The discussion focuses on calculating the moment of inertia (MoI) and angular acceleration of two rockets connected by a rod. The critical equations used include τ = I*α and α = ω/t, with specific values for mass and distances provided. The calculation of the center of mass (CM) is established as 67.102 m, leading to a torque (τ) of 4288680 N. The conversation highlights the importance of accurately determining the MoI using the parallel axis theorem and correcting for the rod's distribution of mass.
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haruspex said:Your answer would have been correct if the CoM had been fixed in space on an axle. What will happen instead?
haruspex said:Yes, thinking about this again I believe your method is correct, but in the calculation you treated the tunnel as a point mass at its centre. That slightly underestimates the moment of inertia of the system. I get 0.166. Is that error enough to explain your answer's rejection?
Use the parallel axis theorem. The MoI about one end is simply a special case of that.Esoremada said:That was correct! So what is the correct way to calculate the moment of a inertia of a rod about a given pivot? I only know that it is 1/12*ML^2 and 1/3*ML^2 for pivots about the centre or end.
Isn't it 12500?Esoremada said:Oh cool I never even noticed that. I get how to derive 1/3*ML^2 now but I try doing it in this problem and get it wrong
1250*(67.102 - 99/2)^2 + 1/12*1250*99^2
That's the moment of the rod rotating about its centre plus the moment of the rod's centre of mass rotating about the systems centre of mass. What am I doing wrong?