Question

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.

Answer 1

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 = $ml^{2}$

                = $0.76 kg \times (0.73)^{2}$

                = 0.405 $kg m^{2}$

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

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

                   K.E = $\frac{1}{2} I \omega^{2}$

                          = $\frac{1}{2} 0.405 kg m^{2} (4.7)^{2}$

                          = 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 = $\frac{1}{3} ml^{2}$

Putting the given values into the above formula as follows.

                  I = $\frac{1}{3} ml^{2}$

                    = $\frac{1}{3} \times 0.76 kg (0.73)^{2}$  

                    = 0.135 $kg m^{2}$

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

                       K.E = $\frac{1}{2} I \omega^{2}$

                              = $\frac{1}{2} \times 0.135 \times (5.7)^{2}$

                              = 2.19 J

Hence, kinetic energy of rod B is 2.19 J.