Very Simple Rotation Inertia of a Thin Rod Problem

[Oijl]

Homework Statement

Calculate the rotational inertia of a meter stick, with mass 0.68 kg, about an axis perpendicular to the stick and located at the 21 cm mark. (Treat the stick as a thin rod.)

Homework Equations

I = integral of radius squared with respect to mass

The Attempt at a Solution

I'm not quite sure how to explain my issue here, but I think all I need is to see this worked out, and then I could probably understand the nature of these problems. Thanks.

 

[aniketp]

You could use the parallel axis theoram if you know how to find the M.I about the centre of mass of the rod (the centre)

 

[nlightner]

As a scientist, it is important to have a solid understanding of the concepts and equations being used in a problem. In this case, the rotational inertia of a thin rod can be calculated using the formula I = mr^2, where m is the mass of the rod and r is the distance from the axis of rotation to the center of mass of the rod.

In this problem, the mass of the meter stick is given as 0.68 kg. To find the distance, we need to consider the center of mass of the rod. Since the rod is uniform, the center of mass is located at the midpoint, which is 50 cm from either end. Therefore, the distance from the axis of rotation to the center of mass is 50 cm - 21 cm = 29 cm = 0.29 m.

Now, we can plug in the values into the formula: I = (0.68 kg)(0.29 m)^2 = 0.056 kgm^2. This is the rotational inertia of the meter stick about an axis perpendicular to the stick and located at the 21 cm mark.

It is important to note that this formula assumes the rod is a thin, uniform rod. If the rod has a non-uniform mass distribution or if it is not a thin rod, a different formula may need to be used. It is always important to carefully consider the assumptions and limitations of any formula being used in a problem.