- #1
hang
- 7
- 1
- Homework Statement
- seesaw
- Relevant Equations
- torque
How Does Using ##m## Instead of ##M = 2m## Affect the Seesaw and Spring System?
Click For Summary
The discussion centers on calculating the torque about the pivot point S in a seesaw and spring system. The user initially calculates the torque from the plate's weight and the spring force but overlooks the impact of using ##m## instead of the correct mass ##M = 2m## for the disk. It is clarified that the weight of the rod does not contribute to torque since it acts directly on the pivot. The confusion arises from the incorrect mass assumption, which affects the overall torque calculation. Accurate mass representation is crucial for determining the system's behavior.
- #2
hang
- 7
- 1
The question ask to find torque about the pivot point S.
I first find the torque due to gravity of the plate and then torque due to spring force.
I understand that the weight of the rod do no torque as it is acting on the pivot S.
Where have I gone wrong?
I first find the torque due to gravity of the plate and then torque due to spring force.
I understand that the weight of the rod do no torque as it is acting on the pivot S.
Where have I gone wrong?
- #3
- 22,817
- 14,874
You used ##m## instead of ##M = 2m## for the mass of the disk.
- #4
hang
- 7
- 1
thank you so muchOrodruin said:
My attempt:
Initial total M.E = PE of hanging part + PE of part of chain in the tube. I've considered the table as to be at zero of PE.
PE of hanging part = ##\frac{1}{2} \frac{m}{l}gh^{2}##.
PE of part in the tube = ##\frac{m}{l}(l - h)gh##.
Final ME = ##\frac{1}{2}\frac{m}{l}gh^{2}## + ##\frac{1}{2}\frac{m}{l}hv^{2}##.
Since Initial ME = Final ME. Therefore, ##\frac{1}{2}\frac{m}{l}hv^{2}## = ##\frac{m}{l}(l-h)gh##. Solving this gives: ## v = \sqrt{2g(l-h)}##. But the answer in the book...
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