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[QUE/ME-11001] ME-PROBLEMNode id: 3969pageShow that the problem of the motion of two charged particles in a uniform electrical field \(E\) can be reduced to the problem of the motion of the centre of mass and that of the motion of a particle in a given potential. |
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22-02-08 19:02:47 |
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[QUE/ME-12013] ME-PROBLEMNode id: 3984pageA mass is attached to the end of a string. The mass moves on a frictionless table, and the string passes through a hole in the table (see Figure P.1.9), under which someone is pulling on the string to make it taut at all times. \hfill \GetQSource
- Is the angular momentum of the mass conserved in this problem? Is energy of the mass conserved? Explain your answer.
- Initially, the mass moves in a circle, with kinetic energy \(E_0\). The string is then slowly pulled, until the radius of the circle is halved. How much work was done?
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22-02-08 19:02:15 |
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[QUE/ME-12012] ME-PROBLEMNode id: 3983pageFind the central force if the orbit of a particle is given by \[ r=a(1+\cos\theta). \] |
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22-02-08 19:02:39 |
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[QUE/ME-12011] ME-PROBLEMNode id: 3982pageA particle moves in two dimensions under the influence of a central force determined by the potential \(V(r)=\alpha r^p +\beta r^q\). Find the powers \(p\) and \(q\) which make it possible to achieve a spiral orbit of the form \( r=c\theta^2\), with \(c\) a constant. |
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22-02-08 19:02:12 |
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[QUE/ME-12010] ME-PROBLEMNode id: 3979pageShow that the radius of circular orbit for energy \(E\) and angular momentum \(L\) is given by \(R=\frac{L^2}{k\mu}\). |
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22-02-08 18:02:42 |
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[QUE/ME-12009] ME-PROBLEMNode id: 3978page$\newcommand{\matrixelement}[3]{\langle#1|#2|#3\rangle}$
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A particle moves in a spherically symmetric potential \[V(r)=\frac{1}{2}\mu \omega^2 + \frac{\lambda^2}{2\mu r^2}\] where \(\mu\) is the mass of the particle, and \(\omega,\lambda\) are real constants.
- Does there exist a circular orbit for zero orbital angular momentum?
- Assume \(L=0, E= \frac{25}{2}\mu\omega^2a^2\), \(\lambda=12\mu\omega a^2\) Use initial conditions \[ r(t)\big|_{t=0} = 4a;\quad \dot{r}(t)\big|_{t=0} = 0 \text{ and } \dot{\theta}(t)\big|_{t=0}=0 \] solve the equations of motion and obtain \(r\), \(\theta\) as function of time. Describe the motion that takes place under conditions specified here.
If angular momentum is zero, we have \(\mu r^2\dot{\theta}=0\) which implies that \(\dot{\theta}=0\). Thus \(\theta\) is constant and motion will be in a straight line. This can also seen by setting \(\vec{r}\times \vec{p}=0\) implies that position and momentum are always parallel (follows from conservation of angular momentum). Therefore the motion is not in a circle. Now we assume that \(\vec{L}=0\). The effective potential in this case is equal to the potential. Since angular momentum is conserved the orbits will lie in a plane. Taking \(r,\theta\) as coordinates, the total energy can be written as \begin{eqnarray} E &=& \frac{\mu}{2} \Big(\dd[r]{t}\Big)^2 + V(r)\\ &=& \frac{\mu}{2} \Big(\dd[r]{t}\Big)^2 +\frac{1}{2}\mu \omega^2r^2 +\frac{\lambda^2}{2\mu r^2} \end{eqnarray} Using the initial conditions at time \(t=0\), the energy is \(E=\frac{25}{2}\mu\omega^2 a^2\), and we get \begin{eqnarray} \frac{\mu}{2}\Big(\dd[r]{t}\Big)^2 = E - \frac{1}{2}\mu\omega^2 r^2 -\frac{\lambda^2}{2\mu\omega^2 r^2}\\ \Big(\dd[r]{t}\Big)^2=25\omega^2 a^2 -\omega^2 r^2 -\frac{\lambda^2}{\mu^2r^2}. \end{eqnarray} Taking square root we get \begin{equation} dt = \frac{dr}{\sqrt{25\omega^2a^2-\omega^2r^2- \frac{\lambda^2}{\mu^2r^2}}}. \end{equation} Substituting the given value \(\lambda=12\mu\omega a^2\), and defining the dimensionless variable \(y=r/a\), after some straightforward algebra we get \begin{eqnarray} dt &=& \int \frac{a dy}{\sqrt{25 \omega^2 a^2 - \omega^2 a^2 y^2 -\frac{144 a^2}{\omega^2 y^2}}}\\ \omega t &=& \int\frac{dy}{\sqrt{25-y^2 - 144/y^2}}\\ &=& \int \frac{ydy}{\sqrt{25y^2-y^4 -144}} \end{eqnarray} Changing the integration variable to \(\sigma\) where \(\sigma=y^2 (=r^2/a^2)\) and with \(d\sigma=2y dy\), \begin{eqnarray} \omega t&=&\int \frac{d\sigma }{2\sqrt{(25\sigma-y^2- 144)}}\\ 2\omega t &=& \int \frac{d\sigma }{\sqrt{49/4 - (\sigma-\sigma_0)^2}},\qquad \text{where } \sigma_0 = 25/2. \end{eqnarray} Therefore \begin{eqnarray} \int\frac{d\sigma}{\sqrt{(7/2)^2 - (\sigma -\sigma_0)^2}} &=& 2\omega t +\phi. \end{eqnarray} Integrating over \(\sigma\) gives \begin{eqnarray} \sin^{-1}\frac{(\sigma -\sigma_0)}{7/2} &=&2\omega t+\phi \quad \mbox{\HighLight{used \(\int\frac{dx}{\sqrt{a^2-x^2}}=\sin^{-1}(x/a)\)}} \end{eqnarray} Thus we get \begin{equation} \sigma =\sigma_0+\frac{7}{2}\sin(2\omega t +\phi) = \frac{25}{2} + \frac{7}{2}\sin(2\omega t+\phi). \end{equation} Substituting \(\sigma=y^2=r^2/a^2\), and applying initial conditions gives the final answer for the orbit. \begin{equation} r^2=\frac{25}{2} a^2 + \frac{7}{2}a^2\cos(2\omega t). \end{equation} The particle oscillates between \(3a\) and \(4a\) along the line \(\theta=0\).
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22-02-08 18:02:10 |
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[QUE/ME-12008] ME-PROBLEMNode id: 3977pageConsider a particle of mass \(\mu\) moving in a potential \[ V(r) = \frac{1}{2}\mu\omega^2 r^2 +\frac{\lambda^2}{2\mu r^2}. \]
- Find condition on energy \(E\) and angular momentum \(L\) for circular orbits to exist.
- Does there exist a circular orbit for \(L=0\)?
- Assume orbital angular momentum \(L=0\), energy \(E= \frac{25}{2}\mu\omega^2a^2\), \(\lambda=12\mu\omega^2\) Use initial conditions \[ r(t)\big|_{t=0} = 4a;\quad \dot{r}(t)\big|_{t=0} = 0 \text{ and } \dot{\theta(t)}\big|_{t=0}=0 \] solve the equations of motion and obtain \(r\), \(\theta\) as function of time. Describe the motion that takes place under conditions specified here.
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22-02-08 18:02:47 |
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[QUE/ME-12006] ME-PROBLEMNode id: 3976pageConsider a particle of mass \(\mu\) moving in a potential \[ V(r) = \frac{1}{2}\mu\omega^2 r^2 +\frac{\lambda^2}{2\mu r^2}. \]
- Find condition on energy \(E\) and angular momentum \(L\) for circular orbits to exist.
- Does there exist a circular orbit for \(L=0\)?
- Assume \(L=0, E= \frac{25}{2}\mu\omega^2a^2\), \(\lambda=12\mu\omega^2\) Use initial conditions \[ r(t)\big|_{t=0} = 4a;\quad \dot{r}(t)\big|_{t=0} = 0 \text{ and } \dot{\theta(t)}\big|_{t=0}=0 \] solve the equations of motion and obtain \(r\), \(\theta\) as function of time. Describe the motion that takes place under conditions specified here.
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22-02-08 18:02:48 |
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[QUE/ME-12005] ME-PROBLEMNode id: 3975page
- A satellite of mass 2000 kg is to be put into a circular orbit around the earth of radius 1100 km. What is the minimum energy required?
- What will the minimum energy required to transfer it to an elliptic orbit having minimum and maximum distances 1100km and 4100km?
\(R_e=6400 \text{km},\quad GM/R_e^2=g\Rightarrow GM=gR_e^2\) Initial energy \(E_i\) of the satellite on the surface of the earth is \begin{eqnarray} E_i &=& -\frac{GMm}{R_e} = -\frac{gR_e^2m}{R_e}\\ &=& -gR_e^2m = (9.8 \text{(m/s}^2))(6400\times 1000 \text{\,m})(2000\text{\,kg})\\ &=& -9.8\times 128 \times10^8\\ &=&-12.5\times 10^{10} J \end{eqnarray} Final energy of the satellite in the orbit at height of 1100 km be \(E_f\). \begin{eqnarray} E_f &=& -\frac{GM,}{2(R+h)} = -\frac{gR_e^2 m}{2(R_e+h)}\\ &=& -\frac{9.8\times(6400\times 6400\times 10^6)\times 2000}{2\times7500\times10^3}\\ &=& \frac{9.8\times64\times 64}{2\times75}\times10^8\\ &=& -5.35\times10^{10} \end{eqnarray} Therefore the energy required to put the satellite in the circular orbit at a height 1100km \\ \(E_f-E_i=(-5.35 +!2.5 )\times 10^{10} \text{J}=7.15\times 10^{10} \text{J}\). \paragraph*{Elliptic orbit} Let \(r_1,r_2,a\) be, respectively, the perigee, apogee and the semi-major axis of the elliptic orbit. Then \(2a=r_1+ r_2\) It is give that \[r_1= 1100 km +R_e, r_2 =4100km+ R_e\] Therefore \[ 2a= 1100+4100 + 2\times 6400 = 1800\text{\,km}\] Hence a=9000km. The energy of the satellite in the elliptic orbit is \begin{eqnarray} E_\text{ell} &=& -\frac{GMm}{2a} = -\frac{gR_e^2m}{2a}\\ &=& -\frac{9.8 \times 6400\times6400\times10^6 \times 2000}{1800\times 10^3}\\ &=& -\frac{9.8\times64^2 \times 2\times 10^5}{18}\\ &=& -4.4 \times 10^{10} J. \end{eqnarray} Therefore energy requires to transfer the satellite from the circular orbit to the elliptic orbit is \((-4.4+5.35)\times 10^{10}\text{ J} = 9.3\times 10^{9}\text{J}.\)
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22-02-08 18:02:38 |
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[QUE/ME-12004] ME-PROBLEMNode id: 3974pageComet Halley approaches the Sun to within 0.570 AU and its period is 75.6 years. How far from the Sun will this comet travel before it starts to return journey. Assume mass of Halley's comet is much smaller than the mass of the Sun. [Given : 1 AU $= 1.5 \times 10^{11}$ m, $G= 6.63\times10^{-11}$ N.m\(^2\) /kg\(^2\), Mass of Sun=\( 1.99\times10^{30}\) kg ]. |
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22-02-08 18:02:58 |
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[QUE/ME-12003] ME-PROBLEMNode id: 3973page
J.S.Plaskett's star is one of the most massive stars known at present. It is a double or a binary star, that is, it consists of two stars bound together by gravity. From spectroscopic studies it is known that the period of revolution of each component is 14.4 days; the velocity of each component is about 220 km/s; The orbit is nearly circular
- [(a)] Argue that the masses of two stars are nearly equal and that they are nearly equidistant from the centre of mass of the system.
- [(b)] Compute the reduced mass and and the separation of the two components.
- Assume that the masses, velocities, radii of the circular orbits of the two stars are \(m_{1,2} v_{1,2}, r_{1,2}\). Let \(F_1, F_2\) be the gravitational pulls on a star due to the other star. Since the orbits are given to be circular \begin{equation}\label{EQ901} \frac{m_1v_1^2}{r_1}= F_1, \qquad \frac{m_2v_2^2}{r_2}= F_2, \end{equation} Now use \(F_!=F_2\) to get \begin{equation}\label{EQ902} \frac{m_1v_1^2}{r_1} =\frac{m_2v_2^2}{r_2} \Longrightarrow \frac{m_1}{r_1} =\frac{m_2}{r_2} \end{equation} the last step follows because it is given that \(v_1=v_2\). For a circular orbit, the speed is constant and time period is equal to (perimeter/ velocity): \begin{equation}\label{EQ903} T_1 = \frac{2\pi r_1}{v_1 }, \qquad T_2 = \frac{2\pi r_2}{v_2 } \end{equation} We are given \(T_1=T_2\) and \(v_1=v_2\), hence we get \(r_1=r_2\), and \EqRef{EQ902} then then gives \(m_1=m_2\).
- The radius of the orbits can be obtained from \eqref{\labelPrefix;EQ903}: \begin{eqnarray} r_1 &=& \frac{T_1 v_1}{2\pi} \\ &=& \frac{\HighLight{\( 14.4 \times 24 \times 60\)} \times 60 \ \text{\,sec})(220\times 10^{3} \text{m/s})}{2\pi}, \text{ use }\HighLight{\((14.4 \times 24)\times 60 \sim (14 \times (25 \times 60)\))} \nonumber\\ &\approx& (\HighLight{\(14\times 1500\)})(10)(220)\times 10^{3} \quad \HighLight{used \((60/2\pi) \sim 10\)} \\ &\approx& (200,00) (22)\times 10^5, \HighLight{Recall \(15^2=225\)}, \, \text{so used } \HighLight{\((14)(15)\sim 200\)}\\ &\approx& 4.4 \times 10^{10} \text{m}.\\ \end{eqnarray} Thus the distance between stars is \(2r_1 \sim 8.8 \times 10^{10}\) \, m. The mass can be computed by using the second law \begin{equation} \frac{mv^2}{r_1}= \frac{G m m }{(2r_1)^2} \Rightarrow m = \frac{4 v^2 r_1}{G} \end{equation} Substituting numerical values \begin{eqnarray} m &=& \frac{4 \times 220^2 \times 10^6 \times 4.4 \times 10^{10}}{6.67\times 10^{-11}} = \frac{4\times 484 \times 4.4 \times 10^{29}}{6.67}\\ &=& 4\times 484 \times \Big(\frac{2}{3}\Big) \times 10^{29} = 4 \times (1.61 \times 2) \times 10^{31} \\ &=& 4 (3.2)\times 10^{31} = 1.28 \times 10^{32} \text{ kg}. \end{eqnarray}
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22-02-08 18:02:45 |
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[QUE/ME-12002] ME-PROBLEMNode id: 3972pageConsider a particle of mass \(\mu\) moving in a potential \[ V(r) = \frac{1}{2}\mu\omega^2 r^2 +\frac{\lambda^2}{2\mu r^2}. \] Does there exist a circular orbit for \(L=0\)? Assume \(L=0, E= \frac{25}{2}\mu\omega^2a^2\), \(\lambda=12\mu\omega^2\) Use initial conditions \[ r(t)\big|_{t=0} = 4a;\quad \dot{r}(t)\big|_{t=0} = 0 \text{ and } \dot{\theta(t)}\big|_{t=0}=0 \] solve the equations of motion and obtain \(r\), \(\theta\) as function of time. Describe the motion that takes place under conditions specified here. |
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22-02-08 18:02:28 |
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[QUE/ME-12001] ME-PROBLEMNode id: 3971pageWhen a certain mass is hung from its bottom end. If this same wire is connected between two points, \(A\) and \(B\), that are a distance \(\ell_0\) apart on the same horizontal level, and the same mass is hung from the middle point of a wire as shown, what is the depression \(y\) of the midpoint and what is the tension in the wire? |
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22-02-08 18:02:25 |
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[QUE/ME-13002] ME-PROBLEMNode id: 3986page
A homogeneous solid circular cylinder of radius \(r\), length \(L\) and mass \(m\) rolls without slipping inside a rough circular track of radius \(R\), as shown in %FigBelow{-20,-12}{80} Choose the centre of the track as the origin of axes, with \(Z\)- axis pointing vertically upwards.
- Obtain expressions for kinetic energy and angular momentum of the cylinder in terms of \(\pmb{\dot{\theta}}\).
- If the cylinder is released from rest at point \(\theta =\pi/2\), determine the angular velocity \(\pmb{\dot{\theta}}\) of the cylinder when in position \(\theta=0\).
- The total kinetic energy, and angular momentum, will be sum of two parts. The first part is translation of centre of mass and the second part is rotation about the centre of mass. The centre of mass rotates about the origin with angul The kinetic energy of the centre of mass at any instant of time is \begin{equation} T_\text{CM} = \frac{1}{2} M \vec{v}^2 = \frac{1}{2}M (R-r)^2 \dot{\theta}^2 \end{equation} where \(|\vec{v}|\) is the absolute value of the velocity given by \(|\vec{v}|= (R-r) \dot{\theta}\). Let \(\omega\)be the angular velocity of rotation about the centre of mass. The kinetic energy of rotational motion about the centre of mass is \begin{equation} T_\text{rf} = \frac{1}{2} I \omega^2 = \frac{1}{4} Mr^2 \omega^2. \end{equation} We now need to find a relation between \(\omega\) and \(\dot{\theta}\). The velocity of any point on the is cylinder is vector sum of translational velocity of the centre of mass and velocity due to rotation about the centre of mass. As the cylinder rolls without slipping the instantaneous velocity of the point of contact is zero. This give a relation between \(\dot{\theta}\) and \(\omega\). \[ (R-r) \dot{\theta} = \omega r\] Thus we get the total kinetic energy as \begin{eqnarray} T&=& \frac{1}{2} I \omega^2 = \frac{1}{4} Mr^2 \omega^2 + \frac{1}{2}M (R-r)^2 \dot{\theta}^2\\ &=& \frac{3}{8} M (R-r)^2 \dot{\theta}^2 \end{eqnarray} The angular momentum angular momentum about the origin is obtained by vector addition of angular momentum of rotation about the centre of the cylinder and the angular momentum of the centre of mass about the center of the track. Thus angular momentum is along the normal out of the plane of paper and has magnitude given by \begin{eqnarray} L&=& M (R-r)^2 \dot{\theta} - I \omega \\ &=& M (R-r) r \dot{\omega} - \frac{1}{4}M r^2 \dot{\omega}\\ &=& M r(\omega) \Big(R-r-\frac{1}{4} r \Big)\\ &=& M r(R-\frac{5r}{4})\omega.\\ &=& M \frac{4R-5r}{4 (R-r)} r \omega. \end{eqnarray}
- The total energy of the system as a function of \(\theta\) is \begin{equation} E= \frac{3}{8} M (R-r)^2 \dot{\theta}^2 + MgR(1- \cos\theta). \end{equation} The bottom point,\(\theta=0\) , has been chosen as a reference point for zero potential energy. At the initial point \(\theta=\pi/2\), \(\dot{\theta}=0\). Using energy conservation and equating the total energy at \(\theta=0\) and \(\theta=\pi/2\), we get \begin{equation} \frac{3}{8} M (R-r)^2 \dot{\theta}^2 = MgR. \end{equation} Thus the angular velocity when the cylinder reaches the bottom point is given by \begin{equation} \dot{\theta} =\sqrt{\frac{8gR}{3(R-r)^2}}. \end{equation}
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22-02-08 08:02:57 |
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[QUE/ME-130001] ME-PROBLEMNode id: 3985pageExpress the kinetic energy, the linear momentum, and the angular momentum of a system of \(N\) particles in terms of the Jacobi coordinates \begin{eqnarray} \vec{R}_j&=&\frac{m_1\vec{r}_1 + \ldots + m_j \vec{r}_j}{m_1+\ldots+m_j}-\vec{r}_{j+1}, \qquad j=1,\ldots,(N-1).\\ \vec{R}_N &=& \frac{m_1\vec{r}_1 + \ldots + m_N \vec{r}_N}{m_1+\ldots+m_N} \end{eqnarray}
%FigBelow{80,10}{70}{0}{em-fig-09008} |
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22-02-08 08:02:48 |
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[QUE/ME-14007] ME-PROBLEMNode id: 3994page
Compute moment of inertia tensor of a space station with solar panels as shown in figure. Let us calculate the inertia tensor for the spacecraft with solar panels illustrated in figure. We will consider the body to be a cylinder and the two panels to be rectangular hexahedra, all of uniform density and of the dimensions indicated in figure.
%FigBelow{10,-35}{70}{0} SolarPanels
Let the \(x\)-axis be the axis of the panel passing through the center of the \(b −c\) face and the \(y\)-axis the symmetry axis of the cylinder. Choose the x-axis to be a principal axis for the cylinder. Let the panels each be inclined at angles \(\pi/2-\theta\) to the generators of the cylinder.
Let B1 be the cylinder and B2 , B3 the (identical) panels. More in Heard's book.\\ William B Heard, "Rigid Body Mechanics" WILEY-VCH Verlag GmbH & Co. KGaA (2006)
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22-02-08 08:02:26 |
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[QUE/ME-14006] ME-PROBLEMNode id: 3993page$\newcommand{\dd}[2][]{\frac{d#1}{d#2}}$
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Two observers \(O\) and \(O{'}\), in frames \(K\) and \(K{'}\) with common origin, find that the components of an arbitrary vector \(\mathbf A\) in their respective frames are related by a time dependent rotation matrix \(R(t)\) \begin{equation} \widetilde{\sf A}{'}= R(t) \widetilde{\sf A}. \end{equation} It is given that components of the vector do not change w.r.t. the frame \(K\). Obtain an expression for angular velocity of frame \(K{'}\) w.r.t. frame \(K\) in terms of the rotation matrix. Verify that your answer gives correct answer correct answer when \(R(t)\) corresponds to a rotation about \(X_3\) axis.
We will use this relation \begin{equation} \dd[{\mathbf A}]{t}\Big|_K = \dd[{\mathbf A}{'}]{t}\Big|_{K{'}} + \pmb{\omega}\times{\mathbf A}{'}. \end{equation} to obtain an expression for the angular velocity. Here \(\vec{\omega}=(\omega_1, \omega_2,\omega_3)\) are the components of the angular velocity of frame \(K{'}\) w.r.t. the frame \(K\).\\ It is given that the vector components of \(\mathbf A\) w.r.t. frame \(K{'}\) do not change with time. Hence \begin{equation} \dd{t} \vec{A}{'} = - \vec{\omega}\times \vec{A}{'}. \end{equation} Writing the above equation in terms of components, we get \begin{eqnarray}\nonumber \dd{t}\begin{pmatrix}{A}_1{'} \\ {A}_2{'} \\ A_3{'}\end{pmatrix} &=&- \begin{pmatrix} \omega_2{A}_3{'} -\omega_3{A}_2{'} \\ \omega_3{A}_1{'} -\omega_1{A}_3{'} \\ \omega_1{A}_2{'} -\omega_2{A}_1{'} \end{pmatrix}\\ &=& \begin{pmatrix} 0 & \omega_3 & -\omega_2\\ -\omega_3 & 0 & \omega_1 \\ \omega_2 & -\omega_1 & 0 \end{pmatrix}\label{EQ04} \begin{pmatrix} {A}_1{'} \\ {A}_2{'} \\ {A}_3{'}\end{pmatrix} \end{eqnarray} Using \(\widetilde{A}{'}\) and \(\widetilde{A}{'}\) to denote the column vectors of components \(\vec{A}\) and \(\vec{A}{'}\), we write the above relation as \begin{equation}\label{EQ06} \dd[\widetilde{A}{'}]{t} = \Omega \cdot \widetilde{A}{'}. \end{equation} where we have introduced the matrix \(\Omega\) defined by \begin{equation} \Omega =\begin{pmatrix} 0 & \omega_3 & -\omega_2\\ -\omega_3 & 0 & \omega_1 \\ \omega_2 & -\omega_1 & 0 \end{pmatrix} \end{equation} The given rotation matrix \(R(t)\) relates the components of vector \(\mathbf A\) in the two frames. \begin{equation} \widetilde{A}{'} = R(t) \widetilde{A} \end{equation} Substituting in \eqRef{EQ06}, we get \begin{equation} \dd[R(t)]{t}\widetilde{A} = \Omega R(t) \widetilde{A} \end{equation} Since vector \(\vec{A}\) is arbitrary, we get the matrix relation \begin{equation} \dd[R(t)]{t} = \Omega R(t). \end{equation} Multiplying by\(R^T\) from the right, this equation implies that \begin{equation} \Omega = \dd[R(t)]{t} R^T(t).\label{EQ10} \end{equation} The last relation is the required relation. Differentiating \(R R^T= \hat{I}\) w.r.t. we get \begin{equation}\label{EQ11} \dd[R]{t} R^T(t) + R\dd[R^T]{t} =0 \Longrightarrow \dd[R]{t} R^T = - R\dd[R^T]{t}. \end{equation} Thus \(\Omega\) in \eqRef{EQ10} can be written as \begin{equation} \boxed{ \Omega = \dd[R(t)]{t} R^T(t)=-R\dd[R^T]{t}.}\label{EQ12} \end{equation} It should be noted that \eqRef{EQ11} implies that \(\Omega=-\Omega^T\) and therefore \(\Omega\) is an antisummetric matrix. \paragraph*{Verification of \eqRef{EQ12} for rotation about \(X_3\)- axis:\\} For rotations about \(X_3\) axis by an angle \(\alpha(t)\) we have \begin{equation} \dd[R]{t} =\dd{t}\begin{pmatrix} \cos\alpha & \sin\alpha& 0\\ -\sin \alpha & \cos\alpha &0\\ 0 & 0& 1 \end{pmatrix} = \begin{pmatrix} -\dot{\alpha}\sin\alpha & \dot{\alpha}\cos\alpha& 0\\ -\dot{\alpha}\cos \alpha & -\dot{\alpha}\sin\alpha &0\\ 0 & 0& 1 \end{pmatrix} \end{equation} Therefore \begin{eqnarray}\nonumber \Omega = \dd[R]{t} R^T &=& \begin{pmatrix} -\dot{\alpha}\sin\alpha & \dot{\alpha}\cos\alpha& 0\\ -\dot{\alpha}\cos \alpha & -\dot{\alpha}\sin\alpha &0\\ 0 & 0& 1 \end{pmatrix} \begin{pmatrix} \cos\alpha & -\sin\alpha& 0\\ \sin \alpha & \cos\alpha &0\\ 0 & 0& 1 \end{pmatrix}\\ &=&\begin{pmatrix} 0 & \dot{\alpha} & 0\\ -\dot{\alpha} & 0 & 0\\ 0 & 0 & 0 \end{pmatrix}\label{EQ14} \end{eqnarray} Comparing with \eqRef{EQ04}, we \(\vec{\omega}=(0,0,\dot{\alpha})\) as expected. This completes the verification of \eqRef{EQ10} for rotations about the \(X_3\) axis. \subsection{$\pmb{\omega}$ in terms of Euler angles} Let \(\phi,\theta,\psi\) denote Euler angles of a axes fixed in body of a rigid body rotating about a fixed point. The fixed point is chosen as the common origin of the space and body fixed axes. Then the rotation matrix \(R\) giving the transformation of vectors from space fixed axes to body fixed axes is given by \begin{equation} R(\phi, \theta, \psi) = R_3(\psi) R_1(\theta) R_(\phi). \end{equation} To obtain an expression for components of the angular velocity vector we use \eqRef{EQ12}. Therefore we must compute \begin{eqnarray}\nonumber \Omega &=& -R \Big(\dd[R^T]{t}\Big)= - R\big(\Dsc R^T\big)\\\nonumber &=& - \big(R_3(\psi) R_1(\theta) R_3(\phi) \Dsc \big[ R_3(\psi) R_1(\theta) R_3(\phi)\big]^T\\\nonumber &=& - \Big\{R_3(\psi) R_1(\theta) R_3(\phi).\Big\}\,\Big\{ \Dsc [R_3^T(\phi)] R_1^T(\theta) R_3^T(\psi) \\ && \qquad + R_3^T(\phi)[\Dsc R_1^T(\theta)] R_3^T(\psi) + R_3^T(\phi) R_1^T(\theta)[\Dsc R^T_3(\psi)]\Big\}\label{EQ16}. \end{eqnarray} where \(\Dsc\) stands for the time derivative \(\dd{t}\). We will use the notation \begin{eqnarray} DR1&=& - \big\{(R_3(\psi) R_1(\theta)\big\} \textcolor{blue}{ [R_3(\phi) [\Dsc R_3^T(\phi)]] } \big\{R_1(\theta)^T R_3(\psi)^T\big\} \\ DR2&=& - \big\{R_3(\psi) R_1(\theta) R_3(\phi)\big\} \big\{R_3^T(\phi) [\Dsc R_1^T(\theta)] R_3^T(\psi)\big\}\\ DR3&=& - \big\{R_3(\psi) R_1(\theta) R_3(\phi)\big\} \big\{R_3^T(\phi)R_1^T(\theta) [\Dsc R_3^T(\psi)]\big\}. \end{eqnarray} Since \(R_1\) and \(R_3\) are orthogonal matrices, we use \(R^T_3 R_3= R_1^T R_1= \text{identity matrix, }\hat{I}\) to simplify \(DR2\) and \(DR3\) as \begin{eqnarray} DR1&=& - \big\{(R_3(\psi) R_1(\theta)\big\} \textcolor{blue}{ [R_3(\phi) \Dsc R_3^T(\phi)] } \big\{R_1^T(\theta) R_3^T(\psi)\big\} \\ DR2&=& - R_3(\psi) \textcolor{blue}{[R_1(\theta) \Dsc R_1^T(\theta)]} R_3^T(\psi)\\ DR3&=& - \textcolor{blue}{[R_3(\psi) \Dsc R_3^T(\psi)]}. \end{eqnarray} Using the form of rotation matrices for rotations about coordinate axes we will have, (see \begin{equation} R_3(\phi)\Dsc R_3^T(\phi) = \begin{pmatrix} 0 & \dot{\phi} & 0\\ -\dot{\phi} & 0 & 0\\ 0 & 0 & 0 \end{pmatrix}, \qquad R_1(\theta)\Dsc R_1^T(\theta)= \begin{pmatrix} 0 & 0 & 0\\ 0 & 0 & \dot {\theta}\\ 0 & -\dot{\theta} & 0 \end{pmatrix} \end{equation} and expression for \( R_3(\psi)\Dsc R_3^T(\psi)\) will be similar to \(R_3(\phi)\Dsc R_3^T(\phi)\).
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22-02-08 08:02:59 |
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[QUE/ME-14005] ME-PROBLEMNode id: 3992page
Find the moment of inertia tensor of a uniform rectangular plate w.r.t. the centre \(O\) of the plate relative to the axes \(K\) shown in figure. Use parallel axes theorem to find moment of inertia tensor relative to a set of axes \(K_2\) with origin taken as the corner \(E\). The \(Z\)- axis for both systems is perpendicular to the plate and out of the plane of paper.
%FigBelow{10,-25}{40}{0} RectangularPlate
We shall do it by evaluating the double integral. Divide the plate by lines parallel to the coordinate axes as in %Figref MI-Plate
Consider one rectangular element of sides \((dx, dy)\) at position \((x,y)\). If the mass of the plate is \(M\), the density of the plate is \(\sigma\), then \(4ab\sigma =M\) and the contribution of the rectangular element to \(I_{xx}\) is \(\sigma (dx\,dy) y^2\). Summing over all elements means integrating over \(x, y\) over their respective ranges. Thus %FigBelow{10,-25}{40}{0} RectangularPlate \begin{eqnarray}\nonumber I_{xx} &=& \int_{-a}^a dx \int_{-b}^b dy \sigma y^2 =\int_{-a}^a dx \sigma \frac{2b^3}{3} = \sigma \frac{2b^3}{3} \int_{-a}^a dx\\ &=& \sigma \frac{2b^3}{3} (2a) = \frac{Mb^2}{3} \qquad \HighLight{\mbox{$\because 4\sigma ab=M$}} \end{eqnarray} % Similarly \begin{eqnarray}\nonumber I_{yy} &=& \int_{-a}^a dx \int_{-b}^b dy \sigma x^2 =\int_{-a}^a dx \sigma x^2 (2b) = \sigma \frac{2b^3}{3} \int_{-a}^a dx\qquad \HighLight{\mbox{$\because 4\sigma ab=M$}}\\ &=& \sigma \frac{2a^3}{3} (2b) = \frac{Ma^2}{3} \end{eqnarray} and appealing to the law of perpendicular axes we get \(I_{zz}= M\frac{a^2+b^2}{3}\) The off diagonal term \(I_{xy}\) is \begin{equation} I_{xy} = - \int_{-a}^a dx \int_{-b}^b dy \sigma (xy) \end{equation} which vanishes due to the symmetry of the problem. Also \begin{equation} I_{xz} = - \int_{-a}^a dx \int_{-b}^b dy \sigma (xz) = 0 \end{equation} because \(z=0\) for all rectangular elements of the plate. Thus the moment of inertia tensor w.r.t.the center of mass is given by \begin{equation} \underline{\Ibb} = \frac{M}{6} \begin{pmatrix} b^2 & 0 & 0\\ 0 & a^2 & 0\\ 0 & 0 & a^2+b^2 \end{pmatrix} \end{equation} \paragraph*{MI Tensor relative to a corner} The parallel axes for the inertia tensor states that \begin{equation} I_{jk} = I^\text{cm}_{jk} + \Delta_{jk} \end{equation} where \(\Delta_{jk} =|\vec{a}|^2 \delta_{jk} -a_ja_k\) and \(\vec{a}\) is the position vector of the corner relative to the origin \(O\). For the corner \(E\) we have \(\vec{a}=(-a,-b,0)\) and \(|\vec{a}|^2=(a^2+b^2)\). Therefore, \begin{eqnarray} &\Delta_{xx} = b^2,\qquad \qquad \qquad \Delta _{yy} = a^2,\quad \qquad \Delta_{zz} = (a^2+b^2)&\\ &\qquad\quad\Delta_{xy}=\Delta_{yx}=-ab\qquad \Delta_{yz}=\Delta_{zy}=0\qquad \Delta_{zx}=\Delta_{xz}=0.\qquad\qquad& \end{eqnarray} Thus the moment of inertia tensor w.r.t.the corner at \(E\) is given by \begin{equation} \underline{\Ibb} = \frac{M}{3} \begin{pmatrix} b^2 + 3a^2 & -3ab & 0\\ -3ab & a^2 + 3b^2 & 0\\ 0 & 0 & 4a^2+4b^2 \end{pmatrix}. \end{equation}
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22-02-08 08:02:37 |
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[QUE/ME-14004] ME-PROBLEMNode id: 3991page
Three equal masses, connected by light rods, are placed at the vertices of a right angle triangle as shown in figure. The origin of the coordinate system is chosen to coincide with the centre of mass of the system. Obtain the values of moment of inertia tensor components.
Method 1:
Let the coordinates of the vertices \(A,B,C\) be \((a,b,0)\), \((a+h,b,0)\) and \((a+h,b+h,0)\). Since the centre of mass is at the origin and all the masses are equal \begin{eqnarray}\nonumber a+(a+h)+(a+h)=0 \Longrightarrow a=-2h/3\\ b+b + b+h=0 \Longrightarrow b=-h/3. \end{eqnarray} Therefore the coordinates of the points \(A,B,C\) are \[\vec{x}_1=(-2h/3, -h/3,0), \vec{x_2}=(h/3,-h/3,0), \vec{x_3}=(h/3, 2h/3). \] The moment of inertia tensor is given by \begin{equation} I_{jk} = \sum m_\alpha ( |\vec{x}_\alpha|^2\delta_{jk} - a_ja_k ). \end{equation} It is obvious that \(I_{13}=I_{31}=I_{23}=I_{32}=0\). Since the inertia tensor is symmetric we need to compute only \(I_{11}, I_{22}, I_{33}\) and \(I_{12}\). Let us write the values needed in a tabular form \begin{equation} \begin{array}{crrcrrrrr} \hline & & & & & \\ \text{Location} & x_1 &\quad x_2 & \quad x_3 & x_1^2 &x_2^2&\quad|\vec{x}|^2 & \quad x_1x_2\\[1mm] A & -\dfrac{2h}{3} & -\dfrac{h}{3} & 0 &\dfrac{4h^2}{9} & \dfrac{h^2}{9}& \dfrac{5h^2}{9}&\dfrac{2h^2}{9} \\[2mm] B & \dfrac{h}{3} & -\dfrac{h}{3} & 0 & \dfrac{h^2}{9}& \dfrac{h^2}{9}&\dfrac{2h^2}{9}&-\dfrac{h^2}{3}\\[2mm] C & \dfrac{h}{3} & \dfrac{2h}{3} & 0 & \dfrac{h^2}{9}&\dfrac{4h^2}{9}& \dfrac{5h^2}{9} & \dfrac{2h^2}{9}\\[3mm] \hline &&&&&&&&\\[-1mm] \sum_\alpha(.) & ... & ... & ... & \dfrac{6h^2}{9} & \dfrac{6h^2}{9} & \dfrac{12h^2}{9} & \dfrac{3h^2}{9} \\[2mm] \hline \end{array} \end{equation} We are now ready to compute the moment of inertia tensor \begin{eqnarray} I_11 &=&m \sum_\alpha \{|x_\alpha|^2 - x_{\alpha1}x_{\alpha1}\} =m\big(\frac{12h^2}{9} - \frac{6h^2}{3}=\frac{2mh^2}{3}\big) = \frac{2mh^2}{3} \\ I_22 &=&m \sum_\alpha \{|x_\alpha|^2 - x_{\alpha2}x_{\alpha2}\} =m\Big(\frac{12h^2}{9} - \frac{6h^2}{3}\Big) = \frac{2mh^2}{3} = \\ I_{33} &=& m\sum_\alpha \{|x_\alpha|^2 - x_{\alpha3}x_{\alpha3}\} =\frac{12mh^2}{9}-0= \frac{4mh^2}{3} \end{eqnarray} \FigBelow{-05,0}{50}{40}{me-fig-14008A}{\(K{'}\) axes} \FigBelow{15,0}{50}{40}{me-fig-14008B}{\(K{'}\) axes}\\
Method 2 :
Choose a convenient set of axes:
We translate the axes from the given origin to vertex \(B\), as shown in
%Figref{me-fig-14008B}. With respect to prime axes, the coordinates of the three vertices \(A,B,C\) are \((-h,0,0),(0,0,0)\) and \((h,0,0)\). Therefore the components of moment of inertia tensor are given by \begin{equation} I_{11}{'}= mh^2, \quad I_{22}=mh^2, \quad I_{33}=2mh^2. \end{equation} All other components are zero. \begin{equation} I{'}_{12}=I{'}_{21}=0, \quad I{'}_{23}=I{'}_{32}=0, \quad I{'}_{31}=I{'}_{13}=0. \end{equation} Now we use parallel axes theorem to get the moment of inertia tensor components w.r.t. the given set with origin at the centre of mass. The coordinates of centre of mass in \(K{'} \) system are given by \(\vec{a}=(h/3,h/3,0)\), and the total mass is \(M=3m\).\\ \noindent To use the parallel axes theorem \begin{equation} I_{jk}= I{'}_{jk} - M ( |a|^2 -a_ja_k), \end{equation} we record the following values computation of inertia tensor. \[|\vec{a}|^2= \frac{2h^2}{9}\qquad |a|^2 -a_1^2= |a|^2-a_2^2= \frac{h^2}{9}, \qquad a_1a_2= \frac{h^2}{9} , \qquad a_1a_3=a_2a_3=0 \] We are now ready to have the answers \begin{eqnarray}\nonumber I_{11} &=&I{'}_{11} - M ( |a|^2 -a_1^2) = mh^2 - (3m) \frac{h^2}{9} = \frac{2mh^2}{3} \\\nonumber I_{22} &=&I{'}_{22} - M ( |a|^2 -a_2^2) = mh^2 - (3m) \frac{h^2}{9} = \frac{2mh^2}{3} \\\nonumber I_{33}&=&I{'}_{33} - M ( |a|^2 -a_3^2) = 2mh^2 -3m \frac{2h^2}{9} = \frac{4mh^2}{3}\\\nonumber I_{12}&=& I{'}_{12} + M ( |a|^2 -a_1a_2) = -(3m)\,\frac{h^2}{9} = -\frac{mh^2}{3}\\\nonumber I_{21}&=&I_{12}=-(3m)\frac{h^2}{9}=-\frac{mh^2}{3}. \end{eqnarray} All other off diagonal components are zero. Thus the inertia tensor is given by \begin{equation} I =\begin{pmatrix} \dfrac{2mh^2}{3} & -\dfrac{mh^2}{3} & 0\\[1.5mm] -\dfrac{mh^2}{3} & \dfrac{2mh^2}{3} & 0\\[1.5mm] 0 & 0 & \dfrac{4mh^2}{3} \end{pmatrix}. \end{equation}
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22-02-07 21:02:54 |
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[QUE/ME-14003] ME-PROBLEMNode id: 3989page
Show that the moment of inertia tensor of a uniform triangular plate, see figure, with centre of mass chosen as origin is given by \begin{equation} I= \frac{mL^2}{18} \begin{pmatrix} 1 & -\frac{1}{2} & 0\\ -\frac{1}{2} & 1 & 0\\ 0 & 0 & 2 \end{pmatrix}\end{equation}
In % Figref{me-fig-14012} OAB is the right angle triangle lamina with two sides of equal length \(L\). We choose the coordinate axes as shown in the figure. The corners of the plate are \((0,0,0),(L,0,0),(l,l,0)\) and the centre of mass has coordinates \((\frac{2L}{3},\frac{L}{3}, 0)\). It is easy to calculate the moment of inertia tensor w.r.t. this set of axes. The inertia tensor relative to the centre of mass will then be found by making use of parallel axes theorem.
% FigBelow{10,-14}{40}{00}{me-fig-14012} Choice of axes
Simplest way to solve the problem is to use double integral. Divide the plate into small rectangular elements as shown, obtain contribution of one such element and perform integrations over a range of \(x\) and \(y\) so as to get sum over all the rectangular elements of the plate. Let the mass per unit area of the plate be \(\sigma\), then \(M=\frac{1}{2}\sigma L^2.\) We first record the expression for moment of inertia tensor for a rigid body consisting of masses \(m_\alpha\) at position \(\vec{r}_\alpha\). \begin{eqnarray} I_{xx} &=& \sum_\alpha m_\alpha \big( \vec{r}\,^2- x_\alpha^2\big) =\sum_\alpha m_\alpha \big( y_\alpha^2+ z_\alpha^2\big)\\ I_{yy} &=& \sum_\alpha m_\alpha \big( \vec{r}\,^2_\alpha- y_\alpha^2\big) =\sum_\alpha m_\alpha \big(x_\alpha^2+ z_\alpha^2\big)\\ I_{zz} &=& \sum_\alpha m_\alpha \big( \vec{r}\,^2_\alpha- z_\alpha^2\big) =\sum_\alpha m_\alpha \big(y_\alpha^2+ z_\alpha^2\big) \end{eqnarray} The contribution of the element \(dx\, dy\) at position \(x,y\) to \(I_{xx}\) is easily written down as, remembering that for points on the plate \(z=0\), we get \begin{eqnarray} dI_{xx}= (\sigma dx\,dy)(\vec{r}^2-x^2) = (\sigma dx\,dy) y^2. \end{eqnarray} For a fixed \(x\), the variable \(y\) takes values from 0 to \(x\) and for the plate \(x\) ranges from 0 to \(L\). Therefore \(I_{xx}\) is given by \begin{eqnarray} I_{xx} &=& \int_0^L dx \int_0^x dy \sigma y^2 = \sigma\int_0^L dx \frac{x^3}{3} = \frac{\sigma L^4}{12} \\ &=& \frac{1}{6}M L^2, \qquad \qquad \because \sigma L^2 =2M. \end{eqnarray} The expression for moment of inertia \(I_{yy}\) \begin{equation} I_{yy} = \sum_\alpha m_\alpha (\vec{r}_\alpha^2 -y_\alpha^2) \end{equation} for the planer lamina problem takes the form \begin{eqnarray} I_{yy} &=& \int_0^L dx \int_0^x dy (\sigma x^2) = \int_0^L dx\, (\sigma {x^2})\times y\Big|_{y=0}^{y=x} = \frac{\sigma L^4}{4}\\ &=& \frac{ML^2}{2}. \end{eqnarray} and \(I_{zz}\) is given by \begin{eqnarray} I_{zz} &=& \int_0^L dx \int_0^x dy \sigma (x^2 + y^2) = \int_0^L \sigma (x^2 y + \frac{y^3}{3})\Big|_{y=0}^{y=x}\\ &=& \int_0^L dx \sigma \frac{4x^3}{3}= \sigma \frac{L^4}{3}\\ &=& \frac{2ML^2}{3}. \end{eqnarray} The off diagonal term \(I_{xy}\) is nonzero and is given by \begin{eqnarray} I_{xy} &=& -\int_0^L dx \int_0^L dy \sigma (\sigma x)\,y = -\int_0^L dx (\sigma x) \times \frac{y^2}{2}\Big|_{y=0}^{y=x}\\ & =& -\int_0^L dx \sigma \frac{x^3}{8} = - \sigma \frac{L^2}{8}\\ &=& - \frac{M L^2}{4}. \end{eqnarray} The parallel axis theorem gives \begin{eqnarray} I^\text{cm}_{jk}= I_{jk} - M (|\vec{a}|^2 \delta_{jk}-a_j a_k) \end{eqnarray} where \(\vec{a}\) is the position vector of the origin \(O\) relative to the centre of mass. In the present case the centre of mass has coordinates \((2L/3,L/3,0)\). Therefore \(\vec{a} = (-2L/3, -L/3,0)\). Thus we get \begin{eqnarray} I^\text{cm}_{xx} &=& I_{xx} - M (L^2/9) = (\frac{1}{6}-\frac{1}{9})ML^2 =\frac{ML^2}{18}\\ I^\text{cm}_{yy} &=& I_{yy} - M (4 L^2/9) = (\frac{1}{2}-\frac{4}{9})ML^2 =\frac{ML^2}{18}\\ I^\text{cm}_{zz}&=& I_{zz} - ML^2{(1+4)L^3}{9} = (\frac{2}{3}-\frac{5}{9})ML^2 = \frac{ML^2}{9}\\ I^\text{cm}_xy &=& I_{xy} -a_xa_y = - \frac{ML^2}{4} -\frac{2ML^2}{9} =(\frac{1}{4}-\frac{2}{9}) ML^2\\ &=& -\frac{ML^2}{36}. \end{eqnarray} Thus the moment of inertia tensor relative to the centre of mass is given by \begin{equation*} \underline{\sf I}=\frac{ML^2}{18} \begin{pmatrix} 1 & -\frac{1}{2} & 0 \\ -\frac{1}{2} & 1 & 0 \\ 0 & 0 & 2 \end{pmatrix} \end{equation*}
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22-02-07 21:02:57 |
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