Which system to apply conservation of momentum to?

AI Thread Summary
The discussion centers on the application of conservation of angular momentum in a scenario involving a person on a swing. It is clarified that the person-swing system should be considered as a whole for conservation, as the angular momentum is conserved when the person stands up. The necessary angular acceleration is provided by the static friction force between the swing and the person, ensuring the net torque remains zero. There is a debate about the role of static friction, with concerns raised about the person being thrown off the swing if not holding on. The key takeaway is that the person-swing system is the appropriate choice for analyzing angular momentum conservation.
chris25
Messages
6
Reaction score
0
Homework Statement
A person stands on the seat of a swing and squats down, so that the distance between their center of mass (CM) and the swing’s pivot is L0. As the swing gets to the lowest point, the speed of their CM is V. At this moment, they quickly stand up, and thus decrease the distance from their CM to the swing’s pivot to L'. Immediately after they finish standing up, their CM speed is v0.
Relevant Equations
Comes from F=ma 2020b
Conservation of Angular Momentum
For this problem I was very confused whether conservation of angular momentum should be applied to the person, the swing or the person-swing system. It seems to me that there is no net torque on any of the three systems I listed above. However, it seems that the angular momentums of the three separates systems I listed cannot all be conserved simultaneously. Which system should I use, and for the systems wthere angular momentum is not conserved, where does the net torque come from? Thanks
 

Attachments

  • Screen Shot 2023-01-24 at 10.56.57 PM.png
    Screen Shot 2023-01-24 at 10.56.57 PM.png
    31.9 KB · Views: 117
Physics news on Phys.org
If you take as your system the person + the swing, then the angular momentum of this system is conserved. As the person stands up, his/her/zes CM at its new radius must acquire a smaller angular velocity for the system's two components to continue moving as one. The torque that provides the needed angular acceleration opposite to the angular velocity comes from the force of static friction exerted by the swing on the person's soles. Of course an equal and opposite torque is exerted by the soles on the swing and the net torque on the two-component system is zero.
 
  • Like
Likes chris25 and Lnewqban
kuruman said:
The torque that provides the needed angular acceleration opposite to the angular velocity comes from the force of static friction exerted by the swing on the person's soles
Not the soles, I think. Unless holding on to the ropes either side, the swinger would be thrown forwards off the seat.
 
Thread 'Minimum mass of a block'
Here we know that if block B is going to move up or just be at the verge of moving up ##Mg \sin \theta ## will act downwards and maximum static friction will act downwards ## \mu Mg \cos \theta ## Now what im confused by is how will we know " how quickly" block B reaches its maximum static friction value without any numbers, the suggested solution says that when block A is at its maximum extension, then block B will start to move up but with a certain set of values couldn't block A reach...
TL;DR Summary: Find Electric field due to charges between 2 parallel infinite planes using Gauss law at any point Here's the diagram. We have a uniform p (rho) density of charges between 2 infinite planes in the cartesian coordinates system. I used a cube of thickness a that spans from z=-a/2 to z=a/2 as a Gaussian surface, each side of the cube has area A. I know that the field depends only on z since there is translational invariance in x and y directions because the planes are...
Thread 'Calculation of Tensile Forces in Piston-Type Water-Lifting Devices at Elevated Locations'
Figure 1 Overall Structure Diagram Figure 2: Top view of the piston when it is cylindrical A circular opening is created at a height of 5 meters above the water surface. Inside this opening is a sleeve-type piston with a cross-sectional area of 1 square meter. The piston is pulled to the right at a constant speed. The pulling force is(Figure 2): F = ρshg = 1000 × 1 × 5 × 10 = 50,000 N. Figure 3: Modifying the structure to incorporate a fixed internal piston When I modify the piston...
Back
Top