## Tuesday, December 6, 2022

### Integration (Center of mass) Problem Solutions

1) A straight, thin rod PQ of length 2a of constant cross sectional area A is of variable density, r.

Given r  = ro (1 + 2x 2 / a 2)  where x is the distance along the rod from the end of P and ro  is a constant.  Find the position of the center of mass.

Solution:
We make use of the sketch below with blackened element dx and distance x from P.

The volume element can then be expressed:

dV=   A dr x

and the mass element:  dm =  Adr

Where  is the density.   Then we can write:

dm =  dx ro (1 + 2x 2 / a 2)

The x -coordinate of the center of mass of  d is x, so therefore:

x =

ò 2a 0   x. Adx ro 1 + 2x 2 / a 2) / ò 2a 0  Adx ro 1 + 2x 2 / a 2)

x =  ro ò 2a 0  1 + 2x 2 / a 2dx /ro ò 2a 0  1 + 2x 2 / a 2dx

=  [x 2 /2  +  2x4  /4a 2] 2a 0 / [x +  2x3  /3a 2] 2a 0

(2a 3 + a 2) / (2a  16a/3)  =   15a/ 11 (from end of P)

2) Find the center of mass of a solid hemisphere of radius r if the density at any point P is proportional to the distance of P from the base of the hemisphere.  (Hint:  Add another lower latitutde layer to the upper latitude circle below to form a thin slice of thickness dy.  Then imagine the whole hemisphere cut into slices of thickness dy, i.e. by planes perpendicular to the y-axis)

Solution:

Below I show the hemisphere again with the lower latitude circle added and slice thickness dimension dy indicated, which will provide the basis for integration:
Note: We imagine the entire solid cut into slices of thickness dy so can write:

dV = A(y) dy  as the volume of the representative slice at distance y above the base of the hemisphere (which lies in the x-z plane).  The area of a face of the slice dV is:

A(y) =  p x   where:  x 2 +  =  r 2

And so:  dV = A(y) dy  =   p (r 2   - 2) dy

With mass element:  dm  =  kp (r 2   - 2) y dy

The center of mass of the 'ground' slice may then be taken at the geometrical center (x, y, z) =  (0, y,  0)  or:

y  =     ò y dm / ò dm

ò 0  kp (r 2   - 2) y dy  / ò 0  kp (r 2   - 2) dy

kp [r 2 y3  /3   -  y5  /5 ] kp [r 2 y2  /2   -  y4  /4 ] 0

y  =  8r/ 15