Nov 13, 2009 · Every reference I see lists the moment of inertia of a half-cylinder about its axis the same as that of a full circle, noting that it is about its centroid: [tex] I_{x_0} = \frac{1}{2}mr^2 [/tex] Try that, using the parallel axis theorem appropriately.

Dec 16, 2015 · Homework Statement Theres an object which makes a pendulum motion.Lets suppose we hang the mass to the ceiling.We released the object with inital angle 0 to the ceiling.(I mean the angle between the object and the ceiling is zero).Whats the moment of the Inertia to the point A. A is a...

Jun 30, 2019 · Ratio of moment of inertia of a circle and that of a square having same area about their centroidal axis is: Sol: both area and square have same area: ##a^2 = \large\frac{π}{4}\normalsize d^{2}; \large\frac{d}{a} =\large \sqrt\frac{4}{p}## Ratio of moment of inertia of a circle and that of a square about their centroidal axis is:

Aug 29, 2004 · For example: The rotational inertia of a circle with a square cut out equals the rotational inertia of the circle minus the rotational inertia of the square. Aug 29, 2004 #3

Oct 11, 2011 · having some major derp problems. I can't seem to remember how to find moment of inertia about the COM of and object or even finding the COM for that matter. I need it to solve a Euler Lagrange problem Can someone give me an idea on the integrals to use and in what coordinate system? Homework Equations The Attempt at a Solution

Oct 12, 2011 · The moment of inertia is the mass of the object times the mass-weighted average of the squared distance from the axis. Because the ring is hollow, all of its mass has to sit at a distance R from the center; hence, you have <r 2 >=R 2 and I = MR 2 .

Mar 7, 2004 · Hurkyl was completely correct but I think what ShawnD meant to say was not "dA for a circle" but "dA in polar coordinates". Doing problems with a circular symmetry, as in finding the area of a circle or finding the moment of inertia of a circle about a diameter, is most often easier in polar coordinates. The "dA" in polar coordinates is "rdrd ...

Apr 21, 2011 · So, the radius for the big circle is r=R, so you'd plug in R for r in the formula, and similar for the other parts. For that matter, let's do something simple, finding the area of the big circle. A = πr 2, of course, and since the big circle is radius R, you'd get A = πR 2. Same idea for area moment of inertia.

May 28, 2011 · Find the moment of inertia of a wire, AB, of mass M and length pi*a, which is bent into a semicircle, about AB. Mr^2/b] [b]The mass of the wire is M=pi*a*m, where m is the mass per unit length of the rod. Then a small element, ds is regarded, of the circumference of the semicircle as being...

Jun 15, 2010 · The center of the hole is a distance 10.8 cm from the center of the disk. Find the moment of inertia of the modified disk about the origin. Homework Equations I=.25mr^2 (I think...!) The Attempt at a Solution I tried finding the moment of inertia for both the hole and the disk and subtract the hole, but that didn't work.