Single Choice

A semicircular portion of radius $$'r'$$ is cut from a uniform rectangular plate as shown in figure. The distance of centre of mass $$'C'$$ of remaining plate, from point $$'O'$$ is:

A$$\dfrac {2r}{(3-\pi)}$$
B$$\dfrac {3r}{2(4-\pi)}$$
C$$\dfrac {2r}{(4+\pi)}$$
D$$\dfrac {2r}{3(4-\pi)}$$
Correct Answer

Solution

Let negative mass in the form of semi circular disk is put on the rectangular slob and both have same man density. [Ref. image 1]

let the mass density be $$\sigma$$
Mass of rectangular slab $$=M_1$$
& semicircular disk $$=M_2$$

$$M_1 = 2r\times r \times \sigma$$

$$M_2 = \dfrac{\pi r^2}{2}\sigma$$
Now
as rectangular slab $$x$$ coordinate of com $$x_1= -\dfrac{r}{2}$$ (as lift side of origin)v [Ref. image 2]
as semi circular disk of (-)ve mass $$x$$ coordinate of com of disk [Ref. image 3]
$$x_2 = - \dfrac{4r}{3\pi}$$

Now Putting the disk on the plate the portion covered by disk have no mass
$$x$$ cordinate of com of combination

$$x_c = \dfrac{x_1m_1+z_2(-m_2)}{m_1+(-m_2)}$$ [Ref. image 4]
as mass of disk is (-)ve

$$x_c = \dfrac{\left(-\dfrac{r}{2}\right) \times 2r^2\sigma + \left(-\dfrac{4r}{3\pi}\right) \left(-\dfrac{\pi r^2}{2}\sigma\right)}{2r^2 \sigma - \dfrac{\pi r^2}{2}\sigma}$$

$$x_c = -\dfrac{2r}{3(4-\pi)}$$

distance of com from $$o$$ is $$|x|$$

$$=\dfrac{2r}{3(4-\pi)}$$

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