- #1

Haveagoodday

- 43

- 1

## Homework Statement

At the Olympic bob-sleigh competition, a new (I admit, fairly boring...) track

has been built, as drawn in the figure. It is composed of straight lines with

lengths as indicated in the figure and 90.0 degree turns, each with the same

radius of curvature r1 = r2 = r3 = r4 = 40,0 m. The track slopes evenly

downwards all the way down (including through the corners), a total altitude

drop of 100 m over a distance of 1.50 km, so between the starting point (A)

and the end point (E). You may assume that the ice is frictionless. There is

gravity, g = 9:80 m/s2.a) Assuming that the sleigh starts from rest at (A), what is its speed when

it reaches (B)? How long does the whole run take (from A to E)? (To keep

it simple, we will ignore that bobsleighs are usually pushed to begin with by

the crew. Here, it is just sliding under the effect of gravity).

b) Consider the point when the sleigh enters the first corner (B). What is the

required centripetal acceleration, that the track has to provide immediately

after (B)? What is the tangential acceleration during the turn? What is the

required centripetal acceleration immediately before reaching (C)?

(c) At what angle with the horizontal does the sleigh need to slide up the

side of the track in order to have the required centripetal force at (B) and

(C)?

d) What is the total acceleration of the sleigh as it comes out of the fourth

corner (immediately before (D))? In units of g?

## The Attempt at a Solution

a) My solutions for a:

In every meter the altitude drops down 0.06666667 m

in 200 m it drops down = 13.3 m

so θ= 4

v=sqrt(2gdsinθ)=16,5m/s from A to B

Time taken from A to E:

t=sqrt(2d/gsinθ)= 66,2 seconds

b) centripetal acceleration immediately after B:

ac=v^2/r= 6,8 m/s^2

tangential acceleration:

a=sqrt(ar^2+at^2)

9.80m/s^2=sqrt(6.8^2+at^2)

at=sqrt(6.8^2-9.80^2)=12m/s^2

centripetal acceleration before reaching C:

ar=sqrt(-9.80^2+12^2)= 15,4 m/s^2

c) angle required:

-mgcosθ=macsinθ

tanθ=-g/ac=-55 degrees

d) total acceleration

a=sqrt(ar^2+at^2)=sqrt(15,4^2+12^2)=19,5 m/s^2

All answer are appreciated!

Last edited: