Question 61: Centre of Mass and Rotational Motion

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 $\sigma=\frac{\sigma_0 x}{a b}$ (where $\sigma_0$ is a constant), would be __________.

Accepted Answer

$$\sigma  = {{{\sigma _0}x} \over {ab}}$$
Let's consider a strip of infinitesimal width dx at x distance away from O.
As $\sigma$ depends only on x i.e. it's constant for a specific value of x so centre of mass of the strip will be $$\left( {x,{b \over 2}} \right)$$.
Now, we know,
$$COM = \left( {{{\int {xdm} } \over {\int {dm} }},{{\int {ydm} } \over {\int {dm} }}} \right)$$
$$ = \left( {{{\int {x\sigma dA} } \over {\int {\sigma dA} }},{{\int {{b \over 2}\sigma dA} } \over {\int {\sigma dA} }}} \right)$$
$$ = \left( {{{\int {xbdx{{{\sigma _0}x} \over {ab}}} } \over {\int {{{{\sigma _0}x} \over {ab}}bdx} }},{{{b \over 2}\int {\sigma dA} } \over {\int {\sigma dA} }}} \right)$$
For whole plate, x varies from o to a.
So, $$ = \overleftarrow C OM = \left( {{{{{{\sigma _0}} \over a}\int_0^a {{x^2}dx} } \over {{{{\sigma _0}} \over a}\int_0^a {xdx} }},{b \over 2}} \right)$$
$$ = \left( {{{{1 \over 3}\left[ {{x^3}} \right]_0^a} \over {{1 \over 2}\left[ {{x^2}} \right]_0^a}},{b \over 2}} \right)$$
$$ = \left( {{2 \over 3}{{{a^3}} \over {{a^2}}},{b \over 2}} \right)$$
$$ \Rightarrow COM = \left( {{2 \over 3}a,{b \over 2}} \right)$$

Answer

$\left(\frac{2}{3} a, \frac{2}{3} b\right)$

Answer

$\left(\frac{2}{3} a, \frac{\mathrm{~b}}{2}\right)$

Answer

$\left(\frac{1}{3} a, \frac{\mathrm{~b}}{2}\right)$

Answer

$\left(\frac{a}{2}, \frac{\mathrm{~b}}{2}\right)$

Centre of Mass and Rotational Motion

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=\frac{\sigma_0 x}{a b} (where \sigma_0 is a constant), would be __________.

Choose the correct answer: