Find the center of mass for a quarter of a uniform solid sphere of radius a

Find the center of mass for a quarter of a uniform solid sphere of radius a

The Correct Answer and Explanation is:

To find the center of mass for a quarter of a uniform solid sphere of radius aaa, we can proceed by applying symmetry and integration.

Step 1: Recognize the Symmetry

The given quarter-sphere is located in the first octant, as suggested by the coordinate system in the diagram. We assume the sphere is uniform in mass, meaning the mass density is constant throughout the volume.

Step 2: Geometry of the Quarter Sphere

The quarter sphere is one-fourth of a full sphere, with radius aaa, centered at the origin, and it occupies the region where x≥0x \geq 0x≥0, y≥0y \geq 0y≥0, and z≥0z \geq 0z≥0.

Step 3: Coordinates of the Center of Mass

Since the quarter sphere is symmetric in terms of the xxx-, yyy-, and zzz-axes, the center of mass will lie at an equal distance from each of these axes. Therefore, the center of mass will have coordinates (x,y,z)(x, y, z)(x,y,z), where all the coordinates are equal. Let this common value be rˉ\bar{r}rˉ, the distance from the origin.

Step 4: Moment of Inertia Integration

The center of mass for a uniform solid sphere is at its geometric center. For a quarter sphere, the center of mass can be found by integrating the positions of all mass elements within the region.

In spherical coordinates:

• x=rsin⁡θcos⁡ϕx = r \sin\theta \cos\phix=rsinθcosϕ
• y=rsin⁡θsin⁡ϕy = r \sin\theta \sin\phiy=rsinθsinϕ
• z=rcos⁡θz = r \cos\thetaz=rcosθ

The volume element is dV=r2sin⁡θ dr dθ dϕdV = r^2 \sin\theta \, dr \, d\theta \, d\phidV=r2sinθdrdθdϕ.

Step 5: Integral for the Center of Mass

To calculate the center of mass, we need to compute the integral for the xxx-coordinate (which is the same as yyy and zzz by symmetry):rˉ=∫Vr dV∫VdV\bar{r} = \frac{\int_V r \, dV}{\int_V dV}rˉ=∫V​dV∫V​rdV​

The bounds for the integration will be:

• rrr ranges from 0 to aaa
• θ\thetaθ ranges from 0 to π2\frac{\pi}{2}2π​
• ϕ\phiϕ ranges from 0 to π2\frac{\pi}{2}2π​

Now, by solving the integrals, we find that the center of mass is located at:rˉ=38a\bar{r} = \frac{3}{8} arˉ=83​a

Thus, the center of mass of the quarter-sphere is at the point:(38a,38a,38a)\left(\frac{3}{8}a, \frac{3}{8}a, \frac{3}{8}a \right)(83​a,83​a,83​a)

Final Answer:

The center of mass for a quarter of a uniform solid sphere of radius aaa is at:(38a,38a,38a)\left(\frac{3}{8}a, \frac{3}{8}a, \frac{3}{8}a \right)(83​a,83​a,83​a)