question_answer

From a uniform circular disc of diameter d, a circular disc of diameter d/6 and having centre at a distance d/4 from the centre of the disc is scooped out. Determine the centre of mass of remaining portion.

Answer:

                Suppose mass per unit area of disc = m                                 \[\therefore \]Mass of disc. \[M=\pi {{\left( d/2 \right)}^{2}}\] \[m=\frac{\pi m{{d}^{2}}}{4}\] Mass of portion removed            \[M'=\pi {{\left( d/12 \right)}^{2}}m=\frac{\pi {{d}^{2}}m}{144}\] Let centre O of disc be taken as the origin. Suppose masses M and M' are concentrated at O and O' respectively. Fig. 5(HT).9. As the portion is removed, taking M' as negative, c.m of remaining portion is at P, at a distance x from O, where \[x=\frac{M{{x}_{1}}-M'{{x}_{2}}}{M-M'}\] \[\frac{(\pi m{{d}^{2}}/4)\times 0-\left( \pi m{{d}^{2}}/144 \right)\left( d/4 \right)}{\pi m{{d}^{2}}/4-\pi m{{d}^{2}}/144}\] \[=-\frac{\pi m{{d}^{2}}}{144}\times \frac{d}{4}\times \frac{144}{35\pi m{{d}^{2}}}=-d/140\] Negative sign indicates c.m is to left of origin O.