← 2013 Paper 2
UPSC 2013 Maths Optional Paper 2 Q8a — Step-by-Step Solution 15 marks · Section B
Lagrange's equations · Mechanics & Fluid Dynamics · asked 9× in 13 yrs · Read the full method →
Question
Two equal rods A B AB A B B C BC B C l l l B B B A A A A A A 2 π n \dfrac{2\pi}{n} n 2 π n 2 = ( 3 ± 6 7 ) g l n^{2}=\left(3\pm\dfrac{6}{\sqrt 7}\right)\dfrac{g}{l} n 2 = ( 3 ± 7 6 ) l g
Technique
Lagrangian mechanics for 2-DOF system; linearisation around stable equilibrium ( θ 1 , θ 2 ) = ( 0 , 0 ) (\theta_1,\theta_2)=(0,0) ( θ 1 , θ 2 ) = ( 0 , 0 ) det ( K − n 2 M ) = 0 \det(K-n^{2}M)=0 det ( K − n 2 M ) = 0
Solution
Setup. Generalised coordinates: θ 1 \theta_1 θ 1 A B AB A B θ 2 \theta_2 θ 2 B C BC B C m m m
Step 1 — Kinetic energy
Rod A B AB A B A A A T 1 = 1 2 I A θ ˙ 1 2 = 1 2 ⋅ m l 2 3 θ ˙ 1 2 = m l 2 θ ˙ 1 2 6 T_1=\dfrac{1}{2}I_A\dot\theta_1^{2}=\dfrac{1}{2}\cdot\dfrac{ml^{2}}{3}\dot\theta_1^{2}=\dfrac{ml^{2}\dot\theta_1^{2}}{6} T 1 = 2 1 I A θ ˙ 1 2 = 2 1 ⋅ 3 m l 2 θ ˙ 1 2 = 6 m l 2 θ ˙ 1 2
Rod B C BC B C G 2 G_2 G 2 ( l sin θ 1 + l 2 sin θ 2 , − l cos θ 1 − l 2 cos θ 2 ) (l\sin\theta_1+\tfrac{l}{2}\sin\theta_2,\,-l\cos\theta_1-\tfrac{l}{2}\cos\theta_2) ( l sin θ 1 + 2 l sin θ 2 , − l cos θ 1 − 2 l cos θ 2 )
∣ G ˙ 2 ∣ 2 = l 2 θ ˙ 1 2 + l 2 4 θ ˙ 2 2 + l 2 cos ( θ 1 − θ 2 ) θ ˙ 1 θ ˙ 2 . |\dot G_2|^{2}=l^{2}\dot\theta_1^{2}+\tfrac{l^{2}}{4}\dot\theta_2^{2}+l^{2}\cos(\theta_1-\theta_2)\dot\theta_1\dot\theta_2. ∣ G ˙ 2 ∣ 2 = l 2 θ ˙ 1 2 + 4 l 2 θ ˙ 2 2 + l 2 cos ( θ 1 − θ 2 ) θ ˙ 1 θ ˙ 2 . Translational KE: 1 2 m ∣ G ˙ 2 ∣ 2 = m l 2 θ ˙ 1 2 2 + m l 2 θ ˙ 2 2 8 + m l 2 cos ( θ 1 − θ 2 ) θ ˙ 1 θ ˙ 2 2 \tfrac{1}{2}m|\dot G_2|^{2}=\tfrac{ml^{2}\dot\theta_1^{2}}{2}+\tfrac{ml^{2}\dot\theta_2^{2}}{8}+\tfrac{ml^{2}\cos(\theta_1-\theta_2)\dot\theta_1\dot\theta_2}{2} 2 1 m ∣ G ˙ 2 ∣ 2 = 2 m l 2 θ ˙ 1 2 + 8 m l 2 θ ˙ 2 2 + 2 m l 2 c o s ( θ 1 − θ 2 ) θ ˙ 1 θ ˙ 2
Rotational KE about G 2 G_2 G 2 1 2 ⋅ m l 2 12 θ ˙ 2 2 = m l 2 θ ˙ 2 2 24 \tfrac{1}{2}\cdot\tfrac{ml^{2}}{12}\dot\theta_2^{2}=\tfrac{ml^{2}\dot\theta_2^{2}}{24} 2 1 ⋅ 12 m l 2 θ ˙ 2 2 = 24 m l 2 θ ˙ 2 2
Total:
T = m l 2 6 θ ˙ 1 2 + m l 2 2 θ ˙ 1 2 + m l 2 8 θ ˙ 2 2 + m l 2 24 θ ˙ 2 2 + m l 2 cos ( θ 1 − θ 2 ) 2 θ ˙ 1 θ ˙ 2 . T=\dfrac{ml^{2}}{6}\dot\theta_1^{2}+\dfrac{ml^{2}}{2}\dot\theta_1^{2}+\dfrac{ml^{2}}{8}\dot\theta_2^{2}+\dfrac{ml^{2}}{24}\dot\theta_2^{2}+\dfrac{ml^{2}\cos(\theta_1-\theta_2)}{2}\dot\theta_1\dot\theta_2. T = 6 m l 2 θ ˙ 1 2 + 2 m l 2 θ ˙ 1 2 + 8 m l 2 θ ˙ 2 2 + 24 m l 2 θ ˙ 2 2 + 2 m l 2 cos ( θ 1 − θ 2 ) θ ˙ 1 θ ˙ 2 . Combine like terms:
T = 2 m l 2 3 θ ˙ 1 2 + m l 2 6 θ ˙ 2 2 + m l 2 2 cos ( θ 1 − θ 2 ) θ ˙ 1 θ ˙ 2 . T=\dfrac{2ml^{2}}{3}\dot\theta_1^{2}+\dfrac{ml^{2}}{6}\dot\theta_2^{2}+\dfrac{ml^{2}}{2}\cos(\theta_1-\theta_2)\dot\theta_1\dot\theta_2. T = 3 2 m l 2 θ ˙ 1 2 + 6 m l 2 θ ˙ 2 2 + 2 m l 2 cos ( θ 1 − θ 2 ) θ ˙ 1 θ ˙ 2 . Step 2 — Potential energy
Take zero PE at the suspension point A A A A A A G 1 G_1 G 1 l 2 cos θ 1 \tfrac{l}{2}\cos\theta_1 2 l cos θ 1 G 2 G_2 G 2 l cos θ 1 + l 2 cos θ 2 l\cos\theta_1+\tfrac{l}{2}\cos\theta_2 l cos θ 1 + 2 l cos θ 2
V = − m g ( l 2 cos θ 1 ) − m g ( l cos θ 1 + l 2 cos θ 2 ) = − m g ( 3 l 2 cos θ 1 + l 2 cos θ 2 ) . V=-mg\!\left(\dfrac{l}{2}\cos\theta_1\right)-mg\!\left(l\cos\theta_1+\dfrac{l}{2}\cos\theta_2\right)=-mg\!\left(\dfrac{3l}{2}\cos\theta_1+\dfrac{l}{2}\cos\theta_2\right). V = − m g ( 2 l cos θ 1 ) − m g ( l cos θ 1 + 2 l cos θ 2 ) = − m g ( 2 3 l cos θ 1 + 2 l cos θ 2 ) . Step 3 — Linearise for small oscillations
cos θ ≈ 1 − θ 2 / 2 \cos\theta\approx 1-\theta^{2}/2 cos θ ≈ 1 − θ 2 /2
V ≈ 3 m g l 4 θ 1 2 + m g l 4 θ 2 2 . V\approx\dfrac{3mgl}{4}\theta_1^{2}+\dfrac{mgl}{4}\theta_2^{2}. V ≈ 4 3 m g l θ 1 2 + 4 m g l θ 2 2 . For T T T cos ( θ 1 − θ 2 ) ≈ 1 \cos(\theta_1-\theta_2)\approx 1 cos ( θ 1 − θ 2 ) ≈ 1
T ≈ 2 m l 2 3 θ ˙ 1 2 + m l 2 6 θ ˙ 2 2 + m l 2 2 θ ˙ 1 θ ˙ 2 . T\approx\dfrac{2ml^{2}}{3}\dot\theta_1^{2}+\dfrac{ml^{2}}{6}\dot\theta_2^{2}+\dfrac{ml^{2}}{2}\dot\theta_1\dot\theta_2. T ≈ 3 2 m l 2 θ ˙ 1 2 + 6 m l 2 θ ˙ 2 2 + 2 m l 2 θ ˙ 1 θ ˙ 2 . Step 4 — Mass and stiffness matrices
Writing T = 1 2 θ ˙ T M θ ˙ T=\tfrac{1}{2}\dot{\boldsymbol\theta}^{T}M\dot{\boldsymbol\theta} T = 2 1 θ ˙ T M θ ˙ V = 1 2 θ T K θ V=\tfrac{1}{2}\boldsymbol\theta^{T}K\boldsymbol\theta V = 2 1 θ T K θ
M = ( 4 m l 2 3 m l 2 2 m l 2 2 m l 2 3 ) , K = ( 3 m g l 2 0 0 m g l 2 ) . M=\begin{pmatrix}\dfrac{4ml^{2}}{3} & \dfrac{ml^{2}}{2}\\[6pt] \dfrac{ml^{2}}{2} & \dfrac{ml^{2}}{3}\end{pmatrix},\quad K=\begin{pmatrix}\dfrac{3mgl}{2} & 0\\[6pt] 0 & \dfrac{mgl}{2}\end{pmatrix}. M = 3 4 m l 2 2 m l 2 2 m l 2 3 m l 2 , K = 2 3 m g l 0 0 2 m g l . Step 5 — Normal-mode characteristic equation
det ( K − n 2 M ) = 0 \det(K-n^{2}M)=0 det ( K − n 2 M ) = 0
det ( 3 m g l 2 − 4 m l 2 n 2 3 − m l 2 n 2 2 − m l 2 n 2 2 m g l 2 − m l 2 n 2 3 ) = 0. \det\!\begin{pmatrix}\dfrac{3mgl}{2}-\dfrac{4ml^{2}n^{2}}{3} & -\dfrac{ml^{2}n^{2}}{2}\\[6pt] -\dfrac{ml^{2}n^{2}}{2} & \dfrac{mgl}{2}-\dfrac{ml^{2}n^{2}}{3}\end{pmatrix}=0. det 2 3 m g l − 3 4 m l 2 n 2 − 2 m l 2 n 2 − 2 m l 2 n 2 2 m g l − 3 m l 2 n 2 = 0. Factor ( m l / 6 ) (ml/6) ( m l /6 )
3 m g l 2 − 4 m l 2 n 2 3 = m l 6 ( 9 g − 8 l n 2 ) , m g l 2 − m l 2 n 2 3 = m l 6 ( 3 g − 2 l n 2 ) . \dfrac{3mgl}{2}-\dfrac{4ml^{2}n^{2}}{3}=\dfrac{ml}{6}(9g-8ln^{2}),\quad\dfrac{mgl}{2}-\dfrac{ml^{2}n^{2}}{3}=\dfrac{ml}{6}(3g-2ln^{2}). 2 3 m g l − 3 4 m l 2 n 2 = 6 m l ( 9 g − 8 l n 2 ) , 2 m g l − 3 m l 2 n 2 = 6 m l ( 3 g − 2 l n 2 ) . Determinant becomes:
m 2 l 2 36 ( 9 g − 8 l n 2 ) ( 3 g − 2 l n 2 ) − m 2 l 4 n 4 4 = 0. \dfrac{m^{2}l^{2}}{36}(9g-8ln^{2})(3g-2ln^{2})-\dfrac{m^{2}l^{4}n^{4}}{4}=0. 36 m 2 l 2 ( 9 g − 8 l n 2 ) ( 3 g − 2 l n 2 ) − 4 m 2 l 4 n 4 = 0. Multiply by 36 / ( m 2 l 2 ) 36/(m^{2}l^{2}) 36/ ( m 2 l 2 )
( 9 g − 8 l n 2 ) ( 3 g − 2 l n 2 ) − 9 l 2 n 4 = 0. (9g-8ln^{2})(3g-2ln^{2})-9l^{2}n^{4}=0. ( 9 g − 8 l n 2 ) ( 3 g − 2 l n 2 ) − 9 l 2 n 4 = 0. Expand:
27 g 2 − 18 g l n 2 − 24 g l n 2 + 16 l 2 n 4 − 9 l 2 n 4 = 27 g 2 − 42 g l n 2 + 7 l 2 n 4 = 0. 27g^{2}-18gln^{2}-24gln^{2}+16l^{2}n^{4}-9l^{2}n^{4}=27g^{2}-42gln^{2}+7l^{2}n^{4}=0. 27 g 2 − 18 g l n 2 − 24 g l n 2 + 16 l 2 n 4 − 9 l 2 n 4 = 27 g 2 − 42 g l n 2 + 7 l 2 n 4 = 0. Step 6 — Solve for n 2 n^{2} n 2
Quadratic in n 2 n^{2} n 2
7 l 2 ( n 2 ) 2 − 42 g l ( n 2 ) + 27 g 2 = 0. 7l^{2}(n^{2})^{2}-42gl(n^{2})+27g^{2}=0. 7 l 2 ( n 2 ) 2 − 42 g l ( n 2 ) + 27 g 2 = 0. n 2 = 42 g l ± ( 42 g l ) 2 − 4 ⋅ 7 l 2 ⋅ 27 g 2 2 ⋅ 7 l 2 = 42 g l ± 1764 − 756 g l 14 l 2 = 42 g l ± 12 7 g l 14 l 2 . n^{2}=\dfrac{42gl\pm\sqrt{(42gl)^{2}-4\cdot 7l^{2}\cdot 27g^{2}}}{2\cdot 7l^{2}}=\dfrac{42gl\pm\sqrt{1764-756}\,gl}{14l^{2}}=\dfrac{42gl\pm 12\sqrt 7\,gl}{14l^{2}}. n 2 = 2 ⋅ 7 l 2 42 g l ± ( 42 g l ) 2 − 4 ⋅ 7 l 2 ⋅ 27 g 2 = 14 l 2 42 g l ± 1764 − 756 g l = 14 l 2 42 g l ± 12 7 g l . (Using 1008 = 144 ⋅ 7 = 12 7 \sqrt{1008}=\sqrt{144\cdot 7}=12\sqrt 7 1008 = 144 ⋅ 7 = 12 7
Simplify:
n 2 = g ( 42 ± 12 7 ) 14 l = g ( 21 ± 6 7 ) 7 l = g l ( 3 ± 6 7 7 ) = g l ( 3 ± 6 7 ) . n^{2}=\dfrac{g(42\pm 12\sqrt 7)}{14l}=\dfrac{g(21\pm 6\sqrt 7)}{7l}=\dfrac{g}{l}\!\left(3\pm\dfrac{6\sqrt 7}{7}\right)=\dfrac{g}{l}\!\left(3\pm\dfrac{6}{\sqrt 7}\right). n 2 = 14 l g ( 42 ± 12 7 ) = 7 l g ( 21 ± 6 7 ) = l g ( 3 ± 7 6 7 ) = l g ( 3 ± 7 6 ) . (Using 6 7 / 7 = 6 / 7 6\sqrt 7/7=6/\sqrt 7 6 7 /7 = 6/ 7
Answer
n 2 = ( 3 ± 6 7 ) g l ; periods T = 2 π n . \boxed{\;n^{2}=\!\left(3\pm\dfrac{6}{\sqrt 7}\right)\dfrac{g}{l};\;\text{periods }T=\dfrac{2\pi}{n}.\;} n 2 = ( 3 ± 7 6 ) l g ; periods T = n 2 π .