Piece-wise Uniform Rod Energy

Classical Mechanics Level 3

Consider a uniform rod of length b b . One half of this rod has a linear mass density of λ 1 = 9 \lambda_1 = 9 while the other half has a linear mass density of λ 2 = 1 \lambda_2 = 1 . The rod is initially placed horizontally in a smooth, fixed hemispherical bowl and the denser side is on the left. It is then released from rest at this initial position. The total energy of the rod at an arbitrary instant of time is given by:

E t o t a l = A b B ( R 2 − b 2 C ) θ ˙ 2 + b g ( D b E sin ⁡ θ − F R sin ⁡ ( β + θ ) ) \boxed{E_{total} = \frac{A b}{B}\left(R^2 - \frac{b^2}{C}\right)\dot{\theta}^2 + bg\left(\frac{D b}{E}\sin{\theta} - F R \sin(\beta+\theta)\right)}

Here, A , B , C , D , E , F A, \ B, \ C, \ D, \ E, \ F are positive integers. The pairs A , B A,B and D , E D,E are co-prime.

Find A + B + C + D + E + F \boxed{A+B+C+D+E+F}

Note:

  • The diagram on the left shows the initial configuration while that on the right shows the configuration at an arbitrary instant of time.

  • Gravity ( g \ g \ ) acts vertically downwards.

  • β \beta is the angle between the radius and the rod as shown in the figure.

  • The horizontal diameter of the hemispherical bowl is taken as the zero gravitational potential energy level.

Hints/Suggestions:

  • The result of this problem might be useful.

  • Be patient.

  • Try to use the expression of the potential energy of the rod and attempt this problem .


The answer is 23.

Karan Chatrath by Karan Chatrath , 27, Netherlands

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1 solution

Steven Chase
Oct 13, 2019

Location of the "heavy" end of the rod:

x + R cos ⁡ ( θ + β ) = 0 y + R sin ⁡ ( θ + β ) = 0 x = − R cos ⁡ ( θ + β ) y = − R sin ⁡ ( θ + β ) x + R \, \cos (\theta + \beta) = 0 \\ y + R \, \sin (\theta + \beta) = 0 \\ x = -R \, \cos (\theta + \beta) \\ y = -R \, \sin (\theta + \beta)

Coordinates of center of mass. Let L L be the distance from the "heavy" rod end to the center of mass:

x c = − R cos ⁡ ( θ + β ) + L cos ⁡ θ y c = − R sin ⁡ ( θ + β ) + L sin ⁡ θ x ˙ c = R sin ⁡ ( θ + β ) θ ˙ − L sin ⁡ θ θ ˙ y ˙ c = − R cos ⁡ ( θ + β ) θ ˙ + L cos ⁡ θ θ ˙ x_c = -R \, \cos (\theta + \beta) + L \cos \theta \\ y_c = -R \, \sin (\theta + \beta) + L \sin \theta \\ \dot{x}_c = R \, \sin (\theta + \beta) \dot{\theta} - L \sin \theta \, \dot{\theta} \\ \dot{y}_c = -R \, \cos (\theta + \beta) \dot{\theta} + L \cos \theta \, \dot{\theta}

Square of COM speed:

v 2 = x ˙ c 2 + y ˙ c 2 = R 2 θ ˙ 2 + L 2 θ ˙ 2 − 2 R L θ ˙ 2 ( sin ⁡ θ sin ⁡ ( θ + β ) + cos ⁡ θ cos ⁡ ( θ + β ) ) = R 2 θ ˙ 2 + L 2 θ ˙ 2 − 2 R L θ ˙ 2 cos ⁡ β = R 2 θ ˙ 2 + L 2 θ ˙ 2 − b L θ ˙ 2 v^2 = \dot{x}_c^2 + \dot{y}_c^2 = R^2 \dot{\theta}^2 + L^2 \dot{\theta}^2 - 2 R L \dot{\theta}^2 \Big( \sin \theta \sin(\theta + \beta) + \cos \theta \cos(\theta + \beta) \Big) \\ = R^2 \dot{\theta}^2 + L^2 \dot{\theta}^2 - 2 R L \dot{\theta}^2 \cos \beta \\ = R^2 \dot{\theta}^2 + L^2 \dot{\theta}^2 - b L \dot{\theta}^2

Total energy (assuming circle center is potential energy reference):

1 2 m v 2 + 1 2 I θ ˙ 2 + m g y c = 1 2 ( 5 b ) ( R 2 θ ˙ 2 + L 2 θ ˙ 2 − b L θ ˙ 2 ) + 1 2 13 b 3 60 θ ˙ 2 + 5 b g ( − R sin ⁡ ( θ + β ) + L sin ⁡ θ ) \frac{1}{2} m v^2 + \frac{1}{2} I \dot{\theta}^2 + m g y_c \\ = \frac{1}{2} (5b) \Big( R^2 \dot{\theta}^2 + L^2 \dot{\theta}^2 - b L \dot{\theta}^2 \Big) + \frac{1}{2} \frac{13 \, b^3}{60} \dot{\theta}^2 + 5 b g (-R \, \sin (\theta + \beta) + L \sin \theta)

Manipulating this expression into a final form:

1 2 ( 5 b ) ( R 2 θ ˙ 2 + L 2 θ ˙ 2 − b L θ ˙ 2 ) + 1 2 13 b 3 60 θ ˙ 2 + 5 b g ( − R sin ⁡ ( θ + β ) + L sin ⁡ θ ) = 1 2 ( 5 b ) ( R 2 + 9 b 2 100 − 3 b 2 10 ) θ ˙ 2 + 1 2 13 b 3 60 θ ˙ 2 + 5 b g ( − R sin ⁡ ( θ + β ) + 3 b 10 sin ⁡ θ ) = 5 b 2 ( R 2 − 21 b 2 100 ) θ ˙ 2 + 13 b 3 120 θ ˙ 2 + b g ( − 5 R sin ⁡ ( θ + β ) + 3 b 2 sin ⁡ θ ) = 5 b 2 ( R 2 − 21 b 2 100 + 13 b 2 300 ) θ ˙ 2 + b g ( − 5 R sin ⁡ ( θ + β ) + 3 b 2 sin ⁡ θ ) = 5 b 2 ( R 2 − b 2 6 ) θ ˙ 2 + b g ( − 5 R sin ⁡ ( θ + β ) + 3 b 2 sin ⁡ θ ) \frac{1}{2} (5b) \Big( R^2 \dot{\theta}^2 + L^2 \dot{\theta}^2 - b L \dot{\theta}^2 \Big) + \frac{1}{2} \frac{13 \, b^3}{60} \dot{\theta}^2 + 5 b g (-R \, \sin (\theta + \beta) + L \sin \theta) \\ = \frac{1}{2} (5b) \Big( R^2 + \frac{9 b^2}{100} - \frac{3 b^2}{10} \Big) \dot{\theta}^2 + \frac{1}{2} \frac{13 \, b^3}{60} \dot{\theta}^2 + 5 b g (-R \, \sin (\theta + \beta) + \frac{3 b}{10} \sin \theta) \\ = \frac{5 b}{2} \Big( R^2 - \frac{21 b^2}{100} \Big) \dot{\theta}^2 + \frac{13 \, b^3}{120} \dot{\theta}^2 + b g \Big(-5 R \, \sin (\theta + \beta) + \frac{3 b}{2} \sin \theta \Big) \\ = \frac{5 b}{2} \Big( R^2 - \frac{21 b^2}{100} + \frac{13 b^2}{300} \Big) \dot{\theta}^2 + b g \Big(-5 R \, \sin (\theta + \beta) + \frac{3 b}{2} \sin \theta \Big) \\ = \frac{5 b}{2} \Big( R^2 - \frac{b^2}{6} \Big) \dot{\theta}^2 + b g \Big(-5 R \, \sin (\theta + \beta) + \frac{3 b}{2} \sin \theta \Big)

Final energy expression:

5 b 2 ( R 2 − b 2 6 ) θ ˙ 2 + b g ( − 5 R sin ⁡ ( θ + β ) + 3 b 2 sin ⁡ θ ) \frac{5 b}{2} \Big( R^2 - \frac{b^2}{6} \Big) \dot{\theta}^2 + b g \Big(-5 R \, \sin (\theta + \beta) + \frac{3 b}{2} \sin \theta \Big)

1 Helpful 1 Interesting 1 Brilliant 0 Confused

Thank you for the detailed solution. Based on your previous comment, I have added a note on the zero gravitational potential energy level.

Karan Chatrath - 1 year, 8 months ago

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