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 L. Obviously, it is the potential energy only:
MEi=mg2L.
Total final mechanical energy is the kinetic energy of the rotating rod with moment of inertia J:
MEf=21Jω2=21⋅31mL2⋅ω2.
Law of conservation of energy states:
MEi=MEf⟺mgL=31mL2ω2,ω=L3g.
Meanwhile the rod passed an angle ϕ=π/2, and the angular acceleration is
α=2ϕω2=Lπ3g=1.2⋅3.143⋅9.8=7.8 s−2.
b) The tangential acceleration of the tip:
a=ω2L=3g=3⋅9.8=29.4 m/s2,
of the center of mass:
acm=3g/2=1.5⋅9.8=14.7 m/s2.
c) The rod would just fall down parallel to the ground if no other forces influence it.
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