please derive the centre of mass of a solid cone and a hollow cone.

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please derive the centre of mass of a solid cone and a hollow
cone

Ruhhi Ralhan · Accepted Answer

Dear Student, For a solid cone, Let (Xc,Yc) are the coordinates of center of mass Then, $Y_{c} = \frac{\int y ⅆ m}{\int d m} a n d X_{c} = \frac{\int x ⅆ m}{\int d m}$ For symmetry consider the center of mass at Xc=0 Divide the cone in horizontal disks of mass then, dm=ÏdV where, Ï is the density of material (constant) In this case, $d V = \frac{1}{3} π r^{2} d y$ where r is the radius of the cone at an arbitrary height dy radius of cone depends upon the height of the cone So, y=0 ; r=R (R is the base radius of the cone) y=h ; r=0 Using point-slope form: $r = - \frac{R}{h} y + R$ Now, the volume $d V = \frac{1}{3} π r^{2} d y$ So, $Y_{c} = \frac{\int y (\frac{1}{3} π r^{2} d y)}{\int \frac{1}{3} π r^{2} d y} = \frac{\int_{0}^{h} y (\frac{1}{3} π {(R (1 - \frac{y}{h}))}^{2} d y)}{\int_{0}^{h} \frac{1}{3} π {(R (1 - \frac{y}{h}))}^{2} d y} = \frac{\frac{1}{3} π R^{2} \frac{h^{2}}{12}}{\frac{1}{2} π R^{2} \frac{h}{3}} = \frac{1}{4} h$ So, the center of mass of solid cone is at the quarter height on the line joining from center of the base to the vertex For hollow cone, Let (Xc,Yc) are the coordinates of center of mass Then, $Y_{c} = \frac{\int y ⅆ m}{\int d m} a n d X_{c} = \frac{\int x ⅆ m}{\int d m}$ For symmetry consider the center of mass at Xc=0 For hollow cone, dm=Ï dA $d A = 2 π r d y Y_{c} = \frac{\int y (2 π r d y)}{\int 2 π r d y} = \frac{\int_{0}^{h} y (2 π R (1 - \frac{y}{h}) d y)}{\int_{0}^{h} 2 π R (1 - \frac{y}{h}) d y} = \frac{\frac{h^{2}}{6}}{\frac{h}{2}} = \frac{1}{3} h$ So, the center of mass of a hollow cone is at one-third of the height on the line joining from center of the base to the vertex