What happens to the center of mass of a disc if a circular portion is removed from the
right side of the center of disc
Answers
Answer:
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)
Explanation:
From a uniform disk 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 centre of gravity of the resulting flat body.
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ANSWER
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)