2011年10月12日 · 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 .

2009年11月17日 · 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.

2024年10月12日 · Fx is the horizontal component, which is the centripetal acceleration, so it is mv^2/R. From the previous part as the sphere rolls without slipping the rotational kinetic energy is considered, and v^2= 20/7 gR when the sphere rolls from 3R to R. substituting it into Fc=mv^2/R gives the horizontal (centripetal) force of 20/7 mg.

2005年1月8日 · For a ring (circle) the moment of inertia is really simple. Since all the mass is at a distance R from the center, r in the integral is constant. So you can bring it outside of the integral and the integral over dm simply gives the mass. [tex]I_{ring}=\int_{C} r^2 dm = R^2 \int_{C}dm=MR^2[/tex] where C is the circle you are integrating over.

2015年12月16日 · 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...

2017年8月13日 · In summary, the conversation discusses calculating the moment of inertia for a circle that is not at its center of gravity, specifically in two dimensional space. The parallel axis theorem may be used to avoid using an integral. The person has spent hours researching this topic but has not been able to find a solution.

2004年3月7日 · 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 ...

2019年6月30日 · 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:

2011年10月12日 · 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

2010年6月14日 · 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.