question_answer

Find the centre of mass of a uniform disc of radius a from which a circular section of .radius b has been removed. The centre of hole is at a distance c from the centre of the disc.

Answer:

                Suppose the circular disc of radius a with centre O is made up of (i) circular section of radius b with centre \[{{O}_{1}}\] and (ii) remaining portion of disc with c.m at \[{{O}_{2}}\]. Taking \[O\] as origin, and \[{{O}_{1}}\], \[{{O}_{2}}\] on X-axis, (y = 0, z = 0), Fig. 5(HT).4, the position of c.m of disc is given by \[{{x}_{cm}}=\frac{{{m}_{1}}{{x}_{1}}+{{m}_{2}}{{x}_{2}}}{{{m}_{1}}+{{m}_{2}}}\]                                                             ?? (i) If \[\sigma \] is surface density of material of the disc, \[{{m}_{1}}=\pi {{b}^{2}}\sigma \], \[{{x}_{1}}=c\] \[{{m}_{2}}=\pi \left( {{a}^{2}}-{{b}^{2}} \right)\sigma \] \[{{x}_{2}}=?\] \[{{m}_{1}}+{{m}_{2}}=\pi {{a}^{2}}\sigma \] \[{{x}_{cm}}=0\]                 From (i),               \[0=\frac{\pi {{b}^{2}}\sigma \times c+\pi \left( {{a}^{2}}-{{b}^{2}} \right)\sigma \times {{x}_{2}}}{\pi {{a}^{2}}\sigma }\]                 \[\therefore \]                                  \[{{x}_{2}}=\frac{-c{{b}^{2}}}{\left( {{a}^{2}}-{{b}^{2}} \right)}\] Hence c.m of rest of the portion of the disc lies on the left of O, at a distance \[{{x}_{2}}\] as shown in the Fig. 5(HT).4.