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⚛️ NEET Physics❓ Question

Centre of mass of semicircular ring vs disc vs hemispheres, which number belongs to which? And explosion in mid-air?

Aarav Kulkarni11th Sample

3 Sept 2026 · 1 min read

My checklist for System of Particles now has a "centre of mass positions" table, and I mixed up the entries. I need someone to help me check it and explain why the numbers differ.

Distances from the centre of the flat face / the centre of the circle (I think):

  • Semicircular ring (wire): 2Rπ\dfrac{2R}{\pi}
  • Semicircular disc (lamina): 4R3π\dfrac{4R}{3\pi}
  • Solid hemisphere: 3R8\dfrac{3R}{8}
  • Hollow hemisphere: ?

I'm not sure about the hollow one (I wrote 3R/83R/8 for both but that felt lazy). Also why is the ring's value bigger than the disc's, intuitively?

Second doubt: a projectile is fired and explodes mid-air into two pieces. NCERT says the centre of mass continues along the original parabola. But one piece may fall straight down and the other goes somewhere else. So where is the CM when the first piece hits the ground? Does CM have to be inside the body? A ring's CM is in the hole, which confused me even more.

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7 answers & replies

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  • [+]Arjun Nair12thSample· 1mo ago

    Your first three are right. Hollow hemisphere (thin shell): R2\dfrac{R}{2}. Check by symmetry: a thin spherical shell's surface area is uniformly spread, and the area of a band is proportional to its height, so the average height of a hemispherical shell is R/2R/2.

    Why the ring is further out than the disc: in the ring all mass is at the rim (distance RR from the centre), while the disc has material spread through the inside, close to the centre, which pulls the CM inward. Same for the shell vs the solid hemisphere (R/2R/2 vs 3R/83R/8).

    • [+]Aarav Kulkarni11thSample· 1mo agoAuthor

      Table updated: ring 2R/π, disc 4R/3π, solid hemisphere 3R/8, hollow hemisphere R/2. Added a bubble that says "CM can be empty space". Thank you!

  • [+]Harsh Gupta12thSample· 1mo ago

    I use "ring/shell is the outer one". 0.64R for the ring, 0.42R for the disc, 0.5R for the shell, 0.375R for the solid hemisphere. Mnemonic: bigger number = more mass at the edge.

  • [+]Dev Agarwal12thSample· 1mo ago

    For the hollow hemisphere it's also 3R/83R/8 because a hollow one is just a solid one with the middle removed, and the CM is the same.

    • [+]Arjun Nair12thSample· 1mo ago

      Not the same, Dev. Removing the middle changes the mass distribution. For a thin hollow hemisphere (a shell) the mass is only at distance RR from the centre, so the CM is further from the flat face than for the solid one: R/2>3R/8R/2 > 3R/8. Note that "same shape" doesn't mean "same CM", it depends on how mass is spread.

      • [+]Dev Agarwal12thSample· 1mo ago

        Yeah, my reasoning confused geometry with mass distribution. Good catch.

  • [+]Vivek Mishra12thSample· 1mo ago

    Explosion: only internal forces act during the explosion, and gravity (external) acts on the system all the time, so the CM of the fragments follows the same parabola the whole projectile would have followed if it had not exploded. So when the first piece is about to land, the CM is still on that original parabola, somewhere above the ground (the other piece is still in the air). The parabola rule holds only while gravity is the only external force. Once a piece touches the ground, the ground's force acts on the system and the CM path changes from that moment.

    CM does not have to lie inside the body. A ring, a bent rod, a boomerang all have CM at an empty point.

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