thin rod of mass m r and length 2 L is allowed to pivot freely about its center,

Question

thin rod of mass mr and length 2L is allowed to pivot freely about its center, as shown in the diagram. A small sphere of mass m1 is attached to the left end of the rod, and a small sphere of mass m2 is attached to the right end. The spheres are small enough that they can be considered point particles. The gravitational force acts downward, with the magnitude of the gravitational acceleration equal to g.

B. Suppose the rod is held at rest horizontally and then released. (Throughout the remainder of this problem, your answer may include the symbol

Explanation / Answer

A) Find the moment of inertia of a rod of a given mass and length about its center. You should have a table in your book that gives you this.

Then add for each mass:
(mass)*(distance from mass to axis)^2

Add em up and that's your answer (I)

B) Torque (N)
= angular accel (alpha) x moment of inertia (I)

Each weight exerts a torque equal to its weight (mg) times the distance from the weight to the axis. Use the convention (clockwise is -, counter is +) to add them up.

Once you have the total torque, solve for alpha.

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