A circular hole of radius 3 cm is cut out from a uniform circular disc of radius 6 cm. The centre of the hole is at 3 cm, from the centre of the original disc. The distance of centre of gravity of th…

Question

A circular hole of radius 3 cm is cut out from a uniform circular disc of radius 6 cm. The centre of the hole is at 3 cm, from the centre of the original disc. The distance of centre of gravity of the resulting flat body from the centre of the original disc is

Accepted Answer

Correct answer: B. (b) Given, radius of circular hole, r=3 cm Radius of circular disc, R=6 cm Mass of disc, M= R^2m Mass of hole, M^ '= r^2m If we assume the centre of mass of hole and disc to be at centre, then The x-coordinate of the centre of mass of the remaining portion of the disc will be x_ CM= Mx_1-M^ 'x_2M-M^ '= M 0-M^ ' 3 R^2m- r^2m= -3M^ ' πm(R^2-r^2)= -3 r^2m πm(R^2-r^2)=- -3r^2R^2-r^2= -3 3^26^2-3^2=-1 cm ⇒x_ CM=-1 cm Hence, distance of centre of gravity of the resulting flat body from the centre of the original disc will be 1 cm left side.

$A circular hole of radius 3 cm is cut out from a uniform circular disc of radius 6 cm . The centre of the hole is at 3 cm, from the centre of the original disc. The distance of centre of gravity of the resulting flat body from the centre of the original disc is$

Solution

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