A circular hole of radius $1m$ is cut off from a disc of radius $6m$. The centre of the hole is $3m$ from the centre of the disc. Find the centre of mass of the remaining disc.
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Let O be the centre of the disc and O' that of the hole. (see Fig.)
To find the centre of mass, we use the fact that a body balances at this point, i.e., the algebraic sum of the moments of the weights about the centre of gravity is zero. The weight $W_1$ of the disc acts at point O. The hole can be regarded as a negative weight $W_2$ acting at O'. If X is the distance of the centre of gravity of the combination from point O, then $\text{X}=\frac{\text{W}_1\times\text{O}+(-\text{W}_2)\times3}{\text{W}_1+(-\text{W}_2)}$ Also $\text{W}_1=\rho\pi\times(6)^2$ $=36\rho\pi;\text{W}_2=\rho\pi\times(1)^2=\rho\pi$ Where $\rho$ is the mass per unit area of the disc. Substituting the values of $W^1$ and $W^2$' we get $\text{X}=\frac{-\rho\pi\times3}{36\rho\pi-\rho\pi}\text{m}=\frac{-3}{35}\text{m}$ The negative sign indicates that the centre of gravity is to the left of the point O.
To find the centre of mass, we use the fact that a body balances at this point, i.e., the algebraic sum of the moments of the weights about the centre of gravity is zero. The weight $W_1$ of the disc acts at point O. The hole can be regarded as a negative weight $W_2$ acting at O'. If X is the distance of the centre of gravity of the combination from point O, then $\text{X}=\frac{\text{W}_1\times\text{O}+(-\text{W}_2)\times3}{\text{W}_1+(-\text{W}_2)}$ Also $\text{W}_1=\rho\pi\times(6)^2$ $=36\rho\pi;\text{W}_2=\rho\pi\times(1)^2=\rho\pi$ Where $\rho$ is the mass per unit area of the disc. Substituting the values of $W^1$ and $W^2$' we get $\text{X}=\frac{-\rho\pi\times3}{36\rho\pi-\rho\pi}\text{m}=\frac{-3}{35}\text{m}$ The negative sign indicates that the centre of gravity is to the left of the point O.
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