given cylindrical rod of length L = 0.240 m , mass m = 1.20 kg & radius r = 0.01
ID: 1701052 • Letter: G
Question
givencylindrical rod of length L = 0.240 m , mass m = 1.20 kg & radius r = 0.015 m
ball of diameter d = 0 .08 m & mass m = 2 kg
a) After rod rotates through 90 degrees its rotational kinetic energy is caculated as
using law of conservation of energy:
KEfinal+ PEfinal = KEinitial + PEinitial + ?E
KEf+ 0 = 0 + mrodg(L/2)+ mballg(Lrod + rball) + 0
KEfinal = the final rotational kinetic energy
KEf = (1.2kg)(9.8 m/s2)(0.24m/2) + 2kg(9.8 m/s2)(0.24m+0 .04m)
= 6.90J
why h is given by L/2 and not L? I thought the total difference in distance is L for the rod.
Explanation / Answer
the potential energy is in the center of mass reference frame
so ball potential enegy for ball is
mgh = mg ( L+r)
rod potential enrgy is mg (L/2)
so total potential energy is= mg ( L/2 ) + mg (L+r)
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