Maximum speed before Rope Breaks

Click For Summary
To determine the maximum uniform speed before a rope breaks when swinging a mass in a vertical plane, the critical point to analyze is at the bottom of the swing. At this position, both gravitational force and tension act in opposite directions, resulting in higher tension in the rope. The equation T - Fg = m(V^2/R) can be used to calculate the maximum speed, where T must not exceed the rope's maximum tension, P. The tension is greatest at the bottom of the swing, confirming that this is the point to evaluate for maximum speed. Understanding this concept is essential for ensuring the rope does not break during the motion.
proness26
Messages
3
Reaction score
0

Homework Statement


A mass of m is placed on the end of a rope of length l and swung in a vertical plane. The rope can withstand a maximum tension of P N(Newtons), after which it breaks. What is the maximum uniform speed that the object can be swung at, before the rope breaks?

Homework Equations



T-Fg=m(V^2/R)

The Attempt at a Solution



At which point on the vertical plane would I check to get maximum uniform speed?
Would I check the top where both Fg and T will be accelerating the object towards the center of the circle? Or would I check the bottom where T and Fg will be acting in opposite directions?
 
Physics news on Phys.org
Welcome to PF!

proness26 said:

Homework Statement


A mass of m is placed on the end of a rope of length l and swung in a vertical plane. The rope can withstand a maximum tension of P N(Newtons), after which it breaks. What is the maximum uniform speed that the object can be swung at, before the rope breaks?

Homework Equations



T-Fg=m(V^2/R)

The Attempt at a Solution



At which point on the vertical plane would I check to get maximum uniform speed?
Would I check the top where both Fg and T will be accelerating the object towards the center of the circle? Or would I check the bottom where T and Fg will be acting in opposite directions?

Well, at which of these two situations would the tension be larger? The rope must not break at any point in the cycle, which means it must be able to withstand the largest tensile force that it will be subjected to.
 
Ohh ok I got it then. Bottom. Seems like a silly question now that I look at it haha
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

Similar threads

Replies
46
Views
4K
  • · Replies 26 ·
Replies
26
Views
3K
Replies
43
Views
4K
  • · Replies 7 ·
Replies
7
Views
5K
  • · Replies 5 ·
Replies
5
Views
5K
  • · Replies 13 ·
Replies
13
Views
3K
Replies
8
Views
2K
  • · Replies 14 ·
Replies
14
Views
2K
  • · Replies 22 ·
Replies
22
Views
1K
Replies
2
Views
2K