second
@p.s.chandravanshi25367 жыл бұрын
Professor, could you please give me some advice on how to solve the same problem if the rod happens to be on rough surface with coefficient of friction 0.7 and we have to find velocity of centre of mass and angular velocity as a function of time???
@lecturesbywalterlewin.they9259
7 жыл бұрын
angular momentum is then not conserved. Solving it analytically would be close to impossible but it can be solved numerically.
@p.s.chandravanshi2536
7 жыл бұрын
Lectures by Walter Lewin. They will make you ♥ Physics. True, I tried integrating the external torque due to friction, as friction is a self adjusting force and it varies along the length,and also for some values of angular impulse(small in magnitude) friction won't let them move i.e. we have to specify limiting value of friction too... And in this way, I really got foiled... But now I'm delighted to hear that it is hardly possible to do it in this way. Thank You for your precious advice.
@feliwx40876 жыл бұрын
I could not understand the integral part . all new terms for me . sir is there any other lectures of yours , in which you explain these terms ..
@sarktaredits106
3 жыл бұрын
Integral Fdt gives us momentum or impulse. Just like force × time.
@COOKIEMONSTER90
2 жыл бұрын
you need to learn Calculus1 to understand derivatives and integrals. There are good lectures on youtube by professor Leonard. He's explaining all this stuff from the scratch.
@AlbertoPulvirenti9 ай бұрын
Dear prof, first of all many thanks forcthe clear explanation, then a question: how does the results change if one assumes to hit the rod with an arbitrary angle, instead of perpendicular one?
@lecturesbywalterlewin.they9259
9 ай бұрын
watch my 8.01 lectures and you should be able to solve the problem if you change the angle.
@THE_HUB_ Жыл бұрын
How to calculate angular velocity about centre of mass by conserving momentum at the point where impulse is applied.... please give me solution of this...
@lecturesbywalterlewin.they9259
Жыл бұрын
Watch 8.01s Module 27.04 and my 8.01 lecture in which I discuss this *+demo* I cannot add to the clarity of my lecture and to Module 27.04
@mohammedkader60216 жыл бұрын
Hello Sir, I am rather confused about the total energy before and after of this system. If hitting the rod away from the COM does not affect its velocity , then where does the rotational energy come from
@lecturesbywalterlewin.they9259
6 жыл бұрын
I hit, thus I have to provide all energy. Impulse is NOT energy. The same impulse through the com requires less energy than that impulse not through the com
@mohammedkader6021
6 жыл бұрын
Thanks for the reply. So can we say that a ball moving with a certain speed would slow down more when it hits away from com than if it hits the com ?
@akshatakshat87265 жыл бұрын
Sir how the net external torque is zero when rod has given the impulse
@lecturesbywalterlewin.they9259
5 жыл бұрын
question unclear also refer to how many minutes into the video
@bandhabkumardas61706 жыл бұрын
how do u decide that the rod is gonna rotate about its center of mass and not any other point?
@lecturesbywalterlewin.they9259
6 жыл бұрын
watch my 8.01 lectures
@bandhabkumardas6170
6 жыл бұрын
Lectures by Walter Lewin. They will make you ♥ Physics. Can u please specify which video i should go for in order to get the idea about what i've asked?
Пікірлер: 34
The BEST EVER illustration to solve these type of problems. Really helped a lot. THANK YOU PROFESSOR WALTER LEWIN...
Subscribed...
Although professor Lewin unfortunately did not make the viewer aware of it, in the video you saw that the exact same impulse J imparts an amount of kinetic energy to the rod THAT DEPENDS ON d!!!! That is remarkable and counter intuitive, isn’t it! The same jolt J imparting more or less energy. It is like getting energy for nothing. How is this possible?! I will tell you. Put a ruler in front of you on a flat slick surface. Push it perpendicular to the direction of its greatest length with your finger at the center of the ruler. Notice and remember the resistance your finger receives from the push. Now push the ruler in the same direction, but close to the top of the ruler. Notice that the ruler seems to put up less of a resistance to your push, like its inertia is lower. This is in fact what is happening. Pushing at the middle of the ruler makes it recede because you make the center of mass start to move. Pushing at the top of the ruler not only makes the center of mass accelerate at exactly the the same rate as when you hit it in the middle, on top of that you start to make it rotate at a rate α=τ/Ι where Alfa is the angular acceleration, tau is the torque and I is the moment of inertia for rotation around the center of mass. This results in an additional acceleration a[rotational]=αL/2 where a[rotational] is the additional acceleration and L/2 is half the length of the ruler, the distance between the application point of the force and the middle of the ruler. So the total acceleration of the point of the rod where it is pushed by the force F for a short but finite time Δt caused is a[total] = a[translational] + a[rotational] = = F/M + τL/2I = F/M + FL²/4I = = F/M + FL² •12/4ML² = F/M + 3F/M = 4F/M This means pushing at the top of the ruler feels the same as pushing a 4 times less massive ruler at the middle! But how does this explain that the same jolt J imparts more kinetic energy to the ruler when it is more off-center? That is because the total transfer of kinetic energy is the total work W that is performed by the force F during the time Δt! That work is equal to W=F•s Where W is the amount of Kinetic energy transferred to the rod and s is the distance that the point of contact moves during the amount of time Δt that the force F is applied. We can compute the displacement s by applying s = at²/2. When the ruler or rod is hit in the middle we get a[total] = a[translational] s = at²/2 = (Δt)² • F/2M with a[total] = a[translational] + a[rotational] s = at²/2 = (Δt)² • 4F/2M s is four times greater when we hit the ruler with the same force during the same time imparting the same jolt J. But because s is four times greater, the amount of kinetic energy is four times greater, because the work W of the jolt is force times distance s, and although F is the same, a is four times greater during the same time interval Δt and because the displacement, though quadratic with time, is only linear with a, the work W is 4 times greater and therefore the kinetic energy transferred to the ruler is four times greater, given the same jolt J, because the displacement during the jolt is four times greater! Just cross checking this with the result from the video. Professor Lewin derived the formula KE = 0.5 + 24 d² at the top of the 0.5 m ruler d = 0.25 m KE = 0.5 + 24 • (0.25)² = 0.5 + 1.5 = 2.0 And that indeed is four times the value of KE the kinetic energy of 0.5 for d=0.
@lecturesbywalterlewin.they9259
5 жыл бұрын
this should NOT surprise you. An impulse changes the momentum by a FIXED amount. Take a satellite, fire its rocket for 5 sec (fixed change of momentum). Do that when the satellite before rocket fire has speed 1 m/sec then do it when the satellite before rocket fire has speed 100 m/s. Then calculate the KE increase in both cases!!!!
@benheideveld4617
5 жыл бұрын
Lectures by Walter Lewin. They will make you ♥ Physics. Thank you professor Lewin. Zeer vereerd met uw respons! We all know you in The Netherlands since your appearance on De Wereld Draait Door. It is indeed counter intuitive that a satellite driven by a rocket engine outputting constant force, when we neglect declining propellant mass for now, and therefore adding the same amount of momentum to the satellite every second, is increasing the kinetic energy of the satellite that is proportional to t² when it starts accelerating at t=0, and thereby the tempo of increase of kinetic energy is increasing proportional to (2t+1), which is increasing with time. This fact only becomes plausible to our intuition when we remember that decelerating the satellite with constant force of friction F will lead to an amount of work W=F•s where s is the braking distance. Because s increases with the square of velocity, we understand that kinetic energy must increase with the square of velocity. Adding kinetic energy to a moving object is a more efficient undertaking than when it is at rest. I found your example of hitting a ruler so enlightening because first of all it hits home that a force applied to the top of the rod accelerates the center of mass at the same rate as hitting it in the middle of the ruler. Second of all the hit at the top is not hitting a moving object, but like an object with reduced inertia, because there are two mechanisms at work that both are making the object recede from the jolt. I though that was an interesting point to explain in full depth. Vriendelijke groet uit Amsterdam!
@lecturesbywalterlewin.they9259
5 жыл бұрын
assume the impulse changes the speed of the satellite by 1 m/s. If the speed was 1 m/s it will now be 2 m/s, if the speed was 10 m/s it will now be 11 m/s 2^2 is 4 1^2=1. Thus the KE has increased by 3. 10^2=100 11^2=121 thus the KE has increased by 21.
@shyam6468
2 ай бұрын
@@benheideveld4617hey, what about the situation when we don't know if the impulse is same or not? Like assume a case in which a small mass m is hitting a rod at the end and then getting stuck to it?
wow, I gather that the Velocity of the Center of Mass is always THE SAME regardless of where the Impulse strikes the Rod???????????.... that is, regardless of the value of "d" ?? that is amazing!!!
@benheideveld4617
5 жыл бұрын
Philip Y You are so right and it is counter intuitive. Also the force you apply to a rod always acts 100% on the center of mass, whatever d is, so how ever far away from the center. Although professor Lewin unfortunately did not make the viewer aware of it, in the video you saw that the exact same impulse J imparts an amount of kinetic energy to the rod THAT DEPENDS ON d!!!! That is remarkable and counter intuitive, isn’t it! The same jolt J imparting more or less energy. It is like getting energy for nothing. How is this possible?! I will tell you. Put a ruler in front of you on a flat slick surface. Push it perpendicular to the direction of its greatest length with your finger at the center of the ruler. Notice and remember the resistance your finger receives from the push. Now push the ruler in the same direction, but close to the top of the ruler. Notice that the ruler seems to put up less of a resistance to your push, like its inertia is lower. This is in fact what is happening. Pushing at the middle of the ruler makes it recede because you make the center of mass start to move. Pushing at the top of the ruler not only makes the center of mass accelerate at exactly the the same rate as when you hit it in the middle, on top of that you start to make it rotate at a rate α=τ/Ι where Alfa is the angular acceleration, tau is the torque and I is the moment of inertia for rotation around the center of mass. This results in an additional acceleration a[rotational]=αL/2 where a[rotational] is the additional acceleration and L/2 is half the length of the ruler, the distance between the application point of the force and the middle of the ruler. So the total acceleration of the point of the rod where it is pushed by the force F for a short but finite time Δt caused is a[total] = a[translational] + a[rotational] = = F/M + τL/2I = F/M + FL²/4I = = F/M + FL² •12/4ML² = F/M + 3F/M = 4F/M This means pushing at the top of the ruler feels the same as pushing a 4 times less massive ruler at the middle! But how does this explain that the same jolt J imparts more kinetic energy to the ruler when it is more off-center? That is because the total transfer of kinetic energy is the total work W that is performed by the force F during the time Δt! That work is equal to W=F•s Where W is the amount of Kinetic energy transferred to the rod and s is the distance that the point of contact moves during the amount of time Δt that the force F is applied. We can compute the displacement s by applying s = at²/2. When the ruler or rod is hit in the middle we get a[total] = a[translational] s = at²/2 = (Δt)² • F/2M with a[total] = a[translational] + a[rotational] s = at²/2 = (Δt)² • 4F/2M s is four times greater when we hit the ruler with the same force during the same time imparting the same jolt J. But because s is four times greater, the amount of kinetic energy is four times greater, because the work W of the jolt is force times distance s, and although F is the same, a is four times greater during the same time interval Δt and because the displacement, though quadratic with time, is only linear with a, the work W is 4 times greater and therefore the kinetic energy transferred to the ruler is four times greater, given the same jolt J, because the displacement during the jolt is four times greater! Just cross checking this with the result from the video. Professor Lewin derived the formula KE = 0.5 + 24 d² at the top of the 0.5 m ruler d = 0.25 m KE = 0.5 + 24 • (0.25)² = 0.5 + 1.5 = 2.0 And that indeed is four times the value of KE the kinetic energy of 0.5 for d=0.
@soniatiwari9986
3 жыл бұрын
@@benheideveld4617 bruh..
@virenderbhardwaj7570
Жыл бұрын
@@benheideveld4617 but isn't the work=F.s where s is the displacement of COM of the rigid body?
old school expo marker... havn't seen one of those in a while
@_John_Sean_Walker
6 жыл бұрын
Yes, and all black and whte video.
If we consider taking angular momentum about the point passing through where the impulse hit the rod ,before collision 0 angular momentum and after collison there is angular momentum due to linear translation of COM (which would be Jd) ..but there was no torque acting along this line and at 19:02 you had mentioned that the angular momentum before and after is 0 and would be conserved.
@michaels333
Жыл бұрын
Here Professor is correct. The angular momentum about the point where the impulse is applied, after the Impulse is zero. I also found this nonintuitive and decided to integrate the angular momenta of tiny masses dm translating about the point where the impulse was applied, and after a page of math the whole thing cancels out to zero. It seems, while the cm of mass is moving forward relative to that point, the rotation of the bar as a whole (especially the parts far from the point of impact) are enough to cancel that out. For anyone curious the integral was L about the point = Integral from -l/2 to l/2 of [(v_cm + ωy)(y-d)*ρ*dy] where ρ is the mass density of the bar.
Force units are Kg*meter/(second^2) not Kg*meter/second
Professor, could you please give me some advice on how to solve the same problem if the rod happens to be on rough surface with coefficient of friction 0.7 and we have to find velocity of centre of mass and angular velocity as a function of time???
@lecturesbywalterlewin.they9259
7 жыл бұрын
angular momentum is then not conserved. Solving it analytically would be close to impossible but it can be solved numerically.
@p.s.chandravanshi2536
7 жыл бұрын
Lectures by Walter Lewin. They will make you ♥ Physics. True, I tried integrating the external torque due to friction, as friction is a self adjusting force and it varies along the length,and also for some values of angular impulse(small in magnitude) friction won't let them move i.e. we have to specify limiting value of friction too... And in this way, I really got foiled... But now I'm delighted to hear that it is hardly possible to do it in this way. Thank You for your precious advice.
I could not understand the integral part . all new terms for me . sir is there any other lectures of yours , in which you explain these terms ..
@sarktaredits106
3 жыл бұрын
Integral Fdt gives us momentum or impulse. Just like force × time.
@COOKIEMONSTER90
2 жыл бұрын
you need to learn Calculus1 to understand derivatives and integrals. There are good lectures on youtube by professor Leonard. He's explaining all this stuff from the scratch.
Dear prof, first of all many thanks forcthe clear explanation, then a question: how does the results change if one assumes to hit the rod with an arbitrary angle, instead of perpendicular one?
@lecturesbywalterlewin.they9259
9 ай бұрын
watch my 8.01 lectures and you should be able to solve the problem if you change the angle.
How to calculate angular velocity about centre of mass by conserving momentum at the point where impulse is applied.... please give me solution of this...
@lecturesbywalterlewin.they9259
Жыл бұрын
Watch 8.01s Module 27.04 and my 8.01 lecture in which I discuss this *+demo* I cannot add to the clarity of my lecture and to Module 27.04
Hello Sir, I am rather confused about the total energy before and after of this system. If hitting the rod away from the COM does not affect its velocity , then where does the rotational energy come from
@lecturesbywalterlewin.they9259
6 жыл бұрын
I hit, thus I have to provide all energy. Impulse is NOT energy. The same impulse through the com requires less energy than that impulse not through the com
@mohammedkader6021
6 жыл бұрын
Thanks for the reply. So can we say that a ball moving with a certain speed would slow down more when it hits away from com than if it hits the com ?
Sir how the net external torque is zero when rod has given the impulse
@lecturesbywalterlewin.they9259
5 жыл бұрын
question unclear also refer to how many minutes into the video
how do u decide that the rod is gonna rotate about its center of mass and not any other point?
@lecturesbywalterlewin.they9259
6 жыл бұрын
watch my 8.01 lectures
@bandhabkumardas6170
6 жыл бұрын
Lectures by Walter Lewin. They will make you ♥ Physics. Can u please specify which video i should go for in order to get the idea about what i've asked?