Question

The center of mass of a thin rectangular plate (fig - x) with sides of length \( a \) and \( b \), whose mass per unit area (\( \sigma \)) varies as \( \sigma = \sigma_0 \frac{x}{ab} \) (where \( \sigma_0 \) is a constant), would be

• A. \( \left(\frac{2}{3} a, \frac{b}{2} \right) \)
• B. \( \left(\frac{1}{3} a, \frac{1}{2} b\right) \)
• C. \( \left(\frac{1}{2} a, \frac{1}{2} b\right) \)
• D. \( \left(\frac{2}{3} a, \frac{1}{2} b\right) \)

Hint:

In problems involving variable density, set up the integral for each coordinate weighted by the density and normalized by the total mass.

Answer 1

The Correct Option is A

Let the rectangular plate be defined by $ 0 \le x \le a $ and $ 0 \le y \le b $. 

The mass per unit area is given as:

$$ \sigma = \frac{\sigma_0 x}{ab} $$

To find the center of mass $ (x_{cm}, y_{cm}) $, we first calculate the total mass $ M $, then use the formulas:

$$ x_{cm} = \frac{1}{M} \iint x \sigma \,dA, \quad y_{cm} = \frac{1}{M} \iint y \sigma \,dA $$

Step 1: Total Mass $ M $

$$ M = \iint \sigma \,dA = \int_0^a \int_0^b \frac{\sigma_0 x}{ab} \,dy\,dx = \frac{\sigma_0}{ab} \int_0^a x \left( \int_0^b dy \right) dx $$
$$= \frac{\sigma_0}{ab} \int_0^a x \cdot b \,dx = \frac{\sigma_0}{a} \int_0^a x \,dx = \frac{\sigma_0}{a} \cdot \frac{a^2}{2} = \frac{\sigma_0 a}{2} $$

Step 2: $ x_{cm} $

$$ x_{cm} = \frac{1}{M} \int_0^a \int_0^b x \cdot \sigma \,dy\,dx = \frac{1}{\frac{\sigma_0 a}{2}} \int_0^a \int_0^b x \cdot \frac{\sigma_0 x}{ab} \,dy\,dx = \frac{2}{\sigma_0 a} \cdot \frac{\sigma_0}{ab} \int_0^a x^2 \cdot b \,dx $$ $$ = \frac{2}{a^2} \int_0^a x^2 \,dx = \frac{2}{a^2} \cdot \frac{a^3}{3} = \frac{2a}{3} $$

Step 3: $ y_{cm} $

$$ y_{cm} = \frac{1}{M} \int_0^a \int_0^b y \cdot \sigma \,dy\,dx = \frac{1}{\frac{\sigma_0 a}{2}} \int_0^a \int_0^b y \cdot \frac{\sigma_0 x}{ab} \,dy\,dx $$
$$= \frac{2}{\sigma_0 a} \cdot \frac{\sigma_0}{ab} \int_0^a x \left( \int_0^b y \,dy \right) dx $$ $$ = \frac{2}{a^2 b} \int_0^a x \cdot \frac{b^2}{2} \,dx = \frac{b}{a^2} \int_0^a x \,dx = \frac{b}{a^2} \cdot \frac{a^2}{2} = \frac{b}{2} $$

Final Answer:

The coordinates of the center of mass are: $ \left(\frac{2}{3}a, \frac{b}{2}\right) $

Answer 2

Step 1: Understand the problem.
We are given a thin rectangular plate of sides \( a \) and \( b \), placed in the \( xy \)-plane with one corner at the origin \( O(0,0) \).
The mass per unit area (surface density) varies along the x-axis according to:
\[ \sigma = \sigma_0 \frac{x}{ab} \] where \( \sigma_0 \) is a constant. We are asked to find the coordinates of the center of mass \( (x_{cm}, y_{cm}) \).

Step 2: Write expressions for mass element.
Consider a small element of area \( dA = dx \, dy \) at a point \( (x, y) \).
The mass of this element is: \[ dm = \sigma \, dA = \sigma_0 \frac{x}{ab} \, dx \, dy \] The total mass of the plate is: \[ M = \int_0^a \int_0^b \sigma_0 \frac{x}{ab} \, dy \, dx \] \[ M = \frac{\sigma_0}{ab} \int_0^a x \, dx \int_0^b dy = \frac{\sigma_0}{ab} \left( \frac{a^2}{2} \right) b = \frac{\sigma_0 a}{2} \]

Step 3: Find the x-coordinate of the center of mass.
The x-coordinate is given by: \[ x_{cm} = \frac{1}{M} \int x \, dm = \frac{1}{M} \int_0^a \int_0^b x \left( \sigma_0 \frac{x}{ab} \right) dy \, dx \] \[ x_{cm} = \frac{\sigma_0}{abM} \int_0^a x^2 \, dx \int_0^b dy = \frac{\sigma_0 b}{abM} \left( \frac{a^3}{3} \right) \] \[ x_{cm} = \frac{\sigma_0 a^2}{3M} \] Substitute \( M = \frac{\sigma_0 a}{2} \): \[ x_{cm} = \frac{\sigma_0 a^2}{3(\sigma_0 a/2)} = \frac{2a}{3} \]

Step 4: Find the y-coordinate of the center of mass.
Since the mass distribution varies only along the x-direction, the y-distribution is uniform. Hence, the y-coordinate of the center of mass is simply the midpoint of the height: \[ y_{cm} = \frac{b}{2} \]

Step 5: Final coordinates of center of mass.
\[ (x_{cm}, y_{cm}) = \left( \frac{2a}{3}, \frac{b}{2} \right) \]

Final Answer:
\[ \boxed{\left( \frac{2a}{3}, \frac{b}{2} \right)} \]