Balls, balls, balls...
Two identical spheres of radius R and mass M are joined by a rod of lenght 4R. Initially one sphere is vertically above the other, at a very unstable equilibrium. A tiny fly hits the upper sphere and destabilizes the equilibrium, making the spheres start to move. If T is the tension at the rod immediately before the upper sphere reaches the ground, Calculate 46 T/(3Mg)
The answer is 15.
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1 solution
Consider a moment when the rod makes an angle theta to the vertical. The net horizontal force that acts on the lower shpere is T sin θ Let's put both spheres on a varying field to the left with magnitude T sin θ / m This way the lower sphere will remain still and the upper sphere will describe a circumference around it. Then we can decompose the forces on the normal direction and we find T = ( M g cos θ − M v ² / 6 R ) / ( 1 + sin ² θ ) So when theta = 90 degrees, T = Mv²/12R. Now let's take out the field (we can do that because at the final configuration the velocity is all vertical, and the field is horizontal, so the final velocity of the balls with the field will be the final velocity of the balls without the field). Without the field the movement is like that: Center of mass is falling with a velocity vy and the rod with the balls are rotating with angular velocity w. As the vertical velocity of the lower ball is zero, vy = w.3R The conservation of energy says that 2 M v y ² / 2 + I w ² / 2 = M g . 6 R where I = 2(2/5 MR² + 9MR²) = 94/5 MR² Substituting we get: vy² = 135/46 gR, and so v = vy + w.3R = 2vy -> T = 45/46 Mg, and the answer is 15 if didn't make any mistake