A uniform meter stick (with a length of 1 m) is pivoted to rotate about a horizontal axis through the 25 cm mark on the stick. The stick is released from rest in a horizontal position. The moment of inertia of a uniform rod about an axis perpendicular to the rod and through the center of mass of the rod is given by 1 12 ML2. Determine the magnitude of the initial angular acceleration of the stick

The initial angular acceleration of the stick is 16.812 radians per square second.

First, we calculate the resulting moment of inertia by Steiner theorem:

(1)

Where:

If we know that , and , then the formula for the moment of inertia is:

(2)

We know that initial angular acceleration (), in radians per square second, is solely due to gravity. By the second Newton's law and the D'Alembert principle, we derive an expression for the initial angular acceleration of the stick:

(3)

Where:

By (2) and (3) we have the following simplified expression:

(4)

If we know that and , then the initial angular acceleration of the stick is:

The initial angular acceleration of the stick is 16.812 radians per square second.

To learn more on moment of inertia, we kindly invite to read this verified question: link

The initial angular acceleration is

Explanation:

From the parallel axis theorem the moment of inertia about 25cm mark is

since & , we have

Now, the gravitational force (equal to the weight of the object) acts as a torque on the center of mass of the rod, which induces angular acceleration according to

since

solving for we get

putting in and we get:

which is the initial angular acceleration.

gravititional force is a force that is acting around you, it is non-contactable and it is a force that pulls you to the core of the earth.

brainliest

i think the answer would be b