Calculate Moment of Inertia?

Question

I know that the moment of inertia for a disk being spun on the z-axis when it lies on the x-y plain is I=0.5M(R^2)

I was wondering how this was calculated. As much explaination to steps will be greatly appreciated. I understand that I= (the integral of) r^2 dm. when i tried to do this my working had dm= pi (pi - 3.14etc) * p(p=density) * dr * dr

Answer 1

You seem to know all the basics to work out the solution. The only place that got you stuck is perhaps your expression for dm. I suppose by dm you mean the mass of a thin ring with radius r and width dr. In your expression you missed a 2, namely, the area of the ring is

2 pi r dr

not pi r dr. (BTW, I assume your first dr is a typo for r.) Did you get 0.25MR^2?

Looks like you know the physics, but made a tiny mistake in geometry. Or, you clear jumped the boulder but tripped over a pebble :)

Answer 2

Take an infinitesimal ring of width dr at distance r from the centre.

Area of the ring = 2 * pi * r * dr

Mass of the ring dM = M * (area of ring/area of disk)

= M * (2 * pi * r * dr/pi * R^2)

dM = 2Mr * dr/R^2

Moment of inertial of the ring dI = r^2 * dM

Or, dI = 2M * r^3 * dr/R^2

Or, dI = (2M/R^2) * r^3 * dr

Moment of inertial of the disk = integral of dI from 0 to R

= integral (2M/R^2) * r^3 * dr from 0 to R

= (2M/R^2) * r^4/4 from 0 to R

= (2M/R^2) * R^4/4

= 0.5 * MR^2

___________________________________

Get help in Science + Mathematics from

http://www.visharadsoftware.com/forum

For further details, send email to

[e-mail]