JEE Main · 2020 · 06 Sep 2020 · Shift 1 · Rotational Mechanics

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Shown in the figure is a hollow icecream cone (it is open at the top). If its mass is M, radius of its top, R and height, H, then its moment of inertia about its axis is:

(A) $\frac{M R ^{2}}{2}$ ✓ Correct
(B) $\frac{M ( R ^{2} + H ^{2} )}{4}$
(C) $\frac{MH ^{2}}{3}$
(D) $\frac{MR ^{2}}{3}$

Answer: (A)

Solution

The cone is a thin shell, so distribute its mass over the slant surface: an element at distance

r

from the axis carries mass proportional to its circumference times a slant step, which works out to

d m = \frac{2 M}{R ^{2}} r d r

r

runs from

0

R

. Its moment of inertia about the symmetry axis is

I = \int r^{2} d m = \frac{2 M}{R ^{2}} \int_{0}^{R} r^{3} d r = \frac{2 M}{R ^{2}} \cdot \frac{R ^{4}}{4} = \frac{M R ^{2}}{2}

. The height

H

does not enter, because the way the mass is spread about the axis is the same as for a disc.

JEE Main · 2020 · 06 Sep 2020 · Shift 1 · Single correct option

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JEE Main · 2020 · 06 Sep 2020 · Shift 1 · Rotational Mechanics

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