Question

❤❤❤❤❤❤❤❤ one question for you guys.....

❓✅ obtain the centre of mass of a uniform semi-spherical ball of radius (r).

Thnku for your help......
❤❤❤❤❤ please solve as fast as u can....

Answer 1

We calculate the centre of mass of a half-ball of radius 11. Without loss of generality we may assume that the ball is made of material with density 11.

Imagine that the ball is sitting on a table, flat side down. By symmetry the centre of mass is on the vertical line through the centre of the ball. The only question is: How far up?

We will calculate the moment of the ball about the plane of the table, and divide by the mass of the half-ball. By a standard formula, the mass of the half-ball is 2π32π3.

Imagine now that the half-ball is an industrial ham. Imagine a very thin slice of that ham, sliced parallel to the table, but left in place. Let the slice be taken from height zz to height z+dzz+dz, where dzdzis extremely small. The slice is almost a cylinder of very small height dzdz.

We first calculate the radius r=r(z)r=r(z) of the slice. By the Pythagorean Theorem, we have r2+z2=1r2+z2=1, so r=1−z2−−−−−√r=1−z2.

Thus the area of the slice is πr2=π(1−z2)πr2=π(1−z2). The thickness is dzdz, so the volume, and therefore the mass, of the slice is approximately π(1−z2)dzπ(1−z2)dz.

The slice is at perpendicular distance zzfrom the table. So the moment of the slice about the plane of the table is approximately π(1−z2)(z)dzπ(1−z2)(z)dz.

"Add up" (integrate) from z=0z=0 and z=1z=1. The full moment of the ball is

∫10π(1−z2)(z)dz.∫01π(1−z2)(z)dz.

Calculate. We get π4π4.

Finally, divide by the mass 2π32π3. We get 3838.

For a ball of radius RR, just multiply by RR. The centre of mass is 3R/8 above the centre of the half-ball.