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 span the region $ 0 \le x \le a $ and $ 0 \le y \le b $. The mass per unit area is given by $ \sigma = \frac{\sigma_0 x}{ab} $. The center of mass $ (x_{cm}, y_{cm}) $ is determined by first calculating the total mass $ M $, and then applying the formulas:

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

Step 1: Calculate 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: Calculate $ 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: Calculate $ 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 center of mass is located at $ \left(\frac{2}{3}a, \frac{b}{2}\right) $.