The length of each rod is 0.550000000000000000000000 m, and the angular speed is 3.60000000000000 rad/s. The m

Question

The length of each rod is 0.55 m, and the angular speed is 3.60 rad/s. The mass of the particle attached to the massless rod A is 0.68 kg. The mass of the uniform rod B is also 0.68 kg.
(a1) What is the algebraic expression for the rotational kinetic energy KER? Express your answer in terms of the moment of inertia I and the angular speed ω.
(b1) What is the rotational kinetic energy KER for rod A with its attached particle?
(c1) What is the rotational kinetic energy KER for rod B?

Answer 1

a1. KER=Iw^2/2, just like KET (translation) = mv^2/2.
You have the formula for KER and you can compute w^2. Now you need the I values to go ahead with the next two questions:
b1. For A, I=mr^2. (r is the distance from the particle to the center of rotation.)
c1. For B, I=(mL^2)/3 if rotating about one end, (mL^2)/12 if rotating about its center. (L is length of rod.)

And for heavens' sake, stop with all the zeros! I mean, the keyboard repeat feature is wonderful, but it's been around for 20 years!