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
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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
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Given that,
A parabolic shape solid object is formed by rotating on a parabola.
[tex]y=2x^2[/tex]
Height = h
Suppose, Area density = σ
We need to calculate the area
Using formula of area
[tex]A=\int_{0}^{h}{2\pi x dy}[/tex]
Put the value of x
[tex]A=\int_{0}^{h}{2\pi \times\dfrac{\sqrt{y}}{\sqrt{2}}}dy[/tex]
[tex]A=\dfrac{2\pi}{\sqrt{2}}(\dfrac{2}{3}(y)^{\frac{2}{3}})_{0}^{h}[/tex]
[tex]A=\dfrac{4\pi}{3\sqrt{2}}h^{\frac{3}{2}}[/tex]
We need to calculate the value of density
Using formula of mass
[tex]\sigma=\dfrac{M}{A}[/tex]
Put the value into the formula
[tex]\sigma=\dfrac{M3\sqrt{2}}{4\pi(h)^{\frac{3}{2}}}[/tex]
We need to calculate the center of mass
Using formula of center of mass
[tex]Y_{cm}=\dfrac{\int_{0}^{h}{y\sigma 2\pi x dy}}{M}[/tex]
Put the value into the formula
[tex]Y_{cm}=\dfrac{\int_{0}^{h}(y\times2\pi x\times(\dfrac{M3\sqrt{2}}{4\pi(h)^{\frac{3}{2}}}))}{M}dy[/tex]
[tex]Y_{cm}=\int_{0}^{h}{y\times2\pi x\times\dfrac{3\sqrt{2}}{4\pi h^{\frac{3}{2}}}}dy[/tex]
[tex]Y_{cm}=\dfrac{3\sqrt{2}}{2h^{\frac{3}{2}}}\int_{0}^{h}{y\times\sqrt{\dfrac{y}{2}}}dy[/tex]
[tex]Y_{cm}=\dfrac{3}{2h^{\frac{3}{2}}}\int_{0}^{h}{y^{\frac{\dfrac{3}{2}}}}dy[/tex]
[tex]Y_{cm}=\dfrac{3}{2h^{\frac{3}{2}}}\times\dfrac{2}{5}\times h^{\frac{5}{2}}[/tex]
[tex]Y_{cm}=\dfrac{3}{5}h[/tex]
Hence, The center of mass is [tex]\dfrac{3}{5}h[/tex]
Given that,
A parabolic shape solid object is formed by rotating on a parabola.
[tex]y=2x^2[/tex]
Height = h
Suppose, Area density = σ
We need to calculate the area
Using formula of area
[tex]A=\int_{0}^{h}{2\pi x dy}[/tex]
Put the value of x
[tex]A=\int_{0}^{h}{2\pi \times\dfrac{\sqrt{y}}{\sqrt{2}}}dy[/tex]
[tex]A=\dfrac{2\pi}{\sqrt{2}}(\dfrac{2}{3}(y)^{\frac{2}{3}})_{0}^{h}[/tex]
[tex]A=\dfrac{4\pi}{3\sqrt{2}}h^{\frac{3}{2}}[/tex]
We need to calculate the value of density
Using formula of mass
[tex]\sigma=\dfrac{M}{A}[/tex]
Put the value into the formula
[tex]\sigma=\dfrac{M3\sqrt{2}}{4\pi(h)^{\frac{3}{2}}}[/tex]
We need to calculate the center of mass
Using formula of center of mass
[tex]Y_{cm}=\dfrac{\int_{0}^{h}{y\sigma 2\pi x dy}}{M}[/tex]
Put the value into the formula
[tex]Y_{cm}=\dfrac{\int_{0}^{h}(y\times2\pi x\times(\dfrac{M3\sqrt{2}}{4\pi(h)^{\frac{3}{2}}}))}{M}dy[/tex]
[tex]Y_{cm}=\int_{0}^{h}{y\times2\pi x\times\dfrac{3\sqrt{2}}{4\pi h^{\frac{3}{2}}}}dy[/tex]
[tex]Y_{cm}=\dfrac{3\sqrt{2}}{2h^{\frac{3}{2}}}\int_{0}^{h}{y\times\sqrt{\dfrac{y}{2}}}dy[/tex]
[tex]Y_{cm}=\dfrac{3}{2h^{\frac{3}{2}}}\int_{0}^{h}{y^{\frac{\dfrac{3}{2}}}}dy[/tex]
[tex]Y_{cm}=\dfrac{3}{2h^{\frac{3}{2}}}\times\dfrac{2}{5}\times h^{\frac{5}{2}}[/tex]
[tex]Y_{cm}=\dfrac{3}{5}h[/tex]
Hence, The center of mass is [tex]\dfrac{3}{5}h[/tex]