Thread Closed

Halliday and Resnick - Boy sliding down ice mound

 
Share Thread Thread Tools
Sep30-08, 07:04 PM   #1
 

Halliday and Resnick - Boy sliding down ice mound


1. The problem statement, all variables and given/known data

This is from Fundamentals of Physics 8th ed, by Halliday and Resnick:
A boy is initially seated on the top of a hemispherical ice mound of radius R = 13.8m. He begins to slide down the ice, with a negligible initial speed. Approximate the ice as being frictionless. At what height does the boy lose contact with the ice?

2. Relevant equations

This is classified under Conservation of Mechanical Energy
There's a picture of a semicircle and a boy sliding down the circumference of the semicircle.

3. The attempt at a solution

I don't understand what conditions would make the boy go off the ice rather than stay on the ice. I thought it had to do with centripetal forces at first (cause he's going in a circle ... for a bit) but I couldn't make that idea work.
PhysOrg.com
PhysOrg
science news on PhysOrg.com

>> King Richard III found in 'untidy lozenge-shaped grave'
>> Google Drive sports new view and scan enhancements
>> Researcher admits mistakes in stem cell study
Sep30-08, 07:10 PM   #2
 
Mentor
Blog Entries: 1
Quote by Farnak View Post
I thought it had to do with centripetal forces at first (cause he's going in a circle ... for a bit) but I couldn't make that idea work.
Try it again--that's the right track. Apply Newton's 2nd law. And conservation of energy.
Nov30-09, 05:48 PM   #3
 
So far what I did was

PEtop= KE bottom
mgh=.5mv^2
gh=.5v^2
(9.81)h=(0.5)v^2

then at the point where the boy just barely stays on the ice mound, I used the equation

Fn= Fw-Fc, which would be 0 at that point
0=mg-(mv^2/r)
mg=(mv^2/r)
g=v^2/r
gr=v^2
(9.81)(13.8)=v^2
v^2= 135.378 m/s

Plug that v^2 back into the initial equation:

(9.81)h= (0.5)(135.378)
h= 6.9m

Does that look right?
Nov30-09, 05:56 PM   #4
 
Mentor
Blog Entries: 1

Halliday and Resnick - Boy sliding down ice mound


Quote by jlee888 View Post
then at the point where the boy just barely stays on the ice mound, I used the equation

Fn= Fw-Fc, which would be 0 at that point
It's true that the normal force will be zero at that point, but only a component of the weight contributes to the centripetal force. Not the full mg, which acts downward. (Find the radial component.)
Thread Closed
Thread Tools


Similar Threads for: Halliday and Resnick - Boy sliding down ice mound
Thread Forum Replies
What's with all these Halliday/Resnick books? Science Textbook Discussion 4
Friction problem from Halliday and Resnick Introductory Physics Homework 4
Halliday/Resnick/Walker - Text Question Science Textbook Discussion 17
Static Equilibrium Problem from Halliday, Resnick... Introductory Physics Homework 9
Halliday & Resnick Science Textbook Discussion 13