Answer to Question #85244 in Mechanics | Relativity for Beast

Question #85244
The length of the flat 2.4 kg rod is 1.2 m. The rod is attached the other end with a joint so that it can rotate freely. In the initial situation, the rod is held stationary horizontally. a) what is the angular acceleration of the rod at α release? b) what are the rods tangential accelerations of tip and center of gravity at? c) if the rod tip would there be a coin free, so what would happen to it at the time of release?
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Expert's answer
2019-02-18T14:27:40-0500

a) Take total initial mechanical energy relative to the lowest possible position of the rod with length LL. Obviously, it is the potential energy only:


MEi=mgL2.ME_\text{i}=mg\frac{L}{2}.


Total final mechanical energy is the kinetic energy of the rotating rod with moment of inertia JJ:


MEf=12Jω2=1213mL2ω2.ME_\text{f}=\frac{1}{2}J\omega^2=\frac{1}{2}\cdot \frac{1}{3}mL^2\cdot \omega^2.


Law of conservation of energy states:


MEi=MEf    mgL=13mL2ω2,ME_\text{i}=ME_\text{f} \iff mgL=\frac{1}{3}mL^2\omega^2,ω=3gL.\omega=\sqrt{\frac{3g}{L}}.


Meanwhile the rod passed an angle ϕ=π/2\phi=\pi/2, and the angular acceleration is


α=ω22ϕ=3gLπ=39.81.23.14=7.8 s2.\alpha=\frac{\omega^2}{2\phi}=\frac{3g}{L\pi}=\frac{3\cdot 9.8}{1.2\cdot 3.14}=7.8\text{ s}^{-2}.


b) The tangential acceleration of the tip:


a=ω2L=3g=39.8=29.4 m/s2,a=\omega^2L=3g=3\cdot9.8=29.4\text{ m/s}^2,


of the center of mass:


acm=3g/2=1.59.8=14.7 m/s2.a_\text{cm}=3g/2=1.5\cdot 9.8=14.7\text{ m/s}^2.


c) The rod would just fall down parallel to the ground if no other forces influence it.


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