Suppose the rod in the previous problem has a mass of 1 kg distributed uniformly over its length.

(a) Find the initial angular acceleration of the rod.


(b) Find the tension in the supports to the blocks of mass 2 kg and 5 kg.


The angular acceleration is given as


The tension on the 2kg and 5kg masses are 27.6N and 29N


Given


The length of the rod is given as 1m and is pivoted at the center with two masses at the center of the rod as 2kg and 5kg and mass is uniformly distributed.


Formula Used


The formula used for the calculation of the angular acceleration in a uniform distributed mass is equivalent to the torque applied on the body by the application of the angular acceleration is



where


τnet is the net torque of the object, Inet is the net moment of Inertia on the rod and α is the angular acceleration.


Explanation


The moment of the inertia is given as



Therefore, the total moment of inertia is given as



The angular acceleration of the rod at initial moment is given as




The tension applied in the block of the given masses


1. For 2 kg mass is




2. For 5 kg mass is



1