Question

Learning Goal: To practice Tactics Box 7.1 Finding the center of gravity.

Even though you can locate an object's center of gravity by suspending it from a pivot, this is rarely a practical technique. More often, we would like to calculate the center of gravity of an object made up of a combination of particles. The following Tactics Box shows how to find the center of gravity of any number of particles.

• Choose an origin for your coordinate system. You can choose any convenient point as the origin.
• Determine the coordinates , , , $\ldots\,$ for the particles of mass $m_1$ , $m_2$ , $m_3$ , $\ldots\,$ respectively.
• The x coordinate of the center of gravity is

$x_{\rm cg}=\frac{x_1 m_1 + x_2 m_2 + x_3 m_3 +\cdots}{m_1+m_2+m_3 +\cdots}$ .

• Similarly, the y coordinate of the center of gravity is

$y_{\rm gc}=\frac{y_1 m_1 +y_2 m_2 + y_3 m_3 +\cdots}{m_1+m_2+m_3 +\cdots}.$

Three identical coins, labeled A, B, and C in the figure, lie on three corners of a square 10.0 $\rm cm$ on a side. Determine the x coordinate of each coin, $x_{\rm A}$ , $x_{\rm B}$ , and $x_{\rm C}$ .

Express your answers in centimeters, separated by commas.

$x_{\rm A}$ , $x_{\rm B}$ , $x_{\rm C}$ =

//

$\rm cm$ , $\rm cm$ , $\rm cm$

Determine the y coordinate of each coin described in Part A: $y_{\rm A}$ , $y_{\rm B}$ , and $y_{\rm C}$ .

Express your answers in centimeters, separated by commas.

$y_{\rm A}$ , $y_{\rm B}$ , $y_{\rm C}$ =		$\rm cm$ , $\rm cm$ , $\rm cm$

Determine the x and y coordinates $x_{\rm cg}$ and $y_{\rm cg}$ of the center of gravity of the three coins described in Part A.

Express your answer in centimeters.

$x_{\rm cg}$ , $y_{\rm cg}$ =

$\rm cm$ , $\rm cm$

Answer 1

Concept and Reason

The given problem can be solved by using the concept of 2-D geometry and expression of the center of mass.

Determine the coordinate of center of gravity of each coin by 2-D geometry and then use that coordinate to find center of gravity of whole system.

Fundamental

2-D Geometry

Any point A on two-dimensional plane can be represented by its coordinate$\left( {x,y} \right)$ in the 2-D space.

Here $x$is the distance of $x$axis of $A$from origin and $y$is the distance of $y$axis of $A$from origin.

Center of gravity of the body is the assumed point in the body, where the whole mass of the body is concentrated and net gravitational force is assumed to act.

For complex bodies center of mass can be calculated by,

$x = \frac{{{M_1}{x_1} + {M_2}{x_1} + \cdots + {M_n}{x_n}}}{{{M_1} + {M_2} \cdots + {M_n}}}$ …... (1)

$y = \frac{{{M_1}{y_1} + {M_2}{y_1}...... + {M_n}{y_n}}}{{{M_1} + {M_2}...... + {M_n}}}$ …... (2)

Here, $(x,y)$ are coordinate of center of gravity and ${M_n}({x_{1,}}{y_2})$are coordinate of center of gravity and ${M_n}$is mass of individual body.

(A)

Consider the diagram given in question.

Choose $B(0,0)$be the reference point and references axis as per the question.

Then, for the coin at $C$the distance from origin in $x - axis$is given by,

${x_C} = 10.0 {\rm{cm}}$

For coin at $B$the distance from origin in $x - axis$is given by

${x_B} = 0.0 {\rm{cm}}$

For coin at $A$the distance from origin in $x - axis$is given by

${x_A} = 0.0 {\rm{cm}}$

(B)

Choose $B(0,0)$be the reference point and references axis as per the question.

Then, for coin at $C$the distance ${y_C}$ from origin in $y - axis$is given by,

${y_C} = 0.0{\rm{cm}}$

For coin at $B$the distance ${y_B}$ from origin in $y - axis$is given by

${y_B} = 0.0{\rm{cm}}$

For coin at $A$the distance ${y_A}$ from origin in $y - axis$is given by

${y_A} = 10.0{\rm{cm}}$

(C)

Let $M$be the mass of each coin.

Then $x$coordinate of center of gravity ${x_{cg}}$ of each of the equation is be given by equation (1) as,

${x_{cg}} = \frac{{M{x_A} + M{x_B} + M{x_C}}}{{M + M + M}},$

Substitute value of $0.0{\rm{ cm}}$for ${x_A}$,$0.0{\rm{ cm}}$for ${x_B}$,$10.0{\rm{ cm}}$for ${x_C}$ in the expression of ${x_{cg}}$.

$\begin{array}{c}\\{x_{cg}} = \frac{{M\left( {0.0 {\rm{cm}}} \right) + M\left( {0.0 {\rm{cm}}} \right) + M\left( {10.0 {\rm{cm}}} \right)}}{{M + M + M}}\\\\ = \frac{{M\left( {10.0 {\rm{cm}}} \right)}}{{3M}}\\\\ = 3.33 {\rm{cm}}\\\end{array}$

Similarly, $y$coordinate of center of gravity of each of the equation is be given by equation (2) as,

${y_{cg}} = \frac{{M{y_A} + M{y_B} + M{y_C}}}{{M + M + M}},$

Substitute value of $10.0 {\rm{cm}}$for ${y_A}$, $0.0 {\rm{cm}}$ for ${y_B}$, $0.0 {\rm{cm}}$ for${y_C}$in the expression of ${y_{cg}}$.

$\begin{array}{c}\\{y_{cg}} = \frac{{M\left( {10.0 {\rm{cm}}} \right) + M\left( {0.0 {\rm{cm}}} \right) + M\left( {0.0 {\rm{cm}}} \right)}}{{M + M + M}}\\\\ = \frac{{M\left( {10.0 {\rm{cm}}} \right)}}{{3M}}\\\\ = 3.33 {\rm{cm}}\\\end{array}$

Ans: Part A

The values of ${x_A},{x_B},{x_C}$ are $0.0 {\rm{cm ,}}0.0 {\rm{cm}},10.0 {\rm{cm}}$.