First a little geometry. Because triangle AOB is isosceles
(note that two of the three sides are the same length), that means that
angle OAB is equal to angle OBA. Since the
three angles of a triangle must add up to 180 degrees, we can easily
find φ in terms of θ.
But let's do it in radians instead of degrees. So the three angles of a triangle
must add up to π radians. Hence, in radians you have
θ + 2φ = π
From that you get
Recall that one of the factors that went to make torque was
sin(φ). By the above you have
sin(φ) = sin(π/2 - θ/2)
And from a basic trig identity, you have
sin(π/2 - θ/2) = cos(θ/2) = sin(φ)
Now we have to determined the length,
s, of the spring. Imagine that
B, is
the point,
(r,0). Then to get from
B
to
A, the train has to go
θ radians
around the track in the counterclockwise (i.e. positive) direction. That puts
A at the point,
(r cos(θ), r sin(θ)).
Using the
Pythagorean formula
to find the distance,
s, between
A and
B,
we find that
s2 = (r - r cos(θ))2 + r2 sin2(θ)
Squaring that out you get
s2 = r2 - 2r2 cos(θ) + r2 cos2(θ) + r2 sin2(θ)
To simplify this a little, let's factor out an
r2
s2 = r2 (1 - 2cos(θ) + cos2(θ) + sin2(θ))
Notice the last two terms in the parentheses. By equation
7.1-2, the above becomes:
s2 = r2 (1 - 2cos(θ) + 1) = r2 (2 - 2cos(θ))
or
___________ ___________________
s = r √2 - 2cos(θ) = 2r √(1/2) - (1/2)cos(θ)
But by the trig identity,
equation 7.1b-8c, this is the
same as
s = 2r sin(θ/2)
Now remember that
f = ks = 2kr sin(θ/2)
and bringing back the equation for torque and substituting for
f:
Q = f r sin(φ) = f r cos(θ/2) = ks r cos(θ/2) =
2kr2 sin(θ/2)cos(θ/2)
And finally, by the trig identity,
equation 7.1b-6b,
this becomes
Q = kr2 sin(θ)
which is about as simplified as it can get.
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