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  • The disc of mass M with uniform surface mass density is shown in the figure. The centre of mass of the quarter disc (the shaded area) is at the position <img alt="\frac{x}{3}\frac{a}{\pi},\frac{x}{3}\fr

The disc of mass M with uniform surface mass density \sigma is shown in the figure. The centre of mass of the quarter disc (the shaded area) is at the position \frac{x}{3}\frac{a}{\pi},\frac{x}{3}\frac{a}{\pi} where x is __________. (Round off to the Nearest Integer). [a is an area as shown in the figure]
Option: 1 4
Option: 2 3
Option: 3 6
Option: 4 8

Answers (1)

best_answer


Mass per unit of quarter disc =
\frac{\frac{M}{\pi R^{2}}}{4}=\frac{4 M}{\pi R^{2}}

Using symmetry, For half disc, ycm = 
                                                             \frac{4 R}{3 \pi}
Similarly, for half disc, along x-axis centre of mass, at x =
                                                                                           \frac{4 R}{3 \pi}
So, the centre of mass of quarter disc is =

                                                                  \left(\frac{4 R}{3 \pi}, \frac{4 R}{3 \pi}\right)

Here, R = a, so - 
\text { C.O.M of quarter disc is at } (\frac{4 a}{3 \pi}, \frac{4 a}{3 \pi})

So, x = 4

Posted by

Deependra Verma

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