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General Physics I:

Section 4.12 Examples E

Center of MassM.

Example 4.12.1.

Find the location of the center of mass of the point masses tabulated below:
Mass (kg) x-coordinate (m) y-coordinate (m)
1 2 3
4 -5 6
7 8 -9
Solution.
\begin{align*} x_{cm} \amp = \frac{\sum\limits_{i=1}^{n}{m_{i}x_{i}}}{\sum\limits_{i=1}^{n}{m_{i}}} = \frac{m_{1}x_{1}+m_{2}x_{2}+m_{3}x_{3}}{m_{1}+m_{2}+m_{3}} \\ \amp = \frac{(1\times 2) +(4\times -5) + (7\times 8)}{1+4+7} = 3.17 m \end{align*}
\begin{align*} y_{cm} \amp = \frac{\sum\limits_{i=1}^{n}{m_{i}y_{i}}}{\sum\limits_{i=1}^{n}{m_{i}}} = \frac{m_{1}y_{1}+m_{2}y_{2}+m_{3}y_{3}}{m_{1}+m_{2}+m_{3}} \\ \amp = \frac{(1\times 3) +(4\times 6) + (7\times -9)}{1+4+7} = -3.00 \,m \end{align*}

Example 4.12.2.

Find the center of mass of the point masses illustrated below:
Solution.
1. Find the coordinates of each mass and specify those coordinate pairs beside each respective mass.
2. Use the center of mass equations to calculate the coordinates of the center of mass.
\(\therefore\) the center of mass lies at point \(D(1.75\,m,4.87\,m).\)

Example 4.12.3.

Find the center of mass of the uniform sheet illustrated at right:
Solution.
1. Break the body down into its two constituent parts.
2. Specify the area of each constituent part.
\((8m \times 10\,m = 80 \,m^{2}\) and \(6\,m \times 4\,m = 24 \,m^{2})\)
3. Locate the coordinates of the center of mass of each constituent part. The center of mass of big surface of surface area \(80 \,m^{2}\) is at (4,5) and that for small surface of surface area \(24 \,m^{2}\) is at (11,2). (Note: For uniform object geometrical center is taken as its center of mass.)
4. Utilize the equations shown above.
Answer.
Note: mass is proportional to area for a uniform surface:
\begin{equation*} x_{cm} =\frac{80\times 4 + 24\times 11}{80+24}=5.62\,m; \end{equation*}
\begin{equation*} y_{cm} = \frac{80\times 5 + 24\times 2}{80+24}=4.31\,m. \end{equation*}

Example 4.12.4.

Find the x-coordinate of a center of mass of a uniform disk of radius 2R from which the mass of a small disk of radius R is missing as shown in given figure.
Solution.
We know that the center of mass of full disk of radius \(2R\) lies at its center (0,0).
\begin{equation*} \therefore\quad x_{f}= \frac{m_{s}x_{s}+m_{r}x_{r}}{m_{s}+m_{r}} =0 \end{equation*}
here \(x_{f}, \quad x_{s},\quad x_{r}\) stand for the x-coordinates of a center of mass of the full disk, small disk, and real disk, and \(m_{s}, \quad m_{r}\) stand for the mass of a small disk which is missing, and the mass of the real disk.
\begin{equation*} \text{or,}\quad x_{r}= -\frac{m_{s}x_{s}}{m_{r}} =-\frac{(\rho \pi R^{2})(-R)}{\rho[\pi(2R)^{2}-\pi R^{2}]}=\frac{R}{3} \end{equation*}
Note: mass is proportional to area for a uniform surface. \(\rho\) is a density of the disk.

Example 4.12.5.

Find the x-coordinate of the center of mass of a rod of length L whose mass per unit length varies as \(\lambda = \alpha x\text{.}\)
Solution.
Consider a rod of length \(L\) of mass \(M\) and has mass per unit length as \(\lambda =\frac{\,dm}{\,dl}= \alpha x \)
\begin{equation*} \text{or,}\quad \,dm = \lambda \,dl = \lambda \,dx \quad (say) \end{equation*}
\begin{equation*} \therefore\quad x_{cm}= \frac{1}{M}\int x\,dm =\frac{1}{M}\int\limits_{0}^{L} x\lambda\,dx =\frac{1}{M}\int\limits_{0}^{L} x(\alpha x)\,dx \end{equation*}
\begin{equation*} = \frac{\alpha}{M}\int\limits_{0}^{L} x^{2}\,dx = \frac{\alpha}{M}\left(\frac{x^{3}}{3}\right)_{0}^{L}=\frac{\alpha L^{3}}{3M} \end{equation*}

Example 4.12.6.

Three bricks, each 2 inch thick by 3 inch wide by 6 inch long, are placed one on top of the other overhanging the edge of a table. What is the maximum distance \(x\) that the top brick can extend beyond the edge of the table without falling off the table?
Solution.
Block(s) can be balanced at the edge of anything if its center of mass lies before edge or at least at the edge of that thing.
Maximum overhung = center of mass of blocks system.
Maximum overhung for blue (I) brick = its center of mass from the free end =\(\frac{l}{2} \text{.}\)
Center of mass for red (II) and blue (I) bricks
\begin{equation*} = \left\{\frac{m\frac{l}{2}+m\left(\frac{l}{2}+\frac{l}{2}\right)}{2m}\right\} =\frac{3l}{4}. \end{equation*}
Center of mass for green (III), red (II) and blue (I) bricks
\begin{equation*} = \left\{\frac{2m\frac{3l}{4}+m\left(\frac{3l}{4}+\frac{l}{2}\right)}{3m}\right\} =\frac{11l}{12}. \end{equation*}
Which is the maximum overhung for three bricks system. Using torque equation Section 4.11, we can get same condition, if we set \(x_{n}(nM) = x_{1}M\text{.}\) Where \(x_{1}\) is C.M. of first object, and \(n\) stands for number of objects. Maximum overhung,
\begin{equation*} x= x_{1}+x_{2}+x_{3}+\cdots. \end{equation*}
For example, in three books system like above: we have for first book
\begin{equation*} x_{1} =\frac{l}{2} \end{equation*}
for second book
\begin{equation*} x_{2}(2M) =\frac{l}{2}M \end{equation*}
\begin{equation*} \therefore x_{2} =\frac{l}{4} \end{equation*}
for third book
\begin{equation*} x_{3}(3M) =\frac{l}{2}M \end{equation*}
\begin{equation*} \therefore x_{3} =\frac{l}{6} \end{equation*}
Hence, maximum overhung,
\begin{equation*} x=\frac{l}{2} + \frac{l}{4} +\frac{l}{6} = \frac{11l}{12} \end{equation*}
The general rule to find the maximum overhung,x for a uniform object is
\begin{equation*} x= l\sum_{n=1}^{n}\left(\frac{1}{2n}\right) \end{equation*}

Example 4.12.7.

A solid sphere is cut into half and stacked on top of each other as shown in figure:
Find its center of mass.
Solution.
The center of mass of a hemisphere of radius \(a\) is given by
\begin{equation*} z_{1} = \frac{\int\limits_{0}^{M} z\,dm}{\int\limits_{0}^{M} \,dm} =\frac{1}{M}\int\limits_{0}^{a} z(\rho \pi r^{2}\,dz) \end{equation*}
\begin{equation*} = \frac{1}{\frac{2}{3}\pi a^{3}\rho}\int\limits_{0}^{a} z(\rho \pi r^{2})\,dz = \frac{3}{2a^{3}}\int\limits_{0}^{a} r^{2} z\,dz \end{equation*}
here \(r\) is the radius of a slice at height \(z\) and thickness \(\,dz\text{.}\) Since \(z^{2}=a^{2}-r^{2} \) we have-
\begin{equation*} 2z\,dz = -2r\,dr \quad \therefore \quad z\,dz=-r\,dr \end{equation*}
\begin{equation*} \therefore \quad z_{1} =-\frac{3}{2a^{3}}\int\limits_{a}^{0} r^{3} \,dr = -\frac{3}{2a^{3}}\left[\frac{a^{4}}{4}\right]_{a}^{0} \end{equation*}
\begin{equation*} = \frac{3}{2a^{3}}\left[\frac{a^{4}}{4}\right] = \frac{3}{8}a \end{equation*}
Hence, the center of mass of the two stacked hemispheres is given by
\begin{equation*} z_{cm} = \frac{m_{1}z_{1}+{m_{2}z_{2}}}{m_{1}+m_{2}} \end{equation*}
\begin{equation*} =\frac{\frac{M}{2}(\frac{3}{8}a)+{\frac{M}{2}(a+\frac{3}{8}a})}{M}=\frac{7}{8}a \end{equation*}

Rotational Equilibrium.

Example 4.12.8.

Write the torque equation about the specified axis for the forces indicated in the figure:
Solution.
\begin{equation*} \sum \Gamma = -(10N)(2m) +10N(x) -F_{1}(1m)-F_{2}y_{1}=0 \end{equation*}

Example 4.12.9.

Write the torque equation about the specified axis for the forces indicated below:
Solution.
\begin{equation*} \sum \Gamma = +f(4m) -N(3m) +w(0)-(T\sin\theta)(3m)+(T\cos\theta )(4m)=0 \end{equation*}

Example 4.12.10.

A symmetrical beam of weight \(w_{1} = 400 \,N\) and length \(2 \,m\) is suspended from the ceiling on one end by a hinge and on the other by a wire also of length \(2 \,m\) so that both beam and wire make an angle of \(30^{o}\) with the ceiling as shown in the figure below. Suspended from the junction of wire and beam is a weight \(w_{2}= 300\, N\text{.}\) What is the tension in the cable and the horizontal and vertical components of the force on the beam at the hinge?
Solution.
  1. List given quantities \(w_{1} = 400 \,N, \quad w_{2} = 400\, N, \quad\theta = 30^{o}, \quad L = 2 \,m\text{.}\)
  2. Draw the forces acting on the beam and resolve them into components.
  3. Choose an axis and specify the perpendicular distances from each line of action to that axis.
  4. Write equations:
    \begin{equation*} \sum F_{x} = T \cos\theta - F_{x} = 0 \end{equation*}
    and
    \begin{equation*} \sum F_{y} = F_{y}+T \sin\theta - w_{1}-w_{2} = 0 \end{equation*}
    \begin{align*} \sum \Gamma \amp = -w_{1}\frac{(L\cos\theta)}{2}- w_{2}(L\cos\theta)+(T\cos\theta)(L\sin\theta) \\ \amp + (T\sin\theta)(L\cos\theta) = 0 \end{align*}
  5. Solve to get answers:
    \begin{equation*} T=\frac{(\frac{w_{1}}{2}+w_{2})L\cos\theta}{2L\sin\theta\cos\theta} =500\,N; \end{equation*}
    \begin{equation*} F_{x}=T\cos\theta =433\,N; \end{equation*}
    \begin{equation*} F_{y}=w_{1}+w_{2}-T\sin\theta =450\,N \end{equation*}

Example 4.12.11.

A uniform rod of length 1 meter and of weight 10 Newtons is resting against a vertical wall with its base 0.5 meters from the bottom of the wall. If a 20 Newton weight is suspended from the rod from a point 3/4 of the way up the rod, what is the minimum coefficient of friction that will keep the rod from slipping on the floor? Assume there is no friction between the rod and the vertical wall.
Solution.
Given: \(L=1\,m,\quad w_{1}=10\,N, \quad w_{2}=20\,N, \quad x=0.5\,m,\)
\begin{equation*} \frac{S}{L} =\frac{3}{4}; \end{equation*}
\begin{equation*} \cos\theta =\frac{x}{L} = \frac{0.5\,m}{1\,m} =0.5 \end{equation*}
\begin{equation*} \therefore\quad \theta=60^{o} \end{equation*}
\begin{equation*} \sum F_{x} = f-N_{2}=0 \end{equation*}
\begin{equation*} \text{and}\quad \sum F_{y} = N_{1}-w_{1}-w_{2}=0 \end{equation*}
\begin{equation*} \sum \Gamma = f\cdot 0+N_{2}L\sin\theta - N_{1}\cdot 0-w_{1}\cdot \frac{x}{2} -w_{2}\left(\frac{S}{L}\right)\cos\theta=0 \end{equation*}
\begin{equation*} \text{but}\quad f=\mu N_{1} \end{equation*}
\begin{equation*} \text{hence,}\quad N_{2}.1.\frac{\sqrt{3}}{2} =10(0.25)+20(\frac{3}{4})(0.5) \end{equation*}
\begin{equation*} or,\quad N_{2}=11.547 \,N \end{equation*}
\begin{equation*} \text{and}\quad f=N_{2}=11.547 N; N_{1}=w_{1}+w_{2}=10+20=30\,N \end{equation*}
\begin{equation*} \therefore\quad \mu=\frac{f}{N_{1}}=\frac{11.547\,N}{30\,N} = 0.385 \end{equation*}

Example 4.12.12.

A uniform rod of length L = 1 meter and mass \(m_{1} = 10 \,kg\) is suspended as shown in the figure below, making an angle of \(\theta = 30^{o}\) with the horizontal and supporting a second unknown mass \(m_{2}\text{.}\) The support wire on the left is horizontal, the support wire on the top makes an angle of \(\theta = 30^{o}\) with the rod, and the other wire on the right is vertical. The top wire is attached at a point S = 0.75 m up the rod from the left.
  1. What is the tension \(T_{1}\) in the wire on the left?
  2. What is the tension \(T_{2}\) in the wire on top?
  3. What is the tension \(T_{3}\) in the wire on the right?
  4. What is the mass \(m_{2}\text{?}\)
Solution.
Given: \(L=1\,m,\quad m_{1}=10 \,kg,\quad \theta=30^{o},\quad S =0.75\,m \)
\begin{equation} \sum F_{x} = T_{2}\sin\phi-T_{1}=0;\tag{4.12.1} \end{equation}
\begin{equation*} \text{and}\quad \sum F_{y} = T_{2}\cos\phi-w_{1}-T_{3}=0 \end{equation*}
\begin{equation} \therefore \quad T_{3}=T_{2}\cos\phi-w_{1}\tag{4.12.2} \end{equation}
\begin{equation} \sum \Gamma = T_{2}S\sin\theta - w_{1}\left(\frac{L}{2}\right)\cos\theta-T_{3}L\cos\theta=0\tag{4.12.3} \end{equation}
\begin{equation} \text{also}\quad \sum F_{y} = T_{3}-w_{2}=0;\tag{4.12.4} \end{equation}
\begin{equation*} \text{and}\quad w_{1} = m_{1}g =98N, \qquad w_{2} = m_{2}g. \end{equation*}
from eqn. (4.12.3)
\begin{equation*} T_{2}S\sin\theta -T_{3}L\cos\theta = w_{1}\left(\frac{L}{2}\right)\cos\theta \end{equation*}
\begin{equation*} T_{2}S\sin\theta -\left(T_{2}\cos\phi-w_{1}\right)L\cos\theta = w_{1}\left(\frac{L}{2}\right)\cos\theta \end{equation*}
\begin{equation*} T_{2}S\sin\theta-T_{2}\cos\theta L\cos\theta=w_{1}\left(\frac{L}{2}\right)\cos\theta-w_{1}L\cos\theta \end{equation*}
\begin{equation*} T_{2}=\left\{\frac{-0.5\times w_{1}\times L\cos\theta}{S\sin\theta-L\cos^{2}\theta}\right\}=113.16N \end{equation*}
\begin{equation*} [\because cos\phi=sin2\theta=cos\theta] \quad at \theta=30^o \end{equation*}
Hence, from eqn. (4.12.2)
\begin{equation*} T_{3}=0, \end{equation*}
from eqn. (4.12.1), we have -
\begin{equation*} T_{1}=T_{2}\sin\phi= T_{2}\sin\theta=56.58\,N, \end{equation*}
\begin{equation*} w_{2}=T_{3}=0, \end{equation*}
\begin{equation*} m_{2}=\frac{w_{2}}{g}=0 \end{equation*}

Example 4.12.13.

A block with a square base measuring \(a \times a\) and height \(h\) is placed on an inclined plane of angle \(\theta\text{.}\) The coefficient of friction is \(\mu \) and the angle of the incline is gradually increased. Find the condition at which block tip over before it begins to slide.
Solution.
The block will not slide if the coefficient of friction \(\mu_{s} \lt \tan\theta\text{.}\) In other words, sliding begins only when \(\theta\) becomes the angle of repose \(\phi\text{.}\) The block will toppled down once the line of action of weight pass throgh the edge of the block, i.e. when normal force acts along the edge (point of rotation) of block as shown in figure below.
From first figure, we have -
\begin{equation*} f_{s}=mg\sin\theta \end{equation*}
\begin{equation*} \text{and}\quad N=mg\cos\theta \end{equation*}
Divide these equations and substitute \(f_{s}=\mu_{s}N\text{,}\) we get-
\begin{equation} \tan\theta = \mu_{s} \tag{4.12.5} \end{equation}
If \(\theta \) is gradually increasing to \(\theta=\phi \) [second figure], then at the instant of toppling, normal force acts along the bottom edge of block which is the point of rotation. Hence the total torque must be zero (i.e., \(\Gamma=0\) at rotational equilibrium).
\begin{equation*} \therefore \quad mg\sin\theta \left(\frac{h}{2}\right) -mg\cos\theta \left(\frac{a}{2}\right) =0 \end{equation*}
\begin{equation*} \text{or,}\quad mg\sin\theta \left(\frac{h}{2}\right) =mg\cos\theta \left(\frac{a}{2}\right) \end{equation*}
\begin{equation} \tan\theta = \frac{a}{h} \tag{4.12.6} \end{equation}
Now, from eqns. (4.12.5) and (4.12.6), if \(\mu_{s} \gt \frac{a}{h}\text{,}\) it will tip over before slide, and if \(\mu_{s} \lt \frac{a}{h}\text{,}\) it will slide before toppling.

Example 4.12.14.

A uniform 10m beam of mass 300kg extends over a ledge. The beam is not attached, but simply rests on the surface. A 60kg man intends to position the beam in such a way that he can walk to the end of it. What is the maximum distance the beam can extend past end of the ledge and still allow him to perform this feat?
Solution.
The beam can be balanced if it is placed on the edge of ledge by its new center of mass, \(P\) (i.e., the center of mass of beam and man system). Here \(o\) is the center of mass of a beam only.
Let \(x\) be the distance of beam overhung past the edge of ledge, then from free -body diagram, we have -
\begin{equation*} \sum F_{y} = N-Mg-mg =0 \end{equation*}
\begin{equation*} \therefore \quad N = (M+m)g \end{equation*}
Since the beam is at static equilibrium. Now,
\begin{equation*} \sum \Gamma = Mg\left(\frac{l}{2}-x\right) -mgx =0 \end{equation*}
This is a total torque about the new center of mass, \(P\) (or the point of rotation).
\begin{equation*} \text{or,}\quad \frac{Ml}{2}-Mx =mx \end{equation*}
\begin{equation*} \therefore \quad x= \frac{Ml}{2(M+m)} = \frac{300\times 10}{2\times(300+60)} =4.17\,m \end{equation*}
Alternative: we can solve this problem using center of mass formula,
\begin{equation*} x_{cm} = \frac{m_{1}x_{1}+m_{2}x_{2}}{m_{1}+m_{2}} = \frac{300\times 5 + 60\times 10}{300+60} = 5.83\,m \end{equation*}
\begin{equation*} \therefore \quad x= l-x_{cm} =10-5.83 = 4.17\,m \end{equation*}