# Lamina Workouts

• Feb 25th 2010, 02:37 PM
Selim
Lamina Workouts
Given a lamina of uniform density $\displaystyle \rho$, bounded by $\displaystyle y = \sqrt{x} , y=0 , x=4.$

1. Set up an integral to find m, the mass of the lamina.
2. Find m, the mass of the lamina
3. Set up Integral to find Mx
4. Find Mx
5. Set up integral to find My
6. Find My
7. Find the center of mass.
• Feb 25th 2010, 02:49 PM
Selim
m = $\displaystyle \rho \int_0^4 \sqrt{x} * dx = \frac{16}{3}$

Mx = $\displaystyle \rho \int_0^4 \frac{\sqrt{x}}{2} * \sqrt{x} * dx = 4$

y = $\displaystyle \frac{.25}{3} = \frac{1}{12}$

Center: (x, $\displaystyle \frac{1}{12}$)

My = ?
• Feb 26th 2010, 09:01 AM
ione
Quote:

Originally Posted by Selim
m = $\displaystyle \rho \int_0^4 \sqrt{x} * dx = \frac{16}{3}$

Mx = $\displaystyle \rho \int_0^4 \frac{\sqrt{x}}{2} * \sqrt{x} * dx = 4$

y = $\displaystyle \frac{.25}{3} = \frac{1}{12}$

Center: (x, $\displaystyle \frac{1}{12}$)

My = ?

$\displaystyle m= \frac{16}{3}\rho$

$\displaystyle M_x=4\rho$

$\displaystyle y = \frac{M_x}{m} = \frac{3}{4}$

$\displaystyle M_y=\rho\int_0^4x\sqrt{x}\,dx$

$\displaystyle x=\frac{M_y}{m}$