Name: Variable Mass System
Uploaded: 2019-07-19
Duration: 12 min 25 s
Description: In this video, we will look at how the laws of motion and their mathematical formulation apply to a variable mass system like a rocket. Don't miss it.

Question

Find the centre of mass of an annular half-disc shown in the figure.

Accepted Answer

In the given annular half-disc has an internal radius $R_{1}$ and external radius $R_{2}$ . Mass is spread uniformly over the disc part.

Suppose the mass of the annular disc is $M$ and having surface mass density $σ Kg m^{- 2}$ . A semicircular ring of radius $r$ and thickness $d r$ is chosen in the annular disc as shown in the figure.

Surface mass density for the annular disc can be written as $σ = \frac{M}{{\frac{π R_{2}}{2}}^{2} - {\frac{π R_{1}}{2}}^{2}} = \frac{2 M}{π {R_{2}}^{2} - π {R_{1}}^{2}}$ and the area of the elemental half ring $d A = π r d r$ .

The mass of the semicircular ring. After putting the value of $σ$ .

We get,

$d m = σ \times d A = \frac{2 M \times π r d r}{π {R_{2}}^{2} - π {R_{1}}^{2}}$ .

Taking the Centre of the half-disk as origin, the coordinate of the centre of mass of the half-ring $x = 0$ , $y = \frac{2 r}{π}$ .

By the symmetry, we can say centre of mass of the annular disc lies along $y$ -axis. The coordinate of centre of mass along $y$ -axis is

$y_{cm} = \frac{\int y d m}{M} . . . (1)$

Putting the value of $d m$ and $y$ in the equation $(1)$ , we get,

$y_{cm} = \frac{1}{M} \times \int \frac{2 M \times π r d r}{π {R_{2}}^{2} - π {R_{1}}^{2}} \times \frac{2 r}{π}$

$y_{cm} = \frac{4}{π {R_{2}}^{2} - π {R_{1}}^{2}} \int r^{2} d r$

Putting the limit from $R_{1}$ to $R_{2}$ . We get,

$y_{cm} = \frac{4}{π {R_{2}}^{2} - π {R_{1}}^{2}} \int_{R_{1}}^{R_{2}} r^{2} d r$

$y_{cm} = \frac{4 ({R_{2}}^{3} - {R_{1}}^{3})}{3 π ({R_{2}}^{2} - {R_{1}}^{2})}$ .

Find the centre of mass of an annular half-disc shown in the figure.

Important Questions on Centre of Mass, Momentum and Collisions

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