From a uniform disc of radius R, a circular hole of radius $$\dfrac{R}{2}$$ is cut out. The centre of the hole is at $$\dfrac{R}{2}$$ from the centre of the original disc. Locate the centre of gravity of the resulting flat body

Asked by Abhisek | 1 year ago |  78

##### Solution :-

Let the mass/unit area of the original disc = σ

Radius of the original disc =2r

Mass of the original disc, m = π(2r2)σ = 4 πr2σ . . . . ... (i)

The disc with the cut portion is shown in the following figure:

Radius of the smaller disc = r

Mass of the smaller disc, m’ = π r2σ

= m’ =$$\dfrac{m}{4}$$        [ From equation ( i ) ]

Let O’ and O be the respective centers of the disc cut off from the original and the original disc. According to the definition of center of mass, the center of mass of the original disc is concentrated at O, while that of the smaller disc is supposed at O′.

We know that :

OO′= $$\dfrac{R}{2}$$ = r2

After the smaller circle has been cut out, we are left with two systems whose masses are:

m’ ($$\dfrac{m}{4}$$) concentrated at O′, and m (concentrated at O).

(The negative sign means that this portion has been removed from the original disc.)

Let X be the distance of the center of mass from O.

We know :

X = $$\dfrac{ (m_1 r_1 + m_2 r_2) }{ (m_1 + m_2)}$$

X =$$\dfrac{m × 0 – m’ \times \dfrac{r}{2}}{( M + (-M‘)}$$

$$\dfrac{-R}{6}$$

(The negative sign indicates that the center of mass is$$\dfrac{R}{6}$$ towards the left of O.)

Answered by Pragya Singh | 1 year ago

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