5. Mass and Center of Mass
Exercises
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A \(6\) cm rod has linear density given by \(\delta(x)=(2x+1)\,\dfrac{\text{gm}}{\text{cm}}\) where \(x\) is measured from one end.
\(M=42\) gm
The mass is \[\begin{aligned} M&=\int_0^6 \delta(x)\,dx=\int_0^6 (2x+1)\,dx \\ &=\left[\dfrac{}{}x^2+x\right]_0^6=36+6=42 \,\text{gm} \end{aligned}\]
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A \(4\) cm rod has linear density given by \(\delta(x)=(5x^3-3x+2)\,\dfrac{\text{gm}}{\text{cm}}\) where \(x\) is measured from one end.
\(M=304\,\text{gm}\)
The mass is \[\begin{aligned} M&=\int_0^4 \delta(x)\,dx=\int_0^4 (5x^3-3x+2)\,dx \\ &=\left[\dfrac{5x^4}{4}-\dfrac{3x^2}{2}+2x\right]_0^4 =320-24+8=304\,\text{gm} \end{aligned}\]
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A chocolate bar which is \(6\) in long has linear density given by \(\delta(x)=\ln(x+2)\,\dfrac{\text{oz}}{\text{in}}\) where \(x\) is measured from one end.
Recall: \(\displaystyle \int \ln u\,du=u\ln u-u+C\).
\(M=(22\ln 2-6)\,\text{oz}\approx 9.25\,\text{oz}\)
The mass is \[ M=\int_0^6 \ln(x+2)\,dx \] To use the formula \(\displaystyle \int \ln u\,du=u\ln u-u+C\), we make the change of variables \(u=x+2\) and \(du=dx\). For the limits, when \(x=0\), we have \(u=2\) and when \(x=6\), we have \(u=8\). So \[\begin{aligned} M&=\int_2^8 \ln u\,du=\left[\dfrac{}{}u\ln u-u\right]_2^8 \\ &=(8\ln8-8)-(2\ln2-2) \\ &=(8\cdot3\ln2-8)-(2\ln2-2) \\ &=(22\ln2-6)\,\text{oz} \approx 9.25\,\text{oz} \end{aligned}\]
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A bar which is \(5\) in long has linear density given by \(\delta(x)=xe^x\) \(\dfrac{\text{oz}}{\text{in}}\) where \(x\) is measured from one end.
Use integration by parts.
\(M=(4e^5+1)\,\text{oz}\approx 595.\,\text{oz}\)
The mass is \[ M=\int_0^5 \delta(x)\,dx=\int_0^5 xe^x\,dx \] We use integration by parts with \[\begin{array}{ll} u=x & dv=e^x\,dx \\ du=dx \quad & v=e^x \end{array}\] So \[\begin{aligned} M&=\left[xe^x-\int e^x\,dx\right]_0^5 =\left[\dfrac{}{}xe^x-e^x\right]_0^5 \\ &=(4e^5+1)\,\text{oz} \approx 595.\,\text{oz} \end{aligned}\]
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A \(6\) cm rod has linear density given by \(\delta(x)=\left(\dfrac{1}{8}+\dfrac{2}{(x+4)^2}\right) \,\dfrac{\text{gm}}{\text{cm}}\) where \(x\) is measured from one end.
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\(M=\dfrac{21}{20}\,\text{gm}\)
The mass is \[ M=\int_0^6 \delta(x)\,dx=\int_0^6\left(\dfrac{1}{8}+\dfrac{2}{(x+4)^2}\right)\,dx \] Using substitution with \(u=x+4\) and \(du=dx\). For the limits, when \(x=0\), we have \(u=4\) and when \(x=6\), we have \(u=10\). So \[ \begin{aligned} M&=\int_4^{10}\left(\dfrac{1}{8}+\dfrac{2}{u^2}\right)\,du =\left[\dfrac{1}{8}u-\dfrac{2}{u}\right]_4^{10}\\ &=\left(\dfrac{1}{8}10-\dfrac{2}{10}\right)-\left(\dfrac{1}{8}4-\dfrac{2}{4}\right)\\ &=\dfrac{5}{4}-\dfrac{1}{5}-\dfrac{1}{2}+\dfrac{1}{2}\\ &=\dfrac{25}{20}-\dfrac{4}{20}=\dfrac{21}{20}\,\text{gm} \end{aligned} \]
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Find the total mass and the center of mass for four discrete masses lying on the \(x\)-axis where the masses and positions indicated below are in grams and centimeters. \[\begin{aligned} m_1&=7 &m_2&=5 &m_3&=2 &m_4&=6 \\ x_1&=-3 &x_2&=-2 &x_3&=4 &x_4&=7 \end{aligned}\]
\(M=20\) gm
\(\bar{x}=\dfrac{19}{20}\,\text{cm}=.95\,\text{cm}\)The total mass is: \[\begin{aligned} M&=m_1+m_2+m_3+m_4 \\ &=7+5+2+6=20\,\text{gm} \end{aligned}\] The first moment is: \[\begin{aligned} M_1&=m_1x_1+m_2x_2+m_3x_3+m_4x_4 \\ &=7(-3)+5(-2)+2(4)+6(7)=19\,\text{gm-cm} \end{aligned}\]
So the center of mass is at: \[ \bar{x}=\dfrac{M_1}{M}=\dfrac{19}{20}\,\text{cm} =.95\,\text{cm} \] The plot shows the location of the center of mass.
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A \(6\) cm rod has linear density \(\delta(x)=(2x+1)\,\dfrac{\text{gm}}{\text{cm}}\) where \(x\) is measured from one end. The mass was found in previous exercise.
\(\bar{x}=\dfrac{27}{7}\,\text{cm}\approx3.86\,\text{cm}\)
From a previous exercise we know the mass is \[ M=42 \] The first moment is \[\begin{aligned} M_1&=\int_0^6 x\delta(x)\,dx=\int_0^6 x(2x+1)\,dx \\ &=\left[\dfrac{2x^3}{3}+\dfrac{x^2}{2}\right]_0^6 =144+18=162\,\text{gm-cm} \end{aligned}\]
So the center of mass is at: \[ \bar{x}=\dfrac{M_1}{M}=\dfrac{162}{42}=\dfrac{27}{7}\,\text{cm} \approx3.86\,\text{cm} \] The plot shows the location of the center of mass.
The density \(\delta(x)=(2x+1)\) is bigger on the right and \(\bar{x}=\dfrac{27}{7}\,\text{cm}\approx3.86\,\text{cm}\) is inside the bar but right of the middle.
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Find the center of mass of a \(4\) cm rod has linear density \(\delta(x)=(5x^3-3x+2)\,\dfrac{\text{gm}}{\text{cm}}\) where \(x\) is measured from one end. The mass was found in previous exercise.
\(\bar{x}=\dfrac{61}{3}\approx 3.21\) gm
From a previous exercise we know the mass is \[ M=304 \] The first moment is \[\begin{aligned} M_1&=\int_0^4 x\delta(x)\,dx=\int_0^4 x(5x^3-3x+2)\,dx \\ &=\left[\dfrac{}{}x^5-x^3+x^2\right]_0^4 \\ &=1024-64+16=976\,\text{gm-cm} \end{aligned}\]
So the center of mass is at: \[ \bar{x}=\dfrac{M_1}{M}=\dfrac{976}{304}=\dfrac{61}{19}\approx 3.21\,\text{cm} \] The plot shows the location of the center of mass.
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Find the mass and the center of mass of a \(30\) cm rod that has linear density \(\delta(x)=\dfrac{1}{2}x^2\) \(\dfrac{\text{gm}}{\text{cm}}\) where \(x\) is measured from one end.
\(M=4500\) gm
\(\bar{x}=\dfrac{45}{2}\,\text{cm}=22.5\,\text{cm}\)The mass is: \[\begin{aligned} M&=\int_0^{30} \delta(x)\,dx=\int_0^{30} \dfrac{1}{2}x^2\,dx \\ &=\left[\dfrac{1}{6}x^3\right]_0^{30} =\dfrac{30^3}{6} =4500 \,\text{gm} \end{aligned}\] The first moment is: \[\begin{aligned} M_1&=\int_0^{30} x\delta(x)\,dx=\int_0^{30} x\dfrac{1}{2}x^2\,dx \\ &=\left[\dfrac{1}{8}x^4\right]_0^{30} =\dfrac{30^4}{8} = 101250\,\text{gm} \end{aligned}\]
So the center of mass is at: \[\begin{aligned} \bar{x}&=\dfrac{M_1}{M}=\dfrac{30^4}{8}\dfrac{6}{30^3} \\ &=\dfrac{45}{2}\,\text{cm} =22.5\,\text{cm} \end{aligned}\] The plot shows the location of the center of mass.
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Find the moment of inertia for four discrete masses lying on the \(x\)-axis and rotating about \(x_0=0\), where the masses and positions indicated below are in grams and centimeters. \[\begin{aligned} m_1&=7 &m_2&=5 &m_3&=2 &m_4&=6 \\ x_1&=-3 &x_2&=-2 &x_3&=4 &x_4&=7 \end{aligned}\]
\(I=409\) gm-cm\(^2\)
Since \(x_0=0\), the moment of intertia is just the second moment. \[\begin{aligned} I&=M_2=m_1x_1^2+m_2x_2^2+m_3x_3^2+m_4x_4^2 \\ &=7(-3)^2+5(-2)^2+2(4)^2+6(7)^2=409\,\text{gm-cm}^2 \end{aligned}\]
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Find the mass, center of mass and moment of inertia of a \(6\) m bar on the \(x\)-axis between \(x=0\) and \(x=6\) rotating about \(x_0=4\),
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with uniform density \(\delta=3\) \(\dfrac{\text{kg}}{\text{m}}\).
\(M=18\,\text{kg}\)
\(\bar{x}=3\,\text{m}\)
\(I=72\,\text{kg-m}^2\)The mass is: \[\begin{aligned} M&=\int_0^6 \delta(x)\,dx=\int_0^6 3\,dx \\ &=\left[\dfrac{}{}3x\right]_0^6=18\,\text{kg} \end{aligned}\] The first moment is: \[\begin{aligned} M_1&=\int_0^6 x\delta(x)\,dx=\int_0^6 3x\,dx \\ &=\left[3\dfrac{x^2}{2}\right]_0^6=54\,\text{kg-m} \end{aligned}\] The second moment is: \[\begin{aligned} M_2&=\int_0^6 x^2\delta(x)\,dx=\int_0^6 3x^2\,dx \\ &=\left[\dfrac{}{}x^3\right]_0^6=216\,\text{kg-m}^2 \end{aligned}\] So the center of mass is at: \[ \bar{x}=\dfrac{M_1}{M}=\dfrac{54}{18}=3\,\text{m} \] Since \(x_0=4\), the moment of intertia is: \[\begin{aligned} I&=M_2-2x_0M_1+x_0^2M \\ &=216-8\cdot54+16\cdot18=72\,\text{kg-m}^2 \end{aligned}\]
Since the density is uniform, the mass, \(M=18\), is just the length times the density.
Since the density is uniform, the center of mass, \(\bar{x}=3\), is at the geomertrical center of the bar which is also called the centroid. -
with linear density \(\delta(x)=x\) \(\dfrac{\text{kg}}{\text{m}}\).
\(M=18\,\text{kg}\)
\(\bar{x}=4\,\text{m}\)
\(I=36\,\text{kg-m}^2\)The mass is: \[\begin{aligned} M&=\int_0^6 \delta(x)\,dx=\int_0^6 x\,dx \\ &=\left[\dfrac{x^2}{2}\right]_0^6=18\,\text{kg} \end{aligned}\] The first moment is: \[\begin{aligned} M_1&=\int_0^6 x\delta(x)\,dx=\int_0^6 x^2\,dx \\ &=\left[\dfrac{x^3}{3}\right]_0^6=72\,\text{kg-m} \end{aligned}\] The second moment is: \[\begin{aligned} M_2&=\int_0^6 x^2\delta(x)\,dx=\int_0^6 x^3\,dx \\ &=\left[\dfrac{x^4}{4}\right]_0^6=324\,\text{kg-m}^2 \end{aligned}\] So the center of mass is at: \[ \bar{x}=\dfrac{M_1}{M}=\dfrac{72}{18}=4\,\text{m} \] Since \(x_0=4\), the moment of intertia is: \[\begin{aligned} I&=M_2-2x_0M_1+x_0^2M \\ &=324-8\cdot72+16\cdot18=36\,\text{kg-m}^2 \end{aligned}\]
Notice that the mass of both bars is \(M=18\). Since the lengths are both \(L=6\), the average densities are the same: \[ \delta_\text{ave}=\dfrac{M}{L}=\dfrac{18}{6}=3 \] Since the non-uniform bar is heavier for \(x>3\), the center of mass, \(\bar{x}=4\), is on that side.
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If we measure \(x\) from one end of the bar, find the net charge on a \(6\) m bar whose linear charge density is:
(In each case, explain why the answer is reasonable by explaining where there is positive or negative charge.)-
\(\delta_e(x)=3-2x\) \(\dfrac{\text{coul}}{\text{m}}\)
\(Q=-18\) coul
The net charge is: \[\begin{aligned} Q&=\int_0^6 \delta_e(x)\,dx=\int_0^6 3-2x\,dx \\ &=\left[\dfrac{}{}3x-x^2\right]_0^6 =18-36=-18\,\text{coul} \end{aligned}\] There is the positive charge for \(0 \le x \lt \dfrac{3}{2}\) but negative charge for \(\dfrac{3}{2} \lt x \le 6\). So there is more negative charge.
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\(\delta_e(x)=x-3\) \(\dfrac{\text{coul}}{\text{m}}\)
\(Q=0\) coul
The net charge is: \[\begin{aligned} Q&=\int_0^6 \delta_e(x)\,dx=\int_0^6 x-3\,dx \\ &=\left[\dfrac{x^2}{2}-3x\right]_0^6 =18-18=0\,\text{coul} \end{aligned}\] There is the positive charge on half the bar and negative charge on the other half.
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\(\delta_e(x)=x-2\) \(\dfrac{\text{coul}}{\text{m}}\)
\(Q=6\) coul
The net charge is: \[\begin{aligned} Q&=\int_0^6 \delta_e(x)\,dx=\int_0^6 x-2\,dx \\ &=\left[\dfrac{x^2}{2}-2x\right]_0^6 =18-12=6\,\text{coul} \end{aligned}\] There is the negative charge for \(0 \le x \lt 2\) but positive charge for \(2 \lt x \le 6\). So there is more positive charge.
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Find the total weight of a \(12\) ft rod with linear density \(\delta(x)=x\sqrt{x^2+1}\) \(\dfrac{\text{lb}}{\text{ft}}\) if \(x\) is measured from one end.
\(M=\dfrac{1}{3}(145^{3/2}-1)\,\text{lb} \approx581.7\,\text{lb}\)
The mass is \[ M=\int_0^{12} \delta(x)\,dx=\int_0^{12} x\sqrt{x^2+1}\,dx \] We make the substitution \(u=x^2+1\) and \(du=2x\,dx\): \[\begin{aligned} M&=\dfrac{1}{2}\int_1^{145} \sqrt{u}\,du =\dfrac{1}{2}\left[\dfrac{2u^{3/2}}{3}\right]_1^{145} \\ &=\dfrac{1}{3}(145^{3/2}-1)\,\text{lb} \approx581.7\,\text{lb} \end{aligned}\]
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Find the total mass and the center of mass for a \(10\) m rod if the linear density is \(\delta(x)=e^x\) \(\dfrac{\text{kg}}{\text{m}}\) where \(x\) is measured from one end.
\(M=e^{10}-1\) kg
\(\bar{x}=\dfrac{9e^{10}+1}{e^{10}-1}\,\text{m} \approx9\,\text{m}\)The mass is: \[ M=\int_0^{10} \delta(x)\,dx=\int_0^{10} e^x\,dx =\left[\dfrac{}{}e^x\right]_0^{10}=e^{10}-1\,\text{kg} \] The first moment is: \[ M_1=\int_0^{10} x\delta(x)\,dx=\int_0^{10} xe^x\,dx \] To compute this integral, we use the parts: \[\begin{array}{ll} u=x & dv=e^x\,dx \\ du=dx \quad & v=e^x \end{array}\] So: \[\begin{aligned} M_1&=xe^x-\int e^x\,dx \\ &=\left[\dfrac{}{}xe^x-e^x\right]_0^{10}=9e^{10}+1\,\text{kg-m} \end{aligned}\] So the center of mass is: \[ \bar{x}=\dfrac{M_1}{M} =\dfrac{9e^{10}+1}{e^{10}-1}\,\text{m} \approx9\,\text{m} \]
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Find the moment of inertia of a \(2\) in bar with linear density \(\delta(x)=\sqrt{x^3+1}\) \(\dfrac{\text{oz}}{\text{in}}\) if \(x\) is measured from one end and the rod rotates about that end.
\(I=\dfrac{52}{9}\,\text{oz-in}^2\)
Since \(x_0=0\), the moment of intertia is just the second moment. \[ I=M_2=\int_0^2 x^2\delta(x)\,dx=\int_0^2 x^2\sqrt{x^3+1}\,dx \] We make the substitution \(u=x^3+1\) and \(du=3x^2\,dx\): \[\begin{aligned} I&=\dfrac{1}{3}\int_1^9 \sqrt{u}\,du =\dfrac{1}{3}\left[\dfrac{2u^{3/2}}{3}\right]_1^9 \\ &=\dfrac{2}{9}(9^{3/2}-1) =\dfrac{52}{9}\,\text{oz-in}^2 \end{aligned}\]
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Find the net charge on a rod with length \(3\pi\) cm and linear charge density \(\delta_e=\sin x\) \(\dfrac{\text{coul}}{\text{cm}}\) if \(x\) is measured from one end.
\(Q=2\) coul
The net charge is \[\begin{aligned} Q&=\int_0^{3\pi} \delta_e(x)\,dx=\int_0^{3\pi} \sin x\,dx =\left[\dfrac{}{}-\cos x\right]_0^{3\pi} \\ &=-\cos(3\pi)--\cos(0)=2\,\text{coul} \end{aligned}\]
PY: All Checked
In each problem, find the total mass.
In each problem, find the center of mass.
Review Exercises
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