Question

From a uniform circular disc of radius R and mass M a small circular disc of radius R/2 is removed in such a way that both have a common tangent. Find the centre of mass of remaining part from the centre of original disc​

Answer 1

Hey,

Let the mass of the disc = M  

Therefore, the mass of the removed part of disc, m = (M/R^2)*(R/2)^2 = M/4  

Now the center of gravity of the resulting flat body,  

R= [M*0 – (M/4)*(R/2)]/(M-M/4)  

= -(MR/8)/(3M/4)  

= -R/6  

Negative sign shows that the center of gravity lies at opposite direction of the original COM at a distance R/6.

HOPE IT HELPS...

:)

Answer 2

Answer:

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