Answer to Question #123775 in Mechanics | Relativity for Ojugbele Daniel

Question

A steel cylinder of 250mm outside and 100mm inside is set in rotation about it's axis. If the cylinder is 900mm long of density 7800kg/m^3. Calculate the moment of inertia of the cylinder about it's polar axis.

Answer 1

$r_1 = 100 mm = 10cm=0.1m$

$r_2=250mm = 25 cm = 0.25m$

$l = 900 mm = 90cm =0.9m$

Polar axis is the axis of rotation. Let rotation axis is axis of cylinder.

Let us assume any small element of mass $dm$ which is cylinder or radius $x$ and thickness $dx$

Now

moment of inertia of this small element is given as

$dm = \frac{M}{A}2\pi xdx$ where M is total mass and A is area of the solid circular region

then $I = \int dmx^2 = \int_{r_1}^{r_2} \frac{M}{A}2\pi x^3 dx$ $= \frac {M \pi }{2A} [ r_2^4 - r_1^4]$

Now put the value,

$M = \pi (r_2^2-r_1^2)l \rho = \pi * ((0.25)^2 - (0.1)^2)*0.9*7800 =1157.8kg$

$A = \pi [ r_2^2 - r_1^2]$

putting value of A, we obtained,

$I = \frac {M \pi }{2\pi (r_2^2 - r_1^2)} [ r_2^4 - r_1^4] = \frac{1}{2} {M } [ r_2^2 + r_1^2]$

so ,

$I = 0.5*1157.8*(0.25^2 + 0.1^2) = 41.97 kgm^2$ (approx)