A rod of length L and mass M_r is attached to a solid sphere of radius R and mas
ID: 1507697 • Letter: A
Question
A rod of length L and mass M_r is attached to a solid sphere of radius R and mass M_a, and the system is rotated about a vertical axis at an angular velocity omega. What is the moment of inertia of only the rod using the Parallel Axis Theorem: I_z = I_cM + Mh^2 given that the moment of inertia a rod for rotations about its center of mass is I_CM = 1/2 M_r^r^2. Use the Parallel Axis Theorem to find the total moment of inertia of the rod+sphere system for rotations about the vertical axis shown above.Explanation / Answer
Parallel axis theorem:
If you ALREADY know the moment of inertia of an object through its own center of mass, and wish to find this value about a parallel axis that doesn't pass through the center of mass, you can calculate the increase in moment of inertia by multiplying mass by the offset distance squared. Remember, you know the value at the center of mass, and you want a shortcut to shift away from the center of mass.
IMPORTANT: the parallel axis theorem can ONLY be used if you are either going to or from the center of mass. To go between axes of which neither pass through the center of mass, you need to use two iterations of the parallel axis theorem to first get to the center of mass, and then go away from the center of mass.
An example of using this is to calculate moment of inertia for a rod about its end point. You already have a formula for moment of inertia of a rod about its center, and that is I0=1/12*m*L^2. How to get it for a rod about its end? Well, you shift away from the center of mass by a distance of L/2. Thus add m*(L/2)^2 to your original center of mass moment of inertia, and get I = 1/12*m*L^2 + m*(L/2)^2.
So I = Icm + Mh^(2), where L = r, L/2 = h, m = Mr
For part b please type the seperate question again
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