Two thin rods of length l are rotating with the same angular speed ω (in rad/s) about axes that pass perpendicularly through one end. rod a is massless but has a particle of mass 0.76 kg attached to its free end. rod b has a mass of 0.76 kg, which is distributed uniformly along its length. the length of each rod is 0.88 m, and the angular speed is 5.2 rad/s. find the kinetic energies of rod a with its attached particle and of rod b.

To solve the exercise it is necessary to apply the concepts related to Kinetic Energy and its respective definition. Angular kinetic energy means that,

Where

I = Inertia moment

\omega = Angular velocity

On the other hand the inertia of a rod is given by

Where,

m = mass

l = length

ROD A: The data for Rod A is given as,

l = 0.88m

m = 0.76Kg

The inertia would be

Therefore the Kinetic Energy would be

ROD B: The data for Rod B is given as

The inertia would be

Note that the moment of inertia is not about its reference system, but about its end.

Therefore the Kinetic Energy would be

number of electrons for neutral cl

the maxium height is 34m/s