Circular Motion Problem w/ Conservation of Energy

Click For Summary
SUMMARY

The discussion centers on a physics problem involving a teacher swinging from a 24m rope, where the rope breaks if the tension exceeds twice the teacher's weight. The relevant equations include kinetic energy (Ek = (1/2)mv^2), gravitational potential energy (Eg = mgh), and centripetal force (Fc = mv^2/r). The participant calculated the velocity at which kinetic energy equals potential energy but needed guidance on applying centripetal force to determine the height at which the rope breaks. The solution involves equating the forces acting on the teacher at the lowest point of the swing.

PREREQUISITES
  • Understanding of conservation of energy principles
  • Familiarity with kinetic and potential energy equations
  • Knowledge of centripetal force and its application in circular motion
  • Ability to manipulate algebraic equations to solve for unknowns
NEXT STEPS
  • Study the relationship between centripetal force and tension in circular motion
  • Learn how to apply conservation of energy in dynamic systems
  • Explore examples of tension in ropes during circular motion scenarios
  • Investigate the effects of varying mass and height on tension and energy conservation
USEFUL FOR

Students studying physics, particularly those focusing on mechanics and energy conservation, as well as educators seeking to enhance their teaching methods in circular motion concepts.

mxwtt
Messages
1
Reaction score
0

Homework Statement


Your favourite physics teacher who is late for class attempts to swing from the roof of a 24m long rope as shown in the picture. The teacher starts from rest (Ek=0) with the rope horizontal, but the rope will break if the tension force in it is twice the weight of the teacher. How high is the swinging physics teacher above the ground when the rope breaks?
(Hint: Use the conservation of energy)
CM2.png



Homework Equations


Ek=(1/2)mv^2
Eg=mgh
Fc=(mv^2)/r

The Attempt at a Solution


My diagram:
CM2-1.png

Eg = mgh
= m(9,8)(24)
= 235.2m
Ek = (1/2)mv^2
= mv^2
(because it says that the rope will break if the tension force in it is twice the weight of the teacher)
then i equate these.
235.2m = mv^2
235.2 = v^2
v = 15.3 m/s.
from here i have no idea what to do. i think i might have to use Fc=(mv^2)/r, but i not sure how.
thanks in advance for you help.




 
Physics news on Phys.org
What you have calculated is the velocity for which the kinetic energy is twice the initial potential energy. It's not that useful or related to your problem.

You already mentioned Fc=mv^2/r. How does this play a role in your problem?
What supplies the centipetal force and what is the condition that the rope breaks?
 

Similar threads

  • · Replies 4 ·
Replies
4
Views
2K
Replies
2
Views
2K
Replies
8
Views
2K
Replies
8
Views
2K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 21 ·
Replies
21
Views
3K
  • · Replies 5 ·
Replies
5
Views
2K