# What is the centripetal acceleration of the bob-sleigh?

1. Oct 13, 2015

### Haveagoodday

1. The problem statement, all variables and given/known data

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?

3. 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

Last edited: Oct 13, 2015
2. Oct 13, 2015

### DEvens

Um... What part are you having trouble with? You seem to have answers for every part.

By the way, it really looks like you just copied down the answers from the "back of the book." I know that isn't what you did, you just forgot to put in your work. So maybe you could edit it and add some of the steps?

3. Oct 13, 2015

### Haveagoodday

Hello, it might look like i did copy the answers, but i can promise you i didn't. The reason we i posted this is just if somebody have the time to correct me if my answers are wrong, plus these problems aren't from the book neither from the internet.

4. Oct 13, 2015

### coffeemanja

I agree with you on your answers in a. The rest is wrong, from my point of view. You forgot that your slope is going down at 4 degrees ALL the time.

5. Oct 13, 2015

### Haveagoodday

Ok, thank you for your answer, but could you give me some guidelines?

6. Oct 13, 2015

### haruspex

First, I would not determine the angle as such. You can go straight from the "1 in 15" slope to whatever else you need without finding the angle or invoking trig functions. Just use Pythagoras. That minimises the accumulated errors.
It is not completely clear whether the 1 in 15 relates to drop per horizontal distance or drop per distance down slope, but the difference is too small to matter here.

What is the distance from B to C?
After C, where, exactly, do you think the 200m shown takes you to?

7. Oct 13, 2015

### coffeemanja

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)?

b) centripetal acceleration immediately after B:
ac=v^2/r= 6,8 m/s^2 -this one is right also...
Here it goes wrong.
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

Make a diagram with forces. Tangential acceleration is in the same direction as speed (look up same for the circle). Radial acceleration will be 0.
By showing sum of forces for x and y directions, you will arrive to an equation that in the end will look like at=g*sintheta.
For the last part of b) ac depends on V. In the point C, what speed did the sleigh get, if it is constantly accelerating at a downwards angle of 4 degrees?

8. Oct 14, 2015

### haruspex

I repeat my advice regarding the angle. By calculating the angle as 4 degrees then using that number as input to trig functions you introduced inaccuracy. The 16.5m/s is a little off. Likewise the time for the whole descent (about 2% off).
What makes you think the total acceleration is g here?
Suppose, for the moment, that the curved corners are level instead of continuing to descend. What would the tangential acceleration be?

9. Oct 14, 2015

### coffeemanja

Was quoting haveagoodday here. These are his functions.

10. Oct 14, 2015

### haruspex

Yes, sorry. Consider my replies as directed to @Haveagoodday

11. Oct 14, 2015

### hooolahooop

I also have a question about this problem, if you find an Acceleration i task a), can you use it in the other tasks? I have done that, but not sure if its right, because wouldn't you get a new acceleration out of every corner? . (srry if my question may seem a bit unclear, was hard to explain)

12. Oct 14, 2015

### haruspex

In practice, the track would have to be banked (increasingly) through each corner so that the sled stays on track without friction. You can consider the normal force in two components at a point on a corner: a horizontal component and a component in the vertical plane. The horizontal component supplies the centripetal acceleration, while the vertical plane component provides the tangential acceleration.
According to the question statement, that vertical plane component will have the same angle to the vertical throughout the track, so the tangential acceleration should be constant.

13. Oct 14, 2015

### hefalomp

how did you get teta = 4, should it not be 3,82 ?? when sin-1(13.33/200)=3.82

14. Oct 14, 2015

### hooolahooop

Thanks!

15. Oct 14, 2015

### Haveagoodday

well i did it this way tan-1(13.3/200)=3.81 and then i rounded it up to 4.

16. Oct 14, 2015

### haruspex

... which was a bad idea, too great an error.
Anyway, as I keep pointing out, there is no benefit in finding the angle. All you need is the fall rate, 1 in 15.

The question is not clear as to whether the 1.5km is horizontal distance or distance along the slope. Probably the second, in which case if yio wanted to find the angle it would be sin-1, not tan-1. Being such a small angle it makes only a tiny difference.