Calculating Moment of Inertia for a System of Small Blocks on a Clamped Rod

[elsternj]

Homework Statement

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.
Express your answer in terms of the given quantities.

Homework Equations

I = $$\sum$$mr²

The Attempt at a Solution

Now I know the answer is 11/16mL² but I am having trouble figuring out how to get to that answer.

the first mass is 1/4L from the axis, the second mass is also 1/4L from the axis and the third mass is 3/4L from the axis.

m(1/4L)²+m(1/4L)²+m(3/4L)²

how does this become 11/16mL²?

 

[praharmitra]

uhh... just open up the brackets and add!

 

[elsternj]

can i see that step by step? my math isn't the greatest but 1/4+1/4+3/4 does not equal 11/16

 

[praharmitra]

Its (1/4)^2 + (1/4)^2 + (3/4)^2

 

[rwisz]

To find the moment of inertia for a system of small blocks on a clamped rod, we can use the formula I = \summr2, where m is the mass of each block and r is the distance of each block from the axis of rotation. In this case, the axis of rotation is perpendicular to the rod and passes through a point one-fourth of the length from one end.

So, for the first block, which is located at a distance of 1/4L from the axis, we have r = 1/4L. Similarly, for the second block, r = 1/4L, and for the third block, r = 3/4L.

Substituting these values into the formula, we get:

I = m(1/4L)2 + m(1/4L)2 + m(3/4L)2

Simplifying, we get:

I = 1/16mL2 + 1/16mL2 + 9/16mL2

Combining like terms, we get:

I = 11/16mL2

So, the moment of inertia for this system of small blocks on a clamped rod is 11/16mL2.