Question

A uniform metal disc of radius R is taken and out of it a disc of diameter R/2 is cut-off from the end.the centre of mass of the remaining part will be

Answer 1

see figure,
Let mass of uniform metal disc of radius R is M.
so, mass of disc of diameter R/2 is M/πR² × π(R/4)² = M/16

now, Let's assume that centre of big disc is located at origin.
so, centre of small disc lies on axis at x = 3R/4

now use formula,
$C.M=\frac{MX-mx}{M-m}$

where M is mass of big uniform disc , m is mass of small uniform disc , X is the centre of big and x is the centre of small disc.

here, X = 0, x = 3R/4 and m = M/16

now, C.M = (M × 0 - M/16 × 3R/4)/(M - M/16)

CM = (-3R/4 × 16)/(15/16)

CM = -R/20

similarly you can use formula $C.M=\frac{MY-my}{M-m}$ to get y coordinate of centre of mass of remaining part.

but Y = 0, and y = 0, so, y coordinate of CM = 0

hence, centre of mass of remaining part is (-R/20, 0)

Answer 2

Answer:

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