moment of inertia of quarter circle derivation

Click here to receive email alerts on new articles. You can find a complete list of all the articles here. Is this a result you expected? It makes sense when you consider that a hemisphere tapers in two axes as it moves from base to tip, so we should expect the centroid to be nearer the base. A portion of the boiler was thrown a quarter of a mile from the scene of the. The centroid is a little nearer the base of the dome. It's a little cleaner than the formula for a semicircle and does not involve π. We can sum up the moment of all these disks and divide by the volume of half the sphere.ĭo you have any intuition about the result? But where is the centroid of a hemi-sphere?ĭo you think this shape will have a different centroid?Īs before we can model this as a pile of infinitesimally thin circular pancakes stacked ontop of each other. If a semicircle is extruded into an arch (of uniform density), then the centroid stays at the 42.44% position from the base. I would not have predicted it would be an irrational percentage. I don't know about you, but this answer surprised me a little. 05 Derivation of Moment of Inertia for Rectangle, Triangle, Circle, Semi-Circle & Quarter Circle from moment of inertia formula circle Watch Video. This is approximately 42.44% of the radius. Moment Of Inertia Of Quarter Circle Derivation Notably, in a full circle, the moment of inertia relative to the x-axis is the same as the y-axis. We just need to determine how far up this point is. The centroid is the location that we can model the shape as if the entire weight acts through this point.įrom symmetry, we can see that the centroid lies along the y-axis, which acts like a mirror to the quarter circles either side. If we sum the moment of all these strips, normalised over the area of the entire half-circle, we can find the centroid. We can model a semicircle as a stack of infinitesimally thin strips. Let’s use some calculus to derive the answer.: If you don’t know the answer, think about it for a couple of seconds. This holds true for all regular polygons.Where is the centroid of a semicircle? If you cut out a semicircle from a sheet of uniform density cardboard, over which point could you, carefully, balance it on the top of a pin? The result is valid for both a horizontal and a vertical axis through the centroid, and therefore is also valid for an axis with arbitrary direction that passes through the origin.

Moment of inertia of quarter circle derivation