Moment of Inertia in slender rod - density varies

[ph123]

A slender rod with length L has a mass per unit length that varies with distance from the left-hand end, where x=0, according to dm/dx = (gamma)x, where (gamma) has units of kg/m^2.

I calculated the total mass of the rod by integrating. I got

M = 0.5(gamma)L^2

In the second part I found the moment of inertia with the axis of rotation to be at x=0, perpendicular to the rod. I don't know how to represent the integration symbol, so I'll use "|." I am integrating from 0 to L

I = |r^2dm
I = |L^2(gamma)L = (gamma)|L^3 = [(gamma)L^4]/4
= [2M/L^2][0.25L^4] = 0.5ML^2

The third part asks me to do the same thing for the rod, this time with the axis of rotation at the opposite end of the rod. I know I have to integrate from L to 0, but shouldn't this just give me the negative of my previous expression? I know it's wrong, but I don't know how the mass expression changes at the other end. Can someone help?

 

[Edward G]

Those first two seem to be fine. For calculating the moment of inertia, you want a little bit of mass multiplied by its distance from the axis, and then you sum this over the length of the rod. For the third part, because the mass is decreasing as you move along the rod, a little piece of mass is

\frac{dm}{dx} dx = \gamma (L-x) dx

So if you integrate this mass times the distance from the axis squared (x^2) between 0 and L, you should get the moment of inertia.

\int_{0}^{L} \gamma (L-x) x^2 dx

Hope that helps

Ed

 

[ph123]

Ok, this makes sense. But they want the answer in terms of M and L. When I substitute my expression for gamma after integration, I get

[(4LMX^3)-(6MX^4)]/(6L^2)

There's no way I can see to get rid of X, so now I'm stuck...again.

 

[Doc Al]

X is your variable of integration; you need to evaluate that definite integral over the range X = 0 to X = L.

 

[ph123]

ahhh. thanks a lot. got (1/6)ML^2

 

[sean_unb]

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