How far away from its location when the string breaks

  • Thread starter Thread starter riggi91
  • Start date Start date
  • Tags Tags
    String
Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
Sign up Log in
2 replies · 2K views
riggi91
Messages
4
Reaction score
0

Homework Statement


A 4-kg rock is tied to the end of a string and twirled overhead, 1.6 m above the ground, in a 1.1-m radius, horizontal circle. The string is known to break if its tension reaches a value of 514 N. The rock it rotated faster and faster until the string breaks. How far away from its location when the string breaks will the rock hit the ground? Your answer should be given in meters, to the nearest centimeter.

Homework Equations


F=mv^2/r
y=-.5gt^2
x=vxo(t)

The Attempt at a Solution



What I did was I found V first by doing sqrt(f*r/m)
Then I found t by doing sqrt(2y/g)
then I multiplied vt.

I've tried this three times and I keep getting this answer wrong. On my last attempts I've been getting between 6.9m and 7.1m based on the configuration of the numbers as they change slightly each time I attempt the problem.

What am I doing wrong?

Thanks
 
Physics news on Phys.org
I think you should go into more detail about your reasoning.
However, the thing that jumps out at me is that you appear to have assumed that the tension in the string is the only force on the rock while it is being twirled around the circle.
i.e. in
What I did was I found V first by doing sqrt(f*r/m)
How did you find "f"?
 
Last edited:
One seems to have the appropriate equations, but the circular trajectory of the rock is horizontal, not necessarily the string.

What are the time to fall and the linear speed at which the string breaks?