Question

From a uniform disc of radius R, a circular hole 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 distance of the gravity of the resulting flat body.

Answer 1

From a big uniform disc of radius R with centre O, a smaller circular hole of radius R/2 with its centre O1 (where OO1 = R/2) is cut out.
Let the centre of gravity of remaining flat body be C and OC= d.

Let mass of big disc = M
So, mass of small disc = M/πR² × π(R/2)² = M/4

from definition of centre of mass,
centre of mass , x = $\frac{m_1x_1-m_2x_2}{m_1-m_2}$

here negative sign is used because mass is removed. otherwise we will use positive sign.

position of centre of mass of m1 { big disc} = origin
position of centre of mass of m2 { small disc} = (-R/2, 0)

so, x = {M × 0 - M/4 × -R/2}/{M - M/4}

or, x = {MR/8}/{3M/4} = R/6

hence, C is at a distance R/6 from the centre of disc on diametrically opposite side of the centre of hole.

Answer 2

Answer:

A man stands on a rotating platform with his arms stretched horizontally holding a 5 kg weight in each hand. The angular speed of the platform is 30 revolutions per minute. The man then bring his arms close to his body with the distance of each weight from the axis changing from 90 cm to 20 cm. The moment of inertia of the man together with the platform may be taken to be constant and equal to 7.6 kg m²a. What is his new angular speed? (Neglect friction)b. Is kinetic energy conserved in the process? If not from where does the change