Chapter 13 Rigid Bodies: Lagrangian Treatment and Euler’s Equations
Chapter 13
Rigid Bodies: Lagrangian Treatment and Euler’s Equations
Theory, worked examples and 10 previous-year questions with full solutions — IIT JAM, GATE, JEST and CSIR NET-JRF, 2020 onwards.
Theory
13.1 Degrees of freedom of a rigid body
A rigid body is a system of particles whose mutual distances do not change. The moments of inertia and the inertia tensor of such bodies were studied in Chapter 7, and their plane motion under forces and torques in Chapter 8. In this chapter we treat rigid bodies by the Lagrangian method of Chapter 9, first for motion in a plane and then for general motion in three dimensions, which leads to Euler’s equations and to the theory of the spinning top.
The first task in any Lagrangian treatment is to count the degrees of freedom. A free rigid body has six. Three coordinates fix the position of one point of the body, conventionally the centre of mass; two angles then fix the direction of a line drawn through that point in the body; and one further angle fixes the rotation of the body about that line. The same number follows from counting constraints. If \(\mathbf r_A\) and \(\mathbf r_B\) are the positions of any two particles of the body, rigidity requires
Three non-collinear particles have nine coordinates and three such distance constraints, leaving six degrees of freedom, and every further particle brings three new coordinates together with three constraints fixing its distances from the first three, so the count stays at six however many particles the body contains.
Differentiating the rigidity constraint, in the form \((\mathbf r_A-\mathbf r_B)^2=\text{constant}\), gives
so the relative velocity of two points of a rigid body is always perpendicular to the line joining them. This is the content of the statement that the relative motion is a rotation, \(\dot{\mathbf r}_A-\dot{\mathbf r}_B=\boldsymbol\omega\times(\mathbf r_A-\mathbf r_B)\), with the same angular velocity \(\boldsymbol\omega\) for every pair of points. A second differentiation gives
The relative acceleration itself does not vanish in general: in a uniform rotation about a fixed axis each point has a centripetal acceleration directed towards the axis, and these accelerations differ from point to point.
When the body is further restricted, each independent holonomic constraint removes one degree of freedom. A body turning about a fixed axis has a single degree of freedom, the angle of rotation; a body that moves parallel to a fixed plane has three, two coordinates of the centre of mass and one angle; and rolling without slipping on a fixed surface, in plane motion, removes one more. A body with one point fixed, such as a top whose tip rests in a small depression, has three rotational degrees of freedom, which in Section 13.5 we describe by the Euler angles.
13.2 The Lagrangian of a rigid body in plane motion
(a) Kinetic and potential energy
When every particle of a rigid body moves parallel to a fixed plane, the motion is called plane motion. The body then rotates about an axis perpendicular to the plane with angular velocity \(\omega=\dot\theta\), where \(\theta\) is the angle that a line fixed in the body makes with a line fixed in space. As shown in Chapter 8, the kinetic energy separates into a translational part and a rotational part,
where \(M\) is the mass, \(v_{\text{cm}}\) the speed of the centre of mass and \(I_{\text{cm}}\) the moment of inertia about the axis through the centre of mass perpendicular to the plane of motion. The first term is the kinetic energy that the body would have if its whole mass were concentrated at the centre of mass, and the second is the kinetic energy of rotation about the centre of mass. The gravitational potential energy depends only on the height \(z_{\text{cm}}\) of the centre of mass,
\[V=Mgz_{\text{cm}},\]
because \(\sum_im_igz_i=Mgz_{\text{cm}}\). To construct the Lagrangian \(L=T-V\) we therefore express the coordinates of the centre of mass and the angle of rotation in terms of the generalized coordinates, using the constraints, and substitute into these two formulae.
(b) The compound pendulum
A rigid body that can swing in a vertical plane about a fixed horizontal axis is called a compound pendulum. Let the axis pass through the point \(O\), let \(C\) be the centre of mass, with \(OC=l\), and let \(\theta\) be the angle that \(OC\) makes with the downward vertical (Figure 13.1). The mass of the body is \(M\) and its moment of inertia about the parallel axis through \(C\) is \(I_{\text{cm}}\).
Figure 13.1 A compound pendulum: a rigid body turning about a fixed horizontal axis through \(O\), with its centre of mass \(C\) at distance \(l\) from the axis.
Take axes through \(O\) in the plane of motion with \(x\) vertically downwards and \(y\) horizontal. The centre of mass is at \(x=l\cos\theta\), \(y=l\sin\theta\), so that
and the body turns with angular velocity \(\omega=\dot\theta\). Measuring the potential energy from the horizontal plane through \(O\), \(V=-Mgl\cos\theta\), and the Lagrangian is
With \(\theta\) as the generalized coordinate, \(\partial L/\partial\dot\theta=I_O\dot\theta\) and \(\partial L/\partial\theta=-Mgl\sin\theta\), and the Lagrange equation \(\frac{d}{dt}\left(\partial L/\partial\dot\theta\right)-\partial L/\partial\theta=0\) gives
This agrees with the result found from the energy equation in Chapter 8, where the equivalent length \((k^2+l^2)/l\) and the condition for the minimum period are also discussed. The Lagrangian method gives nothing new for a system with one degree of freedom, but it extends without change to the systems with several degrees of freedom met below, for which the energy equation alone is not enough.
(c) Constraints in plane motion
Most examination problems on rigid bodies in a plane involve one of three kinds of constraint. A fixed axis through a point \(O\) of the body makes the motion a pure rotation, and the kinetic energy is simply \(\frac12I_O\dot\theta^2\), as for the compound pendulum. A string that passes round a pulley without slipping ties the rotation of the pulley to the displacement of the string: if a point of the string moves through \(x\), a pulley of radius \(a\) turns through \(x/a\). Rolling without slipping ties the rotation of a body to the motion of its centre: the point of the body in contact with a fixed surface is instantaneously at rest, so a wheel of radius \(a\) whose centre moves with speed \(v_{\text{cm}}\) turns with angular speed \(\omega=v_{\text{cm}}/a\), and
For a uniform solid cylinder or disc, \(I_{\text{cm}}=\frac12Ma^2\) and \(T=\frac34Mv_{\text{cm}}^2\); for a ring or thin-walled tube, \(I_{\text{cm}}=Ma^2\) and \(T=Mv_{\text{cm}}^2\). In each case the constraint is used to eliminate the dependent coordinate before the Lagrangian is written down, so that the Lagrange equations involve only the independent generalized coordinates and the forces of constraint never appear.
Exam tip. For a body rolling on a curved track it is the speed of the centre, not of the point of contact, that equals \(a\omega\). A cylinder of radius \(a\) rolling inside a cylinder of radius \(R\) has its centre on a circle of radius \(R-a\), so \(v_{\text{cm}}=(R-a)\dot\theta\) and \(\omega=(R-a)\dot\theta/a\).
13.3 Worked examples: plane motion
Example 13.1 A uniform rod of mass \(M\) and length \(l\) is pivoted at one end \(O\) and is constrained to swing in the vertical \(yz\)-plane, with the \(z\)-axis vertically downwards. (a) Set up the Lagrangian of the system. (b) Find the angular frequency of small oscillations.
Solution. (a) Let \(\theta\) be the angle between the rod and the downward vertical. The centre of mass \(C\) is at the midpoint, at distance \(l/2\) from \(O\), with coordinates \(y=\frac l2\sin\theta\), \(z=\frac l2\cos\theta\), so \(v_{\text{cm}}^2=\frac{l^2}{4}\dot\theta^2\). The moment of inertia about the centre is \(I_{\text{cm}}=Ml^2/12\), and
with \(I_O=Ml^2/12+Ml^2/4=Ml^2/3\), the moment of inertia about the end. The potential energy measured from the horizontal through \(O\) is \(V=-Mg\frac l2\cos\theta\). Hence
For small oscillations \(\sin\theta\simeq\theta\), so the motion is simple harmonic with
\[\boxed{\omega_0=\sqrt{\frac{3g}{2l}}}.\]
The rod therefore swings like a simple pendulum of length \(2l/3\).
Example 13.2 A thin circular loop of mass \(M\) and radius \(R\) hangs from a point \(O\) on its circumference, about which it can turn freely in its own vertical plane. Find the period of small oscillations.
Solution. The centre of mass is the centre \(C\) of the loop, at distance \(R\) from \(O\). By the parallel-axis theorem the moment of inertia about the axis through \(O\) perpendicular to the plane of the loop is
\[I_O=I_C+MR^2=MR^2+MR^2=2MR^2 .\]
With \(\theta\) the angle between \(OC\) and the downward vertical, \(V=-MgR\cos\theta\), and
A hoop hung from a nail thus has the same period as a simple pendulum whose length equals the diameter of the hoop.
Example 13.3 Find the kinetic energy of a uniform solid cylinder of mass \(M\) and radius \(a\) rolling without slipping on the inside of a fixed cylindrical surface of radius \(R\), the axes being horizontal and parallel.
Figure 13.2 A cylinder of radius \(a\) rolling inside a fixed cylindrical surface of radius \(R\). The position is described by the angle \(\theta\) of the line of centres.
Solution. Let \(\theta\) be the angle between the line joining the two axes and the downward vertical (Figure 13.2). The axis of the rolling cylinder moves on a circle of radius \(R-a\) about the fixed axis \(O\), so
\[v_{\text{cm}}=(R-a)\dot\theta .\]
The line of contact is instantaneously at rest because there is no slipping, so the cylinder turns about it with angular speed
Example 13.4 A uniform disc of mass \(M\) and radius \(a\) rolls without slipping on the inside of a fixed vertical circular track of radius \(R\), as in Figure 13.2. Write down the Lagrangian, obtain the equation of motion and find the angular frequency of small oscillations about the lowest point.
Solution. Take the origin at the centre \(O\) of the track, with \(y\) vertically downwards. The centre of the disc is at \(x=(R-a)\sin\theta\), \(y=(R-a)\cos\theta\), so \(\dot x^2+\dot y^2=(R-a)^2\dot\theta^2\), and the rolling condition gives \(\omega=v_{\text{cm}}/a=(R-a)\dot\theta/a\). Exactly as in Example 13.3,
The Lagrange equation \(\frac{d}{dt}\left(\frac32M(R-a)^2\dot\theta\right)+Mg(R-a)\sin\theta=0\) gives
\[\frac32(R-a)\ddot\theta+g\sin\theta=0 .\]
For small oscillations \(\sin\theta\simeq\theta\), and
\[\boxed{\omega_0=\sqrt{\frac{2g}{3(R-a)}}}.\]
A particle sliding without friction in the same track would have \(\omega_0=\sqrt{g/(R-a)}\); the rolling disc oscillates more slowly because one third of its kinetic energy is rotational.
Example 13.5 A uniform rod of mass \(M\) and length \(l\) hangs from the lower end of a vertical spring of force constant \(k\) and natural length \(l_0\), whose upper end is fixed. The lower end of the spring, to which the rod is hinged, can move only vertically, while the rod can swing in a vertical plane. Write down the Lagrangian and the equations of motion, and find the frequencies of small oscillations.
Solution. The system has two degrees of freedom. Take as generalized coordinates the length \(y\) of the spring, measured downwards from the fixed support, and the angle \(\theta\) between the rod and the downward vertical. With \(x\) horizontal and \(y\) vertically downwards, the centre of mass of the rod is at
The momenta are \(p_y=M\dot y-\frac{Ml}{2}\dot\theta\sin\theta\) and \(p_\theta=\frac{Ml^2}{3}\dot\theta-\frac{Ml}{2}\dot y\sin\theta\), and the Lagrange equations are
The equilibrium is \(\theta=0\), \(y=y_0=l_0+Mg/k\). For small displacements the coupling terms are of second order, so to first order \(M\ddot y+k(y-y_0)=0\) and \(\frac{Ml^2}{3}\ddot\theta+\frac{Mgl}{2}\theta=0\). The two small oscillations are independent, with
