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

Question #123775
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.
Expert's answer

"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"


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)

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