Question

0/10 points 1 Provious A light rod of length-1.00 m rotates about an axis perpendicular to connected to the ends of the rod length and passing through its center as in the figure below. Two particles of masses m - 4.35 kg and m2-3.00 kg are (o) Neglecting the mass of the rod, what is the system's kinetic energy when its angular speed is 2.0 radys? Your response differs from the correct answer by more than 10%. Double check your calculations (b) Repeat the problem, assuming the mass of the rod is taken to be 1.60 kg leed Help? Read

Answer 1

kinetic energy=1/2*I*ω^2

now I=summation mr^2

r = l/2 = ½ = 0.5m

for (A)

I=[4.35 * (0.5)^2] + [3.00 * (0.5)^2]

=1.8375 kg m^2

so KE = (1/2)*1.8375*2.10^2 = 4.05 J

for (B)

I=[4.35 * (0.5)^2] + [3.00 * (0.5)^2] + [1/12 * 1.6 * (1)^1/2]

=1.8375 + 0.1333

= 1.9708 kg m^2

so KE= (1/2)*1.9708*2.10^2 = 4.35 J