Question

Obtain Center of Mass:

Obtain the center of mass of an axis of a thin wire with a curve defined by $y=f(x)=x^{3/2}$ from x=0 to x=1 and a lineal density of mass of $p(x)=x^{2}$

Answer 1

Hi there!

To find the center of mass:

$com = \frac{1}{m}\int\ {rdm}$

We can begin by changing "dm" into an available quantity by expressing the lineal density as a function:

$\lambda = x^2 = \frac{m}{x}\\\\x^2 = \frac{m}{x}\\\\x^3 = m$

Differentiate both sides:

$3x^2dx = dm$

Let:

$r = f(x) = x^{3/2}$

Thus:

$com = \frac{1}{m} \int x^{3/2}3x^2dx\\\\com = \frac{1}{m} \int }3x^{7/2}dx$

Evaluate:

$com = \frac{1}{m} \frac{2}{3}x^{9/2}\left \| {{1 \atop {0}} \right.$

We found 'm' above, so plug the expression in:

$com = \frac{1}{x^3} \frac{2}{3}x^{9/2}\left \| {{1 \atop {0}} \right. \\\\com = \frac{1}{x^3} \frac{2}{3}x^{9/2}\left \| {{1 \atop {0}} \right. \\\\com = \frac{2}{3}x^{3/2}\left \| {{1 \atop {0}} \right. = \boxed{\frac{2}{3}}$