Physics, asked by rasheedpokkath1349, 1 year ago

Find the centre of gravity of the arc of the cardioid r=a(1+cos) lying above the initial line.

Answers

Answered by Anonymous
0

Answer:

l=∫√r2+(drdθ)2dθ. r=1+cos θ⇒drdθ=−sinθ. l=2∫π0√(1+cosθ)2+sin2θdθ . l=2∫π0√1+2cosθ+cos2θ+sin2θdθ.

Answered by rishkrith123
0

Complete Question:

Find the center of gravity of the volume generated by the revolution of the cardioid r = a(1 + cos(Ф)) about the x-axis.

Answer:

The center of gravity is 4a/5.

Explanation:

To find,

They're requesting the center of gravity of the volume.

We have a tendency to shall use the subsequent parametrization of the higher half of the cardioid:

\left.\eqalign{x(\phi)&=a(1+\cos\phi)\cos\phi\cr y(\phi) ,y(\phi)=a(1+\cos\phi)\sin\phi\cr}\right\}\qquad(0\leq\phi\leq\pi)\ .

It is obvious that the center of gravity is on the x-axis. We cut the volume into cylindrical plates with radius y(ϕ) and thickness x′(ϕ)dϕ. The total volume V is then

-\int_0^\pi \pi y^2(\phi)\ > x'(\phi)\ > d\phi={8\pi\over3}\,a^3\ .

Writing an overall minus sign here makes everything come out right: For 0 ≤ ϕ ≤ 2π/3 one has

x'(\phi)=-(1+2\cos\phi)\sin\phi\leq0\ ,

and for 2π/3 ≤ ϕ ≤ π one has x′(ϕ) ≥ 0. however the corresponding volume part (the peak of the inward) has got to be deducted.

in an exceedingly second step, we tend to figure the full moment of these cylinders within the x-direction. This ends up in the integral:

-\int_0^\pi  x(\phi)\ > \pi y^2(\phi)\ > x'(\phi)\ > d\phi={32\pi\over15}\,a^4\ .

The x-coordinate ξ of the center of gravity, therefore, is the quotient

\xi={{32\pi\over15}a^4\over{8\pi\over3}a^3}={4a\over5}\ .

Therefore, the center of gravity is 4a/5.

#SPJ2

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