## Center of gravity of unusual circle

### Task number: 1519

Determine position of the center of gravity of the planar object on the picture below.

#### Hint 1

Think about what does mean task “determine position of the center of gravity”. What should be outcome of the task, what exactly determine?

#### Hint 2

Suitably place object in coordinate system. Is possible choose its position so next calculations are simpler, or some isn’t need at all?

#### Description of method

Position vector

\[\vec{r}_T\,=\,\frac{\int{\vec{r}}dm}{{\int}dm}\]*r*_{T}describing position of the center of gravity we usually calculate using integral:Shape of examined object is atypical so gain

*x*-coordinate of the center of gravity using integration is – gently said – enormously difficult. Therefor we try to solve this problem other way below (we restrict to planar problem). Two different perspectives on the same complex are going to help us.**Situation 1:**Imagine for a while that we fill back in small cut circle. Center of gravity T_{1}of completed large circle with mass*m*is in its center, so its coordinate is*x*_{T1}=*R*.**Situation 2:**Now completed circle divide again on:- “bitten” object with mass
*m*_{1}of which center of gravity has unknown coordinate*x*_{T } - small circle with mass
*m*_{2}of which center of gravity T_{2}has coordinate*x*_{T2}= 3*R*/2

Because we describe the same examined object once as a unit and once as system of two parts, it’s obvious that

**center of gravity of the system “bitten” object + small circle has to be in the center of gravity of the completed circle**. This idea will be starting point during next calculations.- “bitten” object with mass
#### Hint 3

Transform idea written bold in previous section “Description of method” into mathematical form. Substitute known coordinates

*x*_{T1}and*x*_{T2}and express unknown*x*_{T}.#### Hint 4

Using areal density we calculate masses

*m*_{1}and*m*_{2}. Expect object is homogenous so its areal density*ρ*is the same everywhere.#### Hint 5

Substitute masses from equations (5) and (6) into equation (4) and express unknown coordinate

*x*_{T}.#### COMPLETE SOLUTION

Determine position of the center of gravity means determine coordinates of the center of gravity in some suitably chosen coordinate system. Because assigned object is planar (the third dimension, “height”, we omit), two coordinates of the center of gravity T [

*x*_{T},*y*_{T}] are outcome of the task. Picture below shows advantageous position of the object in coordinate system – left edge of the object goes through the origin of coordinates.Because object placed like this is axial symmetric by

\[y_T\,=\,0\]*x*axis its center of gravity has to lie on this axis – any point on*x*axis has*y*-coordinate zero so we can immediately write:It’s enough calculate only

\[\vec{r}_T\,=\,\frac{\int{\vec{r}}dm}{{\int}dm}\]*x*-coordinate of the center of gravity*x*_{T}onward. We usually calculate this using integral:where

*r*_{T}is position vector describing position of the center of gravity. Shape of examined object is atypical so gain*x*-coordinate of the center of gravity using integration is – gently said – enormously difficult. Therefor we try to solve this problem other way below (we restrict to planar problem). Two different perspectives on the same complex are going to help us:**Situation 1:**Imagine for a while that we fill back in small cut circle. Center of gravity T_{1}of completed large circle with mass*m*is in its center, so its coordinate is*x*_{T1}=*R*.**Situation 2:**Now completed circle divide again on:- “bitten” object with mass
*m*_{1}of which center of gravity has unknown coordinate*x*_{T } - small circle with mass
*m*_{2}of which center of gravity T_{2}has coordinate*x*_{T2}= 3*R*/2

Because we describe the same examined object once as a unit and once as system of two parts, it’s obvious that

\[x_{T1}\,=\,\frac{m_1x_T\,+\,m_2x_{T2}}{m_1\,+\,m_2}\tag{1}\]**center of gravity of the system “bitten” object + small circle has to be in the center of gravity of the completed circle**. This idea could be expressed in this mathematical form:where right side of equation is based on formula for calculation position center of gravity of the system mass points or bodies – in detail in task Těžiště různě uspořádaných soustav koulí.

Because we know coordinates

\[R\,=\,\frac{m_1x_T\,+\,m_2{\cdot}\frac{3R}{2}}{m_1\,+\,m_2}\tag{2}\] \[R(m_1\,+\,m_2)\,-\,m_2\frac{3R}{2}\,=\,m_1x_T\tag{3}\] \[x_T\,=\,R(1\,-\,\frac{m_2}{2m_1})\tag{4}\]*x*_{T1}and*x*_{T2}(see “Description of method”), we can substitute them into expression (1) and express*x*_{T}:We determine mass of the small circle

\[m_2\,=\,{\pi}\frac{R^2}{4}\varrho\tag{5}\]*m*_{2}as product of small circle’s area and areal density:We can determine mass of “bitten” object

\[m_1\,=\,m\,-\,m_2\,=\,{\pi}R^2\varrho\,-\,{\pi}\frac{R^2}{4}\varrho\,=\,\frac{3}{4}{\pi}R^2\varrho\tag{6}\]*m*_{1}as difference between mass of completed circle*m*and mass of the small circle*m*_{2}:In the end substitute from equations (5) and (6) into equation (4):

\[x_T\,=\,R(1\,-\,\frac{m_2}{2m_1})\,=\,R(1\,-\,\frac{{\pi}\frac{R^2}{4}\varrho}{2{\cdot}\frac{3}{4}{\pi}R^2\varrho})\,=\,\frac{5}{6}R\tag{7}\]Coordinate of the center of gravity “bitten” object

*x*_{T}is equal 5*R*/6.- “bitten” object with mass
#### ALTERNATIVE SOLUTION

Another method of solving is based on knowledge of work with moments. Using fact that if have (in the plain) forces acting on system of mass points or planar objects, we can determine moments to any point of plain.

**These forces have the same tendency to rotate the system as their resultant force acting in the center of gravity of the system.**Determine individual moments at the start:**Determination of moment of gravity force – consider object as unit:**Origin of gravity force

*F*_{G}is in the center of gravity T examined object – lever-arm of gravity force is equal to unknown quantity*x*_{T}(see picture):For moment of gravity force’s magnitude

\[M_1\,=\,F_Gx_T\,=\,mgx_T\,=\,{\varrho}Sgx_T\tag{8}\]*F*_{G}to origin of coordinates stand:where

*ρ*is areal density of object,*S*is its area and*g*is gravity acceleration.We can determine area of the object according to its geometrical quality as:

\[S\,=\,{\pi}R^2\,-\,{\pi}(\frac{R}{2})^2\,=\,\frac{3}{4}{\pi}R^2\tag{9}\]By substitute from equation (9) into (8) we gain:

\[M_1\,=\,\frac{3}{4}{\pi}R^2{\varrho}g{\cdot}x_T\tag{10}\]**Determine moments – object divide in two parts:**Now we consider large circle without any cuts. Gravity force

\[M_2\,=\,F_{G1}R\] Because by completion large circle with cut part we increase its mass, we’re considering effective mass of small circle negative – so we gain final mass of the object. (As if small circle lightens large completed circle.) Effective gravity force on small circle*F*_{G1}acting this circle and its origin is in the center of circle and for moment of gravity force*F*_{G1}to origin of coordinates stand:*F*_{G2}is heading upright and its origin is in the center of small circle. For moment of gravity force*F*_{G2}to origin of coordinates stand: \[M_3\,=\,F_{G2}{\cdot}\frac{3R}{2}\]Lever-arms are easy to see in the picture:

We gain resultant moment of forces

\[M\,=\,M_2\,-\,M_3\,=\,R(F_{G1}\,-\,\frac{3}{2}F_{G2})\tag{11}\]*F*_{G1}and*F*_{G2}as difference partial moments (difference because forces have opposite tendency to rotate the system). For resultant moment’s magnitude*M*stand:Now adjust equation (11) using areal density

\[M\,=\,Rg(m_1\,-\,\frac{3}{2}m_2)\,=\,Rg{\varrho}(S_1\,-\,\frac{3}{2}S_2)\tag{12}\]*ρ*similarly as equation (1):where

\[M\,=\,Rg{\varrho}({\pi}R^2\,-\,\frac{3}{2}{\pi}({\frac{R}{2}})^2)\,=\,\frac{5}{8}R^3g{\varrho}{\pi}\tag{13}\]*S*_{1}is area of large circle and*S*_{2}area of small circle. Considering geometry of the object we can write:**Comparison above calculate moments:**According to base idea of our method stand:

\[M_1\,=\,M\tag{14}\]Substitute into equation (14) from equations (10) and (13) and express

\[\frac{3}{4}{\pi}R^2{\varrho}g{\cdot}x_T\,=\,\frac{5}{8}R^3g{\varrho}{\pi}\] \[\frac{3}{4}x_T\,=\,\frac{5}{8}R\] \[x_T\,=\,\frac{20}{24}R\,=\,\frac{5}{6}R\]*x*_{T}:Coordinate of the center of gravity

*x*_{T}is equal 5*R*/6.#### Answer

Coordinates of the center of gravity of assigned object in by us chosen coordinate system are:

\[x_T\,=\,\frac{5}{6}R\] \[y_T\,=\,0\]