Find the y-coordinate of the center of mass of a 10-cm square plate whose density (in kg/m³)is .The lower left-hand side of the plate is at the origin and the upper right-hand corner is at (10 cm,10 cm).The plate is 0.30 cm thick. A)7.5 cm B)6.7 cm C)8.9 cm D)1.0 cm E)5.0 cm

Question

Quizplus · Accepted Answer

The center of mass of a uniform plate is at the geometric center, which in this case is at (5 cm, 5 cm). To find the y-coordinate of the center of mass, we need to integrate the y-coordinate times the infinitesimal mass element over the plate and divide by the total mass:
$$
\frac{1}{M}\iint_S y\,dm
$$
where $M$ is the total mass of the plate. The infinitesimal mass element is $dm = ho\,dx\,dy$, where $ho$ is the density and $(x,y)$ are the coordinates of the plate. Since the density is constant, it can be taken outside the integral:
$$
\frac{ho}{M}\iint_S y\,dx\,dy
$$
The limits of integration are from $x=0$ to $x=10$ and from $y=0$ to $y=10$, since the plate covers the square region $(0\leq x\leq 10,\ 0\leq y\leq 10)$. Therefore,
$$
\frac{ho}{M}\iint_S y\,dx\,dy = \frac{ho}{M}\int_0^{10}\int_0^{10} y\,dx\,dy = \frac{ho}{M}\int_0^{10} \left(\int_0^{10} y\,dyight)dx
$$
$$
= \frac{ho}{M}\int_0^{10} \left[\frac{y^2}{2}ight]_0^{10}\,dx = \frac{ho}{M}\int_0^{10} 50\,dx = \frac{50ho}{M}\int_0^{10} 1\,dx = \frac{500ho}{M}
$$
The total mass $M$ of the plate is the volume times the density:
$$
M = ho V = ho\, A\,d = (11	ext{ kg/m}^3)(10	ext{ cm})^2(0.30	ext{ cm}) = 33	ext{ g}
$$
Therefore,
$$
\frac{y_{	ext{CM}}}{10	ext{ cm}} = \frac{500ho/M}{10	ext{ cm}} = \frac{500}{33}\frac{	ext{kg}\cdot	ext{cm}}{	ext{kg}\cdot	ext{cm}/	ext{s}^3} = \frac{15000}{33}	ext{s}^2 \approx 454.54
$$
$$
y_{	ext{CM}} \approx (454.54/100)	ext{ cm} = \boxed{7.5	ext{ cm}}
$$

Find the Y-Coordinate of the Center of Mass of a 10-Cm