Question

This is the diagram that was provided.

3. It can be shown that the rotational inertia (moment of inertia) for a uniform rod about an axis that's perpendicular to the rod and passes through one of its ends is: Where M is the rod's total mass and L is its total length. (a) (10 points) Use the Parallel Axis Theorem to find the moment of inertia of a uniform rod about an axis that's perpendicular to the rod and passes through a point that's ¼ (0.25) down the length of the rod (so L/40.25L from one of its ends) in terms of M and L (b) (10 points) A uniform rod (which tells you something about the location of its center of mass) of length 0.40m and mass 3.0kg is attached to a pole on its left by a frictionless hinge located LI4 from its left end and is also attached to the ceiling by a light cord that's connected to its right end and forms an angle of 26 with the vertical as depicted in the diagram below. The rod is stationary (not moving). Find the magnitude of the tension in the cord as well as the horizontal and vertical components of the force exerted on the rod by the hinge. 0 (c) (10 points) The cord is now cut. Find the angular acceleration of the rod as it rotates about the hinge immediately after the cord is cut.

Answer 1

月 스 Ml® Using Ponauel ctes タheonun 3 그,-M止 12 Yow ogain 12 l 6 l 6 6

me Bo pancing 大on que abo.d hinge Bco26 When condcuet 16