Mechanical Engineering - Engineering Mechanics - Discussion
Discussion Forum : Engineering Mechanics - Section 4 (Q.No. 25)
25.
A circular hole of 50 mm diameter is cut out from a circular disc of 100 mm diameter as shown in the below figure. The centre of gravity of the section will lie
Discussion:
2 comments Page 1 of 1.
Hritik Chapagain said:
2 years ago
In a horizontal axis,
CG of bigger circle(X1) = 50mm.
CG of smaller circle(X2) = 50 + 25 = 75mm.
Area of Bigger Circle(A1) = PI * 100^2/4 = 2500π
Area of Reduced circle (A2)) = -PI*50^2/4 = 625π
∑A = A1 + A2 = 1875π
A1X1 = 125000π
A2X2 = -46875π
∑AX = 78125π
Total CG= ∑AX/ ∑A = 41.66.
so, it shifts towards the left side.
CG of bigger circle(X1) = 50mm.
CG of smaller circle(X2) = 50 + 25 = 75mm.
Area of Bigger Circle(A1) = PI * 100^2/4 = 2500π
Area of Reduced circle (A2)) = -PI*50^2/4 = 625π
∑A = A1 + A2 = 1875π
A1X1 = 125000π
A2X2 = -46875π
∑AX = 78125π
Total CG= ∑AX/ ∑A = 41.66.
so, it shifts towards the left side.
Omar Farooq said:
1 decade ago
It should be towards right in the hole to balance out the disturbed weight distribution of the circular disc?
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