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

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SUMMARY

The discussion centers on calculating the distance Mike Harris slides down the frictionless roof of a sky dome stadium, which has a circular cross-section with a radius of 125 meters. Using the principles of centripetal force and energy conservation, the angle of descent was determined to be 48 degrees. The arc length formula was then applied, resulting in a total sliding distance of 105 meters along the curve of the roof. This solution effectively combines physics concepts with geometric calculations to arrive at the final answer.

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Homework Statement


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.

Homework Equations


FUN=ma
ETi=ETf
a=(mv2)/r

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
 

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You are correct for the angle and can use

angle (in rad) = arc length / radius
 
mukundpa said:
You are correct for the angle and can use

angle (in rad) = arc length / radius

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

So, if I used that formula:

angle (in rad) x radius = arc length
arc length = (48 x ∏/180) x 125
arc length = 105m

So the total distance Mike Harris travels on the sky dome rooftop is 105m? :smile:
 

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