Take a 1kg point mass at the end of a 1 meter massless rod, free to rotate about a pivot at the other end of the rod. If I apply 1 unit of force to the point mass at a right angle to the rod, the point mass accelerates tangentially with 1 unit of acceleration and 1 unit of angular acceleration. However if I apply the same unit of force at 1/2 the length of the rod the point mass now accelerates at 1/2 a unit of acceleration and 1/2 a unit of angular acceleration. It’s the same force being applied but 1/2 of the momentum that force would usually supply is being lost somehow just by changing the location that it is being applied at. Where is that other 1/2 of the momentum going ?
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$\begingroup$ This may help - Toppling of a cylinder on a block $\endgroup$– mmesser314Commented Dec 7, 2023 at 0:04
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1$\begingroup$ People love to downvote without any answer on this site. $\endgroup$– Blue5000Commented Dec 7, 2023 at 0:44
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$\begingroup$ I cannot argue that point. It isn't always your fault. I don't see why this question deserves a down vote. $\endgroup$– mmesser314Commented Dec 7, 2023 at 3:35
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$\begingroup$ this link may help hyperphysics.phy-astr.gsu.edu/hbase/conser.html $\endgroup$– anna vCommented Dec 7, 2023 at 5:15
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$\begingroup$ This has nothing to do with conservation of momentum. A force acts in two different situations and produces different effects. $\endgroup$– nasuCommented Dec 10, 2023 at 15:57
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2 Answers
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Since there is a hinge at the end of the ruler, linear momentum is never conserved because the hinge is applying force on the ruler when acted by a force. However, angular momentum is conserved about the hinge because. the torque due to the hinge is zero because it passes through it.
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$\begingroup$ When the force is applied at the point mass, the magnitude of the change in momentum is that of the linear case the only difference is the centripetal force by the rod is changing the direction of the tangential velocity. The magnitude of the acceleration of the rod is F/M = a. However once you move the force half way down the acceleration becomes 1/2 a. How does a force from a hinge result in this? $\endgroup$– Blue5000Commented Dec 7, 2023 at 9:04
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$\begingroup$ @Blue5000 The acceleration of the CM of the system (ruler & mass) is determined by the net force on the system. If there's a force from the hinge with a component opposing the applied force, the acceleration will be lower. $\endgroup$ Commented Dec 7, 2023 at 12:56
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$\begingroup$ I forgot to mention It’s a frictionless hinge, I can’t see how it would oppose the force on the mass. And if there is an opposing force why does it only become apparent when the force is applied along some point on the rod. As the tangential acceleration of the mass is F/M when the force is applied on the mass, exactly the same as the linear case. $\endgroup$– Blue5000Commented Dec 7, 2023 at 13:34
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In the second case you apply half the torque so the change in angular momentum is also halved. The moment of inertia is only a quarter so the change in angular speed is double. Its product with r is therefore the same as before halving r. Thus the tangential momentum is conserved.
Only the total linear momentum is conserved, which includes the system that the mass is connected to.
