# Find distance when slipping down the roof of a sky dome stadium

1. Apr 10, 2013

### totallyclone

1. The problem statement, all variables and given/known data
Mike Harris stands at the very top of the sky dome stadium roof to pitch the opening ball for the Blue Jays game. He slips a bit left of centre and slides down along the frictionless roof surface. For what distance, measured along the curve, will he slide before leaving the roof? Assume the roof has a circular cross section of radius 125m.

2. Relevant equations
FUN=ma
ETi=ETf
a=(mv2)/r

3. The attempt at a solution
I think, at the point before Mike Harris falls off the roof, FN=0
So I wonder if this question would involve centripetal forces because the roof is shaped like a circle.

I have attached my diagram and here's how I decided to tackle the question.

FUN=ma
Fg$\bot$-FN=(mv2)/r
mgcosθ-0=[m(2gr(1-cosθ))]/r
cosθ=2(1-cosθ)
cosθ=2-2cosθ
cosθ=2/3
θ=48°

I found the angle, don't quite know how to proceed

#### Attached Files:

File size:
2.9 KB
Views:
70
• ###### physics-41.png
File size:
3.9 KB
Views:
72
Last edited: Apr 10, 2013
2. Apr 10, 2013

### mukundpa

You are correct for the angle and can use

3. Apr 10, 2013

### totallyclone

OOH, so the distance the question is called the arc length? Didn't know that!

So, if I used that formula: