Rotational Dynamics of 2 Particles and 2 Rods

[cd80187]

In Fig. 10-35, two particles, each with mass m = 2.30 kg, are fastened to each other, and to a rotation axis at O, by two thin rods, each with length d = 0.670 m and mass M = 0.360 kg. The combination rotates around the rotation axis with angular speed ω = 0.211 rad/s. Measured about O, what are the combination's (a) rotational inertia and (b) kinetic energy?
(I'm not sure how to post pictures on here, because the picture is on wileyplus, and to view it, you need a password)

But for this problem, I am not even sure where to start. I know that I need to find the total inertia, but how many parts should i split it into? Should I take each rod seperately and each particle seperately, and use the summation formula for inertia, or should i use the parallel axis theorem? I am just so lost on where to start.

 

[OlderDan]

If the configuration is axis-rod-particle-rod-particle, then you will want to use the parallel axis theorem for the rods and add the contributions from the masses. The parallel axis theorm has already been used for the rod rotating at one end in most places where you look up moments of inertia. Since the second rod is not touching the axis, you will have to compute it yourself.

 

[kyle8921]

As a scientist, my response would be to approach this problem by breaking it down into smaller, manageable parts. In this case, since we are dealing with rotational dynamics, it would be helpful to consider the individual rotational inertia of each component (particles and rods) and then combine them to find the total rotational inertia of the system.

To find the rotational inertia of each particle, we can use the formula I = mr^2, where m is the mass and r is the distance from the rotation axis. In this case, both particles have the same mass (m = 2.30 kg) and are located at a distance of d/2 = 0.335 m from the rotation axis. Therefore, the rotational inertia of each particle is I = (2.30 kg)(0.335 m)^2 = 0.387 kg*m^2.

Next, we can use the parallel axis theorem to find the rotational inertia of each rod. This theorem states that the rotational inertia of an object about an axis parallel to its center of mass is equal to the rotational inertia about its center of mass plus the product of its mass and the square of the distance between the two axes. In this case, the center of mass of each rod is located at its midpoint, so the distance between the rotation axis and the center of mass is d/2 = 0.335 m. Therefore, the rotational inertia of each rod is I = (1/12)(M)(d^2) + (M)(0.335 m)^2 = 0.024 kg*m^2.

To find the total rotational inertia of the system, we can simply add the individual inertias of the particles and rods. So, the total rotational inertia of the system is I = (2)(0.387 kg*m^2) + (2)(0.024 kg*m^2) = 0.846 kg*m^2.

To find the kinetic energy of the system, we can use the formula KE = (1/2)(I)(ω^2), where I is the rotational inertia and ω is the angular speed. Plugging in the values, we get KE = (1/2)(0.846 kg*m^2)(0.211 rad/s)^2 = 0.0090 J.

In summary, to solve this problem, we used the formulas for rotational inertia and kinetic energy, as well as the parallel axis theorem, to find the