Centripetal acceleration: angle at which one will leave a dome

  • #1
dawn_pingpong
50
0

Homework Statement


see attached.


Homework Equations


centripedal acceleration: a=v^2/r


The Attempt at a Solution


Okay I found the solution for part 1.

http://www.vic.com/~syost/baylor/phy1422s07/final-answers2.pdf [Broken]

question 2. I know the change in energy part. But I don't really understand how

"Newton’s law states that the centripetal force must be the net radial force
on the skater, which is the normal force FN directed radially
outward, and the radial component of his weight, mg cos θ
directed inward."

and the equation that follows.

I think I can do part 2 and 3 by vectors and projectile motion once the 1st part is solved.
 

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Answers and Replies

  • #2
voko
6,054
391
I don't really understand how

"Newton’s law states that the centripetal force must be the net radial force
on the skater, which is the normal force FN directed radially
outward, and the radial component of his weight, mg cos θ
directed inward."

The law states F = ma, and that is also true for any projection of the vectors involved. If we take the radial projection, we end up with the quoted statement.
 
  • #3
dawn_pingpong
50
0
okay thanks now I understand. Thanks so much for always coming to my rescue!
 

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