# Moment of inertia

• princessfrost

#### princessfrost

Small blocks, each with mass m, are clamped at the ends and at the center of a rod of length L and negligible mass

Compute the moment of inertia of the system about an axis perpendicular to the rod and passing through a point one-fourth of the length from one end.

Since I=1/2mL^2 for a system about an axis perpendicular to the rod and passing through the center of the rod, can't I just add 1/4mL^2 + 1/2mL^2 and get 3/4mL^2=I for the equation?

Since I=1/2mL^2 for a system about an axis perpendicular to the rod and passing through the center of the rod, can't I just add 1/4mL^2 + 1/2mL^2 and get 3/4mL^2=I for the equation?

Mass moment of Inertia of a rigid uniform rod of length L and mass m about an axis passing through its centre of mass and perpendicular to the length of the rod is given by I = m(L^2)/12. (and not, I=1/2mL^2, as u have quoted.)

Plz note, in the given problem, rod is of negligible mass.. so u can't apply the above formula. Instead, since point masses are involved, you should use the fundamental formula for mass moment of inertia.

is it 11/16mL^2 then?

is it 11/16mL^2 then?

Think of it this way, if you look at the link Astronuc provided, you see that we can use I = mR^2 if we have a massless rod, and a mass at the end of the rod. What you essentially have is two massless rods put into one. The first massless rod will have the length L/4, and the second rod will have the length 3L/4. These are also where the two masses are located relative to the axis of rotation. If you find the moment of inertia for each rod, again, using the equation Astronuc provided, you can add the rotational inertias together to get the total inertia you're looking for.

is it 11/16mL^2 then?

$$I = mr^2$$