The problem states: Assume that the wheel consists of two solid and uniformcylin
ID: 1675072 • Letter: T
Question
The problem states: Assume that the wheel consists of two solid and uniformcylinders fused together by sides. The smaller cylinder (withradius .1m) has a mass of 2 kg. The larger cylinder (withradius .25m) has a mass of 5 kg. Find the moment of inertiaof the wheel.
Ok, so I know that moment of inertia for a solid cylinder isgiven by I = 1/2 * M * R^2
However, am I allowed to use this equation for the givenproblem? If I plug in the number I get I = 1/2(7)(.25^2) = .21875
But I feel like since the two cylinders are possiblydifferent, I should take their moment of inertias separately. So I would have to use the equation for a hollowcylinder for the blue cylinder and add the moment of inertia forthe green cylinder, so it would look something like this:
I1 = 1/2*MR^2 = 1/2(2)(.1)^2 = .01 I2 = 1/2*M*(R1^2+R2^2) = 1/2(5)(.25^2+.1^2) = .18125
Itot = I1 + I2 = .01 + .18125 = 0.19125
However this result is different from the result calculatedusing the combined masses to find the moment of inertia. Somy question is, which method is correct?
Thanks
Explanation / Answer
You would use your second method. You cannot simply combinethe masses and treat the object as a single cylinder because themoment of inertia depends not only upon mass but also thedistribution of the mass. The inner cylinder and the outercylinder have different densities.
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.