Mechanical Engineering - Engineering Mechanics - Discussion
Discussion Forum : Engineering Mechanics - Section 5 (Q.No. 31)
31.
The centre of gravity of a quadrant of a circle lies along its central radius (r) at a distance of
Discussion:
12 comments Page 1 of 2.
Rabindra Chakrabarty said:
8 years ago
The center of gravity of quadrant of a circle lies at a distance (4r/3*3.14) from both axes. And the question asks us to its cg lies along its central radius so we have considered it a right triangle to get the answer. And the hypotenuse will be the answer.
So let's start doing
(4r/3 *3.14)*2 + (4r/3 * 3.14) * 2 = X * 2. So, X = 0.6r.
So let's start doing
(4r/3 *3.14)*2 + (4r/3 * 3.14) * 2 = X * 2. So, X = 0.6r.
(2)
Diptangshu said:
7 years ago
Actually, it is the diagonal of the square made by the CGs in X and Y direction. That's why the radial distance of the CG from the centre will be the length of the diagonal of the square that is 2x(4r/3π)=0.6r.
Hardik said:
5 years ago
The center of gravity of a circular sector making semi-vertical angle α is at a distance of (2/3) are. Sinα/α from the center of the sector measured along the central axis.
2a=90.
A=45.
C.G=0.599999.
2a=90.
A=45.
C.G=0.599999.
(1)
Vyom said:
9 years ago
Distance form central radius i.e. along radial direction.
In both directions from origin it is 4r/3φ but for radial direction (4r/3φ) /cos45 which is nearly equal to 6.
In both directions from origin it is 4r/3φ but for radial direction (4r/3φ) /cos45 which is nearly equal to 6.
Manish said:
8 years ago
CG IN Y AXIS =4R/3π.
THEN CG ALONG R =(4R/3π)/SINE45°.
=.6R.
X, Y AND R CREATE A TRIANGLE IN QUADRANT CIRCLE.
THEN CG ALONG R =(4R/3π)/SINE45°.
=.6R.
X, Y AND R CREATE A TRIANGLE IN QUADRANT CIRCLE.
Vinay BEL said:
5 years ago
Simple if CG of the semicircle is 4r/3π.
Then quadrant vl b 4r/6π = 0.6r.
Then quadrant vl b 4r/6π = 0.6r.
(1)
Karuna said:
8 years ago
Simple if CG of the semicircle is 4r/3pi.
Then quadrant vl b 4r/6pi = 0.6 r.
Then quadrant vl b 4r/6pi = 0.6 r.
(1)
Piyush said:
6 years ago
4r/3π = 0.42r.
r-0.42r = 0.6r (approx).
r-0.42r = 0.6r (approx).
(6)
Surya pratap singh said:
1 decade ago
Please any one can explain this question.
Rajesh Khutdar said:
9 years ago
I think it should get by using 4r/3vpi.
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