Rotaional Inertia of a Thin Rod

[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 r squared with respect to mass

The Attempt at a Solution

?

 

[Pythagorean]

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 r squared with respect to mass

The Attempt at a Solution

?

so you want to find the mass of the object as a function of the distance from the axis. We can assume the the density is constant, so we could write:

dm = pdV

where dm is the mass differential from the integral, p the density (a constant) and dV the volume.

But some parts of the volume are constant, so you can further manipulate this equation. Can you see where to go from here?

 

[nlightner]

The rotational inertia of a thin rod can be calculated using the formula I = (1/12) * m * L^2, where m is the mass of the rod and L is the length of the rod. In this case, the length of the meter stick is 1 meter, and the mass is given to be 0.68 kg. Plugging these values into the formula, we get I = (1/12) * (0.68 kg) * (1 m)^2 = 0.0057 kg*m^2.

However, since the axis of rotation is not at the center of mass of the meter stick, we need to use the parallel axis theorem to account for the distance from the center of mass to the axis of rotation. This theorem states that I = Icm + md^2, where Icm is the rotational inertia about the center of mass, m is the mass of the object, and d is the distance between the center of mass and the axis of rotation.

In this case, the center of mass of the meter stick is at the 50 cm mark, so the distance from the center of mass to the axis of rotation at the 21 cm mark is 29 cm. Therefore, the total rotational inertia of the meter stick about the given axis is I = 0.0057 kg*m^2 + (0.68 kg) * (0.29 m)^2 = 0.0074 kg*m^2.

It is important to note that the unit for rotational inertia is kg*m^2, which is different from the unit for moment of inertia (m^4). This is because rotational inertia is a property of an object's mass distribution, while moment of inertia takes into account the distribution of mass and the distance from the axis of rotation.