In the figure below, two particles, each with mass m = 0.82 kg, are fastened to each other, and to a rotation axis at O, by two thin rods, each with length d = 10 cm and mass M = 2.0 kg. The combination rotates around the rotation axis with angular speed ù = 0.34 rad/s.
(a) Measured about O, what is the combination's rotational inertia?
(b) What is the combination's kinetic energy?
The figure looks like this:
Rotation axis-rod-particle-rod-particle
At first I used (1/3)(M)(L^2) to find the moment of inertia but that is not working. I also tried finding the individual kinetic energies and the total kinetic energy (I converted omega to velocity using radius of .2m)but neither worked. I've tried some other methods but just cant seem to get the right answer.
To find the rotational inertia of the combination in this problem, you will need to consider the moment of inertia of the rods and the particles separately, since they have different distributions of mass.
(a) Measured about O, the rotational inertia of the combination is the sum of the rotational inertias of the rods and the particles.
For the rods:
The moment of inertia of each rod can be computed using the parallel axis theorem, which states that the moment of inertia of an object about an axis parallel to and a distance d away from an axis through its center of mass is equal to the sum of the moment of inertia about the center of mass and the product of the object's mass and the square of the distance between the two axes.
For a thin rod rotating about its center of mass, the moment of inertia is given by (1/12) * M * L^2, where M is the mass of the rod and L is its length.
Applying the parallel axis theorem for both rods, the moment of inertia of the combination due to the rods is:
I_rods = (1/12) * M * L^2 + (1/12) * M * L^2 = (1/6) * M * L^2
For the particles:
Each particle's moment of inertia can be considered as a point mass rotating about the rotation axis, which results in a moment of inertia of m * r^2, where m is the mass of the particle and r is the distance from the rotation axis to the particle.
Since the rods are connected (perpendicularly) to the rotation axis at a distance d = 10 cm, the distance of each particle to the rotation axis is the perpendicular distance from the rod to the rotation axis, which is d/2 = 5 cm = 0.05 m.
Therefore, the moment of inertia of the combination due to the particles is:
I_particles = m * r^2 + m * r^2 = 2 * (m * r^2) = 2 * m * (0.05^2) = 0.005 m kg^2
The total rotational inertia of the combination measured about O is the sum of the rotational inertias of the rods and the particles:
I_total = I_rods + I_particles = (1/6) * M * L^2 + 0.005 m kg^2
(b) The kinetic energy of the combination can be calculated using the formula:
KE = (1/2) * I * ω^2
Substituting the given value, ω = 0.34 rad/s, and the previously calculated value of I_total, you can find the kinetic energy of the combination.