Figure-4.11 shows a circular a disc of radius R from which a small disc is cut such that the periphery of the small disc touch the large disc and whose radius is `R//2`. Find the centre of mass of the remaining part of the disc.

A thin homogeneous lamina is in the form of a circular disc of radius R. From it a circular hole exactly hall the radius of the lamina and touching the lamina's circumference is cut off. Find the centre of mass of the remaining part.

A disc of radis R is cut out from a larger disc of radius 2R in such a way that the edge of the hole touches the edge of the disc. Locate the centre of mass of the residual disc.

A disc of radius r is cut from a larger disc of radius 4r in such a way that the edge of the hole touches the edge of the disc. The centre of mass of the residual disc will be a distance from centre of larger disc :-

From a uniform circular dis of radius R , a circular dis of radius R//6 and having centre at a distance R//2 from the centre of the disc is removed. Determine the centre of mass of remaining portion of the disc.

A circular hole is cut from a disc of radius 6 cm in such a way that the radius of the hole is 1 cm and the centre of 3 cm from the centre of the disc. The distance of the centre of mass of the remaining part from the centre of the original disc is

From a uniform disc of radius R , a circular section of radius R//2 is cut out. The centre of the hole is at R//2 from the centre of the original disc. Locate the centre of mass of the resulating flat body.

Given a uniform disc of mass M and radius R . A small disc of radius R//2 is cut from this disc in such a way that the distance between the centres of the two discs is R//2 . Find the moment of inertia of the remaining disc about a diameter of the original disc perpendicular to the line connecting the centres of the two discs

From the circular disc of radius 4R two small discs of radius R are cut off. The centre of mass of the new structure will be at

A circular disc of radius (R)/(4) is cut from a uniform circular disc of radius R. The centre of cut portion is at a distance of (R)/(2) from the centre of the disc from which it is removed. Calculate the centre of mass of remaining portion of the disc.