Question

A flat thin rectangular lamina lies in the x-y plane as shown in Figure 2. It has a width W and a length L. If the surface density mu = mu_0(y/L + 1): (a) What is the total mass of the lamina (ignore the thickness T of the object). (b) Where is the centre-of-mass located on the object?

Answer 1

The density depends on the position over the lamina so to calculate the mass of the lamina we need to integrate the next function:

$dM=\mu dA$

The area differential and the surface density:

$dA=dxdy$

$\mu=\mu_{0}(\frac{y}{L}+1)$

Substituting:

$M=\int\int(\mu_{0}(\frac{y}{L}+1)) dxdy$

Now, we need to determinate the limits of each variable:

$M=\int^{L}_{0}\int^{W}_{0}(\mu_{0}(\frac{y}{L}+1)) dxdy$

$M=\int^{L}_{0}\mu_{0}(\frac{y}{L}+1))xdy\rightarrow M=\int^{L}_{0}\mu_{0}(\frac{y}{L}+1))(W-0)dy$

$M=W\int^{L}_{0}\mu_{0}(\frac{y}{L}+1))dy\rightarrow M=\mu_{0}W(\frac{y^{2}}{2L}+y)$

$M=\mu_{0}W(\frac{L^{2}}{2L}+L)$

$M=\mu_{0}W(\frac{L}{2}+L)=\frac{3}{2}\mu_{0}WL$

To calculate the center-of-mass we need to solve the next two integrals:

$x_{CM}=\frac{1}{M}\int xdM=\frac{1}{M}\int^{L}_{0}\int^{W}_{0}x(\mu_{0}(\frac{y}{L}+1)) dxdy$

$x_{CM}=\frac{1}{M}\int^{L}_{0}(\mu_{0}(\frac{y}{L}+1)) \frac{x^{2}}{2}dy$

$x_{CM}=\frac{1}{M}\frac{W^{2}}{2}\int^{L}_{0}(\mu_{0}(\frac{y}{L}+1)) dy$

$x_{CM}=\frac{1}{M}\frac{W^{2}}{2}\mu_{0}(\frac{y^{2}}{2L}+y)=\frac{1}{M}\frac{W^{2}}{2}\mu_{0}(\frac{L^{2}}{2L}+L)$

$x_{CM}=\frac{1}{\frac{3}{2}\mu_{0}WL}\frac{3}{4}\mu_{0}W^{2}L=\frac{1}{2}W$

$y_{CM}=\frac{1}{M}\int^{L}_{0}\int^{W}_{0}y(\mu_{0}(\frac{y}{L}+1)) dxdy=\frac{1}{M}W\int^{L}_{0}(\mu_{0}(\frac{y^{2}}{L}+y)) dy$

$y_{CM}=\frac{1}{M}W\mu_{0}(\frac{y^{3}}{3L}+\frac{y^{2}}{2})=\frac{1}{M}W\mu_{0}(\frac{L^{3}}{3L}+\frac{L^{2}}{2})$

$y_{CM}=\frac{1}{\frac{3}{2}\mu_{0}WL}\frac{5}{6}\mu_{0}L^{2}W=\frac{5}{9}L$

So the center of mass;

$(x_{CM},y_{CM})=\frac{1}{2}W,\frac{5}{9}L$

Answer 2

The density depends on the position over the lamina so to calculate the mass of the lamina we need to integrate the next function:

$dM=\mu dA$

The area differential and the surface density:

$dA=dxdy$

$\mu=\mu_{0}(\frac{y}{L}+1)$

Substituting:

$M=\int\int(\mu_{0}(\frac{y}{L}+1)) dxdy$

Now, we need to determinate the limits of each variable:

$M=\int^{L}_{0}\int^{W}_{0}(\mu_{0}(\frac{y}{L}+1)) dxdy$

$M=\int^{L}_{0}\mu_{0}(\frac{y}{L}+1))xdy\rightarrow M=\int^{L}_{0}\mu_{0}(\frac{y}{L}+1))(W-0)dy$

$M=W\int^{L}_{0}\mu_{0}(\frac{y}{L}+1))dy\rightarrow M=\mu_{0}W(\frac{y^{2}}{2L}+y)$

$M=\mu_{0}W(\frac{L^{2}}{2L}+L)$

$M=\mu_{0}W(\frac{L}{2}+L)=\frac{3}{2}\mu_{0}WL$

To calculate the center-of-mass we need to solve the next two integrals:

$x_{CM}=\frac{1}{M}\int xdM=\frac{1}{M}\int^{L}_{0}\int^{W}_{0}x(\mu_{0}(\frac{y}{L}+1)) dxdy$

$x_{CM}=\frac{1}{M}\int^{L}_{0}(\mu_{0}(\frac{y}{L}+1)) \frac{x^{2}}{2}dy$

$x_{CM}=\frac{1}{M}\frac{W^{2}}{2}\int^{L}_{0}(\mu_{0}(\frac{y}{L}+1)) dy$

$x_{CM}=\frac{1}{M}\frac{W^{2}}{2}\mu_{0}(\frac{y^{2}}{2L}+y)=\frac{1}{M}\frac{W^{2}}{2}\mu_{0}(\frac{L^{2}}{2L}+L)$

$x_{CM}=\frac{1}{\frac{3}{2}\mu_{0}WL}\frac{3}{4}\mu_{0}W^{2}L=\frac{1}{2}W$

$y_{CM}=\frac{1}{M}\int^{L}_{0}\int^{W}_{0}y(\mu_{0}(\frac{y}{L}+1)) dxdy=\frac{1}{M}W\int^{L}_{0}(\mu_{0}(\frac{y^{2}}{L}+y)) dy$

$y_{CM}=\frac{1}{M}W\mu_{0}(\frac{y^{3}}{3L}+\frac{y^{2}}{2})=\frac{1}{M}W\mu_{0}(\frac{L^{3}}{3L}+\frac{L^{2}}{2})$

$y_{CM}=\frac{1}{\frac{3}{2}\mu_{0}WL}\frac{5}{6}\mu_{0}L^{2}W=\frac{5}{9}L$

So the center of mass;

$(x_{CM},y_{CM})=\frac{1}{2}W,\frac{5}{9}L$