Question

how to find the centre of gravity of the cardioid r=a(1+cos$)​

Answer 1

Answer:

Step-by-step explanation:

dA=12r2dθ

A=12∫θ2θ1r2dθ

 

integral_009-centroid-cardioid.gif

 

A=2[12∫π0a2(1+cosθ)2dθ]

A=a2∫π0(1+cosθ)2dθ

A=a2(32π)

A=32πa2

 

By symmetry

YG=0

 

Solving for XG

AXG=∫ba23rcosθdA

32πa2XG=∫θ2θ123rcosθ(12r2dθ)

32πa2XG=13∫θ2θ1r3cosθdθ

32πa2XG=13[2∫π0a3(1+cosθ)3cosθdθ]

32πa2XG=23a3∫π0(1+cosθ)3cosθdθ

32πa2XG=23a3(15π8)

32πa2XG=54πa3

XG=56a

 

Centroid is at (5a/6, 0)           answer

Answer 2

Concept Introduction:-

It might resemble a word or a number representation of the quantity's arithmetic value.

Given Information:-

We have been given that $r=a(1+cos\ )$

To Find:-

We have to find that the centre of gravity of the cardioid $r=a(1+cos\ )$

Solution:-

According to the problem

Due to symmetry, the CG lies on the $x$-axis, whose coordinate is

$x_{c}=\frac{\int_{V} x d x d y d z}{\int_{V} d x d y d z}\\=\frac{2 \pi \int_{0}^{\pi} \int_{0}^{a(1+\cos \ )} r \cos \ r^{2} \sin \ d \ }{2 \pi \int_{0}^{\ } \int_{0}^{a(1+\cos \ )} r^{2} \sin \ d \ }\\=\frac{\frac{16}{15} a^{4}}{\frac{4}{3} a^{3}}\\=\frac{4}{5} a$

Final Answer:-

The correct answer is $\frac{4a}{5}$.

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