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Circular Motion Problem w/ Conservation of Energy

  • Thread starter mxwtt
  • Start date
  • #1
1
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.

Homework Statement





Homework Equations





The Attempt at a Solution


Homework Statement





Homework Equations





The Attempt at a Solution


Homework Statement





Homework Equations





The Attempt at a Solution


Homework Statement





Homework Equations





The Attempt at a Solution


Homework Statement





Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
Galileo
Science Advisor
Homework Helper
1,989
6
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?
 

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