Find the moment of inertia of the rod of mass m about the axis shown below by direct integration. (Use the following a necessary: m and L.) I = += X 4L 5

Moment of inertia Derive the formula for the moment of inertia of a uniform thin rod of length L and mass M about an axis through its center, perpendicular to its face. Repeat the calculation, only now assume the rod has a density that increases uniformly from a value of po on one end to 2po on the other end. Suggestion For the second integration, it is important to define the density function correctly: It will be a straight line that includes the two points (-L/2, po) and (L/2, 2po). Find the slope of the line as rise of run, and the x-intercept, p(0). Total mass of the rod, M, will be the integral of the density function from -L/2 to L/2. If you solve the second integration correctly, including the value of the total mass, M, in terms of p, you should see a very interesting relationship between the two moments of inertia.

A wagon wheel heading down south from the land of the pines consists of a thin ring having a mass of me and six spokes made from slender rods with each having a mass of m,.. Tw Variable Value mc 7 kg mr 1.3 kg Tw 0.85 m a.) What is the moment of inertia through the center of the wheel in kg - m²? b.) What is the moment of inertia about point A in kg - m²?? A

A rod of mass 3m and length L can pivot about an axis through its center and perpendicular to the length of the rod as shown below. The rod has 2 point masses attached to either end, each of mass m. What is the moment of inertia of the rod + masses system about the axis through the center?               Needs Complete typed solution with 100 % accuracy.