Suppose that the rod is held at rest horizontally and then released. (throughout the remainder of this problem, your answer may include the symbol , the moment of inertia of the assembly, whether or not you have answered the first part correctly.) what is the angular acceleration of the rod immediately after it is released? take the counterclockwise direction to be positive. express in terms of some or all of the variables , , , , , and .

For better understanding refer the below solution.

Solution :

Let mass of rod is and inrertia be is given by

Let moment of inertia of particle whose mass is and is given by

Let moment of inertia of particle whose mass is and is given by

Now for the system toatal moment of inertia is

Torque on the pivot can be written as

Now net torque is

--- (1)

Now put the value of in equation (1)

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Explanation:

Mass of rod is and its moment of inertia is given by

Moment of inertia of particle of mass is given by

Moment of inertia of particle of mass is given by

Total moment of inertia for the system is given by

Torque on the pivot is given by

Net torque is given by

Making or simply  

positive reinforcement? ?

see the picture for the vector

the magnitude of the vector by pythagorean theorem

r = √( (-10)^2 + (3)^2 ) = √109

magnitude = √109 units

the direction angle between vector and x-axis is

tan^(-1)(3/10) = 16.699

so the direction is approximately 16.7 degrees north of west