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

Given r = r

_{o}(1 + 2x^{ 2}/ a^{ 2}) where x is the distance along the rod from the end of P and r_{o}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 = Adx r

Where r is the density. Then we can write:

dm = A dx r

_{o}(1 + 2x^{ 2}/ a^{ 2})The x -coordinate of the center of mass of dm is x, so therefore:

__x__=

ò

^{2a }_{0}_{ }x. Adx r_{o}( 1 + 2x^{ 2}/ a^{ 2}) / ò^{2a }_{0}_{ }Adx r_{o}( 1 + 2x^{ 2}/ a^{ 2})__x__= A r

_{o}ò

^{2a }_{0}_{ }( 1 + 2x

^{ 2}/ a

^{ 2}) dx /A r

_{o}ò

^{2a }_{0}_{ }( 1 + 2x

^{ 2}/ a

^{ 2}) dx

= [x

^{ 2}/2 + 2x^{4}/4a^{ 2}]^{2a }_{0 }/ [x^{ }+ 2x^{3}/3a^{ 2}]^{2a }_{0}

**=**(2a

^{ 3}

**+**8 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

^{2 }where: x^{ 2}+ y^{2 }= r^{ 2}And so: dV = A(y) dy = p (r

^{2 }- y^{2}) dyWith mass element: dm = kp (r

^{2 }- y^{2}) y dyThe 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

^{r }_{0}_{ }kp (r^{2 }- y^{2}) y dy / ò^{r }_{0}_{ }kp (r^{2 }- y^{2}) dy= kp [r

^{ 2}^{ }y^{3}/3 - y^{5}/5 ]^{r }_{0 }/ kp [r^{ 2}^{ }y^{2}/2 - y^{4}/4 ]^{r }_{0}

_{}

__y__= 8r/ 15

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