question_answer

A circular plate of uniform thickness has a diameter of 56 cm. A circular portion of diameter 42 cm is removed from the edge of the plate as shown in Fig. 5(HT).3. Find the position of centre of mass of the remaining portion.

Answer:

                Suppose mass per unit area of circular plate = m. \[\therefore \] Total mass of circular plate, \[M=m\times \pi {{\left( d/2 \right)}^{2}}=\pi m{{\left( \frac{56}{2} \right)}^{2}}=784\pi m\] This is supposed to be concentrated at the centre O of the disc, Mass of cut off portion,  \[{{M}_{1}}=m\times \pi {{\left( \frac{{{d}_{1}}}{2} \right)}^{2}}=\pi m{{\left( 42/2 \right)}^{2}}=441\pi m\] This is supposed to be concentrated at \[{{O}_{1}}\], where  \[{{O}_{1}}O=AO-A{{O}_{1}}=28-21=7cm\]. Mass of remaining portion of disc, \[{{M}_{2}}=M-{{M}_{1}}=784\pi m-441\pi m=343\,\pi m\,\] Let it be concentrated at \[{{O}_{2}}\] where \[O{{O}_{2}}=x\]. As                           \[{{x}_{cm}}=\frac{{{M}_{1}}{{x}_{1}}+{{M}_{2}}{{x}_{2}}}{{{M}_{1}}+{{M}_{2}}}\]           and                 \[{{x}_{cm}}=0(at\,O)\]                                 \[\therefore \] \[{{M}_{1}}{{x}_{1}}+{{M}_{2}}{{x}_{2}}=0\] \[441\pi m\times 7+343\pi mx=0\]           or            \[x=\frac{-441\pi m\times 7}{343\,\pi m}=-9cm\] Hence, centre of mass of remaining portion of disc is at 9 cm from centre of disc on the right side of 0.