A sphere of diameter r is cut from a solid sphere of radius r such tha...
Let x = distance between COM from the original center of the sphere, after the small sphere is cut off from it.
Let the density of the sphere be = d
Mass of big sphere (full) = 4/3 * π r^3 * d
Mass of the small sphere cut off from it = 4/3 * π (r/2)^3 * d
Mass of the remaining object = 7/6 π r^3 * d
Position of the COM of the complete sphere = 0
=> - x * 7/6 π r^3 d + r/2 * 4/3 π (r/2)^3 * d = 0
=> x = r / 14
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A sphere of diameter r is cut from a solid sphere of radius r such tha...
Answer is option 3 but I am unable to post the derivation because it is notsupported by my device
A sphere of diameter r is cut from a solid sphere of radius r such tha...
The problem states that a sphere of diameter r is cut from a solid sphere of radius r in such a way that the center of mass of the remaining part will be at the maximum distance from the original center. We need to determine the maximum distance.
To solve this problem, we can break it down into smaller steps:
1. Find the center of mass of the remaining part:
- Let's assume that the original solid sphere has mass M and density ρ.
- The volume of the original solid sphere is (4/3)πr^3.
- The volume of the cut-out sphere is (4/3)π(r/2)^3 = (1/8)πr^3.
- Therefore, the volume of the remaining part is (4/3)πr^3 - (1/8)πr^3 = (31/24)πr^3.
- The mass of the remaining part is M' = ρ * (31/24)πr^3.
- The center of mass of the remaining part will be at a distance d from the original center.
2. Find the maximum distance of the center of mass:
- To find the maximum distance, we need to maximize d.
- We can use the parallel axis theorem to find the distance of the center of mass from the center of the remaining part.
- The parallel axis theorem states that the moment of inertia of a body about an axis parallel to and at a distance d from an axis through its center of mass is given by I = Icm + Md^2, where Icm is the moment of inertia about the center of mass and M is the total mass.
- In our case, the moment of inertia about the center of mass of the remaining part is Icm = (2/5)M'r^2.
- The moment of inertia about the original center is I = (2/5)M'r^2 + M'd^2.
- Since the remaining part is symmetric, the center of mass will lie on the line connecting the centers of the two spheres.
- Therefore, the distance d will be equal to the radius of the remaining part, which is r/2.
- Substituting these values into the equation, we get I = (2/5)M'r^2 + M'(r/2)^2 = (9/40)M'r^2.
- Since M' = ρ * (31/24)πr^3, we can rewrite the equation as I = (9/40)ρ(31/24)πr^5.
- The moment of inertia is proportional to r^5, so to maximize it, we need to maximize r.
3. Determine the maximum distance:
- Since the distance d is equal to r/2, the maximum distance is r/2.
Therefore, the maximum distance of the center of mass of the remaining part from the original center is r/2. So, the correct answer is option (2) r/2.
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