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 0.73 m, and the angular speed is 4.7 rad/s. find the kinetic energies of rod a with its attached particle and of rod b.

Explanation:

For rod A:

      Length = 0.73 m

      Mass of particle attached, m = 0.76 kg

Moment of inertia of this system about the given axis is as follows.

              I =

                =

                = 0.405

Angular speed of the rod () = 4.7 rad/s

Now, the relation between kinetic energy, moment of inertia and angular speed is as follows.

                   K.E =

                          =

                          = 4.47 J

Therefore, kinetic energy of rod A is 4.47 J.

For rod B:

          Length = 0.73 m

          Mass of the rod, m = 0.76 kg

Moment of inertia will be as follows.

                    I =

Putting the given values into the above formula as follows.

                  I =

                    =  

                    = 0.135

Angular speed is given as 4.7 rad/s. Therefore, calculate the value of kinetic energy as follows.

                       K.E =

                              =

                              = 2.19 J

Hence, kinetic energy of rod B is 2.19 J.

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