Three small balls of masses 1 k g , 2 k g and 3 k g are moving in a plane and their velocities are 1 m s - 1…

Three small balls of masses $1\mathrm{ }\mathrm{k}\mathrm{g},\mathrm{ }2\mathrm{ }\mathrm{k}\mathrm{g}$ and $3\mathrm{ }\mathrm{k}\mathrm{g}$ are moving in a plane and their velocities are $1 \mathrm{m} {\mathrm{s}}^{-1}$, $2 \mathrm{m} {\mathrm{s}}^{-1}$ and $3 \mathrm{m} {\mathrm{s}}^{-1}$, respectively as shown. The total angular momentum of the system of three balls about point $\mathrm{P}$ at the given instant of time is

1. $\mathrm{8 }\mathrm{k}\mathrm{g }{\mathrm{m}}^2 {\mathrm{s}}^{-1}$
2. $\mathrm{9 }\mathrm{k}\mathrm{g }{\mathrm{m}}^2 {\mathrm{s}}^{-1}$
3. $\mathrm{36 }\mathrm{k}\mathrm{g }{\mathrm{m}}^2 {\mathrm{s}}^{-1}$
4. $\mathrm{7 }\mathrm{k}\mathrm{g }{\mathrm{m}}^2 {\mathrm{s}}^{-1}$

Solution

The particle of mass 2 kg is situated perpendicular to the axis of rotation. So, the angle will be $90°$. Similarly, the particle of mass 1 kg is situated opposite side of the circular path. So, the angle will be $90°$ and angular momentum will act in opposite direction.

According to the diagram, total angular momentum

$=2\times 2\times 2\sin 90°+3\times 3\times 3\sin 0°-1\times 1\times 1\sin 90°$

$=8-1$

$=7\frac{\text{kg}{\text{m}}^{\text{2}}{\text{ s}}^{-1}}{\text{s}}$ $[\because \sin 0°=0 \mathrm{a}\mathrm{n}\mathrm{d}\sin 90°=1]$

Asked in: JEE Mains - Rotational Motion - Test 4

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