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I calculated the centre of mass of solid cone and hollow cone . Can anybody help me find the centre of mass of a frustrum?

I calculated the centre of mass of solid cone and hollow cone . Can anybody help me find the centre of mass of a frustrum?

Grade:11

1 Answers

Arun
25750 Points
5 years ago
Dear student

In accord with your question, let the frustum have its axis along the y-axis, and it's base on the x-z plane. Let the radius of the base be R, the height of the cone from which the frustum is cut be H, the height of the frustum be h (h

The radius of the frustum as a function of distance along the y axis is given by: 

r(y) = R- y*R/H = (H-y)*R/H, 0

The mass loading (the linear mass density) as a function of distance along the y-axis is given by: 

M(y) = D*pi*(r(y))^2 = D * pi * (H-y)^2 * (R/H)^2 

The total mass between two values of y is given by the integral of M(y) dy between the two points. We want to find a point y = C such that: 

Integral from 0 to C of {D * pi * (H-y)^2 * (R/H)^2 dy} = Integral from C to h of {D * pi * (H-y)^2 * (R/H)^2 dy}  

That is, essentially want to find the value y = C for which the volume to one side of C is equal to the volume on the other side of C (we can use volume instead of mass because we've assumed that the density is uniform). 

Cancelling like factors on either side of the above equation, we have that: 

Integral from 0 to C of {(H-y)^2 dy} = Integral from C to h of {(H-y)^2 dy}  

-1/3 * (H-C)^3 + 1/3 * H^3 = -1/3 * (H-h)^3 + 1/3 * (H-C)^3 

H^3 - 2*(H-C)^3 + (H-h)^3 = 0 

This has only one real root: 

C =H + 0.5*(12*h*H^2 + 4*h^3 -12*H*h^2 - 8*H^3)^(1/3) 

The center of mass of a right circular frustum with uniform density and with its axis along the y axis is at x = 0, y = C, z = 0, where C is given above. 
 

 

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