Centripetal acceleration: angle at which one will leave a dome

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dawn_pingpong
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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

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.
 
Attachments
  • Diagram of a domed observatory on a tall cylindrical base of height 3R, with radius R marked to the dome top and angle phi measured from the vertical at the centre, plus a skater on a horizontal ground line with a double-headed arrow showing distance.
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dawn_pingpong said:
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.
 
okay thanks now I understand. Thanks so much for always coming to my rescue!