# Question #3a15a

Dec 19, 2017

The centre-of-mass is at a distance of $\setminus \quad {x}_{c m} = \left(\setminus \frac{A + \frac{2}{3} B L}{2 A + B L}\right) L$ from the less-heavy end.

#### Explanation:

Total Mass: We can write the total mass $M$ of the rod in terms of the coefficients $A$ and $B$, of the linear mass density, and its length $L$.

$M = \setminus {\int}_{0}^{L} \setminus \quad \setminus \lambda \left(x\right) . \mathrm{dx} = \setminus {\int}_{0}^{L} \setminus \quad \left(A + B x\right) . \mathrm{dx}$
$M = A \left(L - 0\right) + \frac{B}{2} \left({L}^{2} - 0\right) = A L + B {L}^{2} / 2$ ...... (Eq 1)

Centre-of-Mass:

${x}_{c m} = \frac{1}{M} \setminus \int \setminus \quad x . \mathrm{dm}$

If $\setminus \lambda \left(x\right)$ is the linear mass density (density per unit length) at a distance $x$ from the less-heavy end,

$\mathrm{dm} = \setminus \lambda \left(x\right) \mathrm{dx} = \left(A + B x\right) . \mathrm{dx}$

${x}_{c m} = \frac{1}{M} \setminus {\int}_{0}^{L} x . \left(A + B x\right) . \mathrm{dx}$
${x}_{c m} = \frac{1}{M} \left[\frac{A}{2} \left({L}^{2} - {0}^{2}\right) + \frac{B}{3} \left({L}^{3} - {0}^{3}\right)\right] ,$
${x}_{c m} = \frac{1}{M} \left[\frac{A}{2} {L}^{2} + \frac{B}{3} {L}^{3}\right]$......(Eq 2)

Substituting for $M$ using Eq 1 in Eq 2,

${x}_{c m} = \setminus \frac{\frac{A}{2} {L}^{2} + \frac{B}{3} {L}^{3}}{A L + \frac{B}{2} {L}^{2}} = \left(\setminus \frac{\frac{A}{2} + \frac{B}{3} L}{A + \frac{B}{2} L}\right) L$

${x}_{c m} = \left(\setminus \frac{A + \frac{2}{3} B L}{2 A + B L}\right) L$