TAM 212 Worksheet 3 Solutions

Name: Group members:
TAM 212 Worksheet 3
Solutions
This worksheet is concerned with designing the track for the Chicago O’Hare Airport Transit System (ATS):
We want to design one curve on the track between two perpendicular straight line segments. One easy option
would be to use a quarter-circle:
direction of travel
1
O
r = 100 m
θ
êθ
P
êr

1. If the train moves around the track at constant speed v = 10 m/s, what is its angular velocity ω about O
while on the circular segment?
ω = 0.1 rad/s
2. Plot the radial ar and angular aθ components of the train’s acceleration a = ar êr + aθ êθ versus θ as the
train moves around the circular segment. Clearly mark all significant points on the vertical axis.
1
0
1
ar
Π
4
Π
2
Θ
3. On airport trains people are normally standing. As the lead engineer for the ATS system design, do you
anticipate any problems for the passengers on this train as it goes along this track?
The radial acceleration will suddenly jump from zero on the straight-line segments to −1 m/s 2 on the curve.
This will be uncomfortable for the passengers who will feel this as a sudden jerk sideways.
4. To give the passengers a smoother ride, we want to make the track have a radius r that varies with the
angle θ, so r = r(θ), while still keeping a constant angular velocity ˙ θ = ω. Use the chain rule to find ˙r and
¨r and so write the acceleration in the polar basis in terms of r ′ (θ), r ′′ (θ) and ω. Here r ′ (θ) is the derivative
with respect to θ.
The chain rule gives ˙r = r ′ ˙ θ and ¨r = r ′′ ˙ θ 2 + r ′ ¨ θ. Also ˙ θ = ω and ¨ θ = 0, so
a = (¨r − r ˙ θ 2 )êr + (r ¨ θ + 2 ˙r ˙ θ)êθ = (r ′′ − r)ω 2 êr + 2r ′ ω 2 êθ
2
1
0
1
aΘ
Π
4
Π
2
Θ

5. From now on we will keep ω fixed at the constant value from Question 1 and we will use the functional
form:
2
r(θ) = Aθ θ − π
2 + 100 m
2
where θ is in radians and A is an unknown constant. This function looks like:
100
r
0
0
Π
4
Using this function, what are the radial and angular components of the acceleration when the train just
enters the curved track segment?
Evaluating the above acceleration expression with this r(θ) gives:
2
r(θ) = Aθ θ − π
2
r ′
(θ) = 2Aθ θ − π
r ′′
(θ) = 2A θ − π
2
The acceleration at θ = 0 is thus:
2
2
2
2
+ 100 r(0) = 100
2
+ 2Aθ θ − π
2
r ′
+ 8Aθ θ −
(0) = 0
π
+ 2Aθ
2
2
r ′′ (0) = Aπ2
2
2 Aπ
a =
200
ar = Aπ2
− 1
200
aθ = 0
− 1 êr
6. What value of A will give an acceleration profile that does not abruptly jump when moving onto the curve
from the straight-line segment?
To have zero initial acceleration on the curve we need A = 200/π 2 ≈ 20.3 m.
3
Π
2
Θ