Which system to apply conservation of momentum to?

Click For 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: 122
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.
 
If have close pipe system with water inside pressurized at P1= 200 000Pa absolute, density 1000kg/m3, wider pipe diameter=2cm, contraction pipe diameter=1.49cm, that is contraction area ratio A1/A2=1.8 a) If water is stationary(pump OFF) and if I drill a hole anywhere at pipe, water will leak out, because pressure(200kPa) inside is higher than atmospheric pressure (101 325Pa). b)If I turn on pump and water start flowing with with v1=10m/s in A1 wider section, from Bernoulli equation I...

Similar threads

Replies
17
Views
1K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 30 ·
2
Replies
30
Views
3K
  • · Replies 17 ·
Replies
17
Views
685
Replies
23
Views
2K
  • · Replies 71 ·
3
Replies
71
Views
3K
Replies
10
Views
2K
Replies
67
Views
4K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 9 ·
Replies
9
Views
1K