Question

pls help me with this
a parabolic shape solid object is formed by rotating on a parabola Y is equal to 2 x square about y axis as shown in the figure if the height of the body is Hach then the position of centre of mass from its origin is​

Answer 1

Given that,

A parabolic shape solid object is formed by rotating on a parabola.

$y=2x^2$

Height = h

Suppose, Area density = σ

We need to calculate the area

Using formula of area

$A=\int_{0}^{h}{2\pi x dy}$

Put the value of x

$A=\int_{0}^{h}{2\pi \times\dfrac{\sqrt{y}}{\sqrt{2}}}dy$

$A=\dfrac{2\pi}{\sqrt{2}}(\dfrac{2}{3}(y)^{\frac{2}{3}})_{0}^{h}$

$A=\dfrac{4\pi}{3\sqrt{2}}h^{\frac{3}{2}}$

We need to calculate the value of density

Using formula of mass

$\sigma=\dfrac{M}{A}$

Put the value into the formula

$\sigma=\dfrac{M3\sqrt{2}}{4\pi(h)^{\frac{3}{2}}}$

We need to calculate the center of mass

Using formula of center of mass

$Y_{cm}=\dfrac{\int_{0}^{h}{y\sigma 2\pi x dy}}{M}$

Put the value into the formula

$Y_{cm}=\dfrac{\int_{0}^{h}(y\times2\pi x\times(\dfrac{M3\sqrt{2}}{4\pi(h)^{\frac{3}{2}}}))}{M}dy$

$Y_{cm}=\int_{0}^{h}{y\times2\pi x\times\dfrac{3\sqrt{2}}{4\pi h^{\frac{3}{2}}}}dy$

$Y_{cm}=\dfrac{3\sqrt{2}}{2h^{\frac{3}{2}}}\int_{0}^{h}{y\times\sqrt{\dfrac{y}{2}}}dy$

$Y_{cm}=\dfrac{3}{2h^{\frac{3}{2}}}\int_{0}^{h}{y^{\frac{\dfrac{3}{2}}}}dy$

$Y_{cm}=\dfrac{3}{2h^{\frac{3}{2}}}\times\dfrac{2}{5}\times h^{\frac{5}{2}}$

$Y_{cm}=\dfrac{3}{5}h$

Hence, The center of mass is $\dfrac{3}{5}h$

Answer 2

Given that,

A parabolic shape solid object is formed by rotating on a parabola.

$y=2x^2$

Height = h

Suppose, Area density = σ

We need to calculate the area

Using formula of area

$A=\int_{0}^{h}{2\pi x dy}$

Put the value of x

$A=\int_{0}^{h}{2\pi \times\dfrac{\sqrt{y}}{\sqrt{2}}}dy$

$A=\dfrac{2\pi}{\sqrt{2}}(\dfrac{2}{3}(y)^{\frac{2}{3}})_{0}^{h}$

$A=\dfrac{4\pi}{3\sqrt{2}}h^{\frac{3}{2}}$

We need to calculate the value of density

Using formula of mass

$\sigma=\dfrac{M}{A}$

Put the value into the formula

$\sigma=\dfrac{M3\sqrt{2}}{4\pi(h)^{\frac{3}{2}}}$

We need to calculate the center of mass

Using formula of center of mass

$Y_{cm}=\dfrac{\int_{0}^{h}{y\sigma 2\pi x dy}}{M}$

Put the value into the formula

$Y_{cm}=\dfrac{\int_{0}^{h}(y\times2\pi x\times(\dfrac{M3\sqrt{2}}{4\pi(h)^{\frac{3}{2}}}))}{M}dy$

$Y_{cm}=\int_{0}^{h}{y\times2\pi x\times\dfrac{3\sqrt{2}}{4\pi h^{\frac{3}{2}}}}dy$

$Y_{cm}=\dfrac{3\sqrt{2}}{2h^{\frac{3}{2}}}\int_{0}^{h}{y\times\sqrt{\dfrac{y}{2}}}dy$

$Y_{cm}=\dfrac{3}{2h^{\frac{3}{2}}}\int_{0}^{h}{y^{\frac{\dfrac{3}{2}}}}dy$

$Y_{cm}=\dfrac{3}{2h^{\frac{3}{2}}}\times\dfrac{2}{5}\times h^{\frac{5}{2}}$

$Y_{cm}=\dfrac{3}{5}h$

Hence, The center of mass is $\dfrac{3}{5}h$