a circular part of diameter L is cut out from uniform rectangular piece of steel of length and breath 2Las in the figure above . O located the location of the center of mass of the leftover piece with the origin O located at the lower left hand corner of the rectangular sheet
Answers
Let mass per unit area of the original disc=σ
Thus mass of original disc=M=σπR
2
Radius of smaller disc=R/2.
Thus mass of the smaller disc=σπ(R/2)
2
=M/4
After the smaller disc has been cut from the original, the remaining portion is considered to be a system of two masses. The two masses are:
M(concentrated at O), and -M(=M/4) concentrated at O'
(The negative sign indicates that this portion has been removed from the original disc.)
Let x be the distance through which the centre of mass of the remaining portion shifts from point O.
The relation between the centres of masses of two masses is given as:
x=(m
1
r
1
+m
2
r
2
)/(m
1
+m
2
)
=(M×0−(M/4)×(R/2))/(M−M/4)=−R/6
(The negative sign indicates that the centre of mass gets shifted toward the left of point O)