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Document Title
Fundamentals of Stress and Vibration Chapter Title
[A Practical guide for aspiring Designers / Analysts] 2. Engineering Mechanics

Differentiating the above expression with respect to theta we get:

dR v 2 1 2 ∗ sinθ ∗ cos θ
2
= − sin θ sinθ + k + sin θ + cos θ cos θ + ∗ = 0
2
dθ g 2 k + sin θ
Let us simplify the terms inside the bracket first and then combine the rest, as follows:

1 2 ∗ sin θ ∗ cosθ
2
− sinθ sin θ + k + sin θ + cos θ cosθ + ∗ = 0
2 k + sin θ
2

sinθ
2
2
− sinθ sin θ + k + sin θ + cos θ 1 + = 0
2
k + sin θ

k + sin θ + sinθ
2
2
2
− sinθ sinθ + k + sin θ + cos θ = 0
2
k + sin θ

2
2
2
2
− sin θ k + sin θ sinθ + k + sin θ + cos θ ( k + sin θ + sinθ)
= 0
k + sin θ
2

Taking k + sin θ + sin θ common in the above expression, we get:
2

2
2
− sin q k + sin θ + cos q]
2
sinθ + k + sin θ = 0
2
k + sin θ

Since the first term of the above expression gives sinθ = − k + sin θ , and the sign of the
2
acute angle (angle of launch of the bullet) cannot be negative, only the second term is equated
to zero. Therefore, we have:

2
2
− sin q k + sin θ + cos q] 2 2
k + sin θ = 0 = − sinq k + sin θ + cos θ = 0
2

Upon further simplification we have:

2
2
cos q = sinq k + sin θ

Squaring both sides and simplifying we get

(1 − sin θ) = sin θ (k + sin θ)
2
2
2
2
Expanding the above expression, we get:
1 + sin θ − 2 sin θ = k sin θ + sin θ
2
4
2
4
Page 30 QP No. SSC/Q4401, Version 1.0, NSQF Level 7, Compliant with Aero and Auto Industries,