Lagrangian Formulation Questions and Answers

Q: The Lagrangian of a particle of mass \(m\) in spherical polar co-ordinates is given by :\[ \begin{aligned} L &= \tfrac{1}{2}m(\dot{r}^{2} + r^{2}\dot{\theta}^{2} + r^{2}\sin^{2}\theta\,\dot{\phi}^{2}) \\ &\quad - mgl\cos\theta \end{aligned} \]The quantity that is conserved is :

The conserved quantity is \(\partial L/\partial\dot{\phi}\).\(\phi\) does not appear in L, only \(\dot{\phi}\) does, so \(\phi\) is a cyclic coordinate. Lagrange's equation for it gives\[ \begin{aligned} \frac{d}{dt}\frac{\partial L}{\partial\dot{\phi}} &= \frac{\partial L}{\partial\phi} \\ &= 0 \end{aligned} \]so \(p_{\phi} = mr^{2}\sin^{2}\theta\,\dot{\phi}\), the angular momentum about the vertical axis, is constant.\(r\) and \(\theta\) both appear explicitly in L (through \(r^{2}\dot{\theta}^{2}\), \(\sin^{2}\theta\) and \(\cos\theta\)), so \(\partial L/\partial\dot{r}\) and \(\partial L/\partial\dot{\theta}\) are not conserved, and \(\partial L/\partial r\) is a generalised force, not a momentum.

Q: For a Lagrangian \(L(q, \dot{q}, \ddot{q}, t)\), the equation motion is of the form :

The equation of motion is the \(\ddot{q}\) term, minus the \(\dot{q}\) term, plus \(\partial L/\partial q\), equal to zero: the Euler-Lagrange (Ostrogradsky) equation for a Lagrangian that depends on \(\ddot{q}\).Varying the action with \(L(q, \dot{q}, \ddot{q}, t)\) and integrating by parts once for the \(\dot{q}\) term and twice for the \(\ddot{q}\) term gives the Ostrogradsky equation:\[ \begin{aligned} &\frac{\partial L}{\partial q} - \frac{d}{dt}\frac{\partial L}{\partial \dot{q}} \\ &\quad + \frac{d^{2}}{dt^{2}}\frac{\partial L}{\partial \ddot{q}} = 0 \end{aligned} \]The signs alternate because each integration by parts brings a minus sign. Rearranged with the second-derivative term first, this is option B. Option A drops the \(\dot{q}\) term, and options C and D get the alternating signs wrong.

Q: A dynamical system consists of three particles \(m_{1}, m_{2}, m_{3}\) in motion in space. The distance between \(m_{1}\) and \(m_{2}\), \(m_{2}\) and \(m_{3}\), \(m_{1}\) and \(m_{3}\) remain constant throughout the motion. The motion will be governed by \(n\) Euler Lagrange equations. Then :

The number of equations is \(n = 6\).Three free particles in space have \(3 \times 3 = 9\) coordinates. The three fixed distances \(m_{1}m_{2}\), \(m_{2}m_{3}\) and \(m_{1}m_{3}\) are three holonomic constraints, leaving\[ 9 - 3 = 6 \]degrees of freedom. The system is a rigid body (three non-collinear points fix it), which indeed has 3 translational and 3 rotational degrees of freedom. There is one Euler-Lagrange equation per generalised coordinate, so 6 equations. 9 ignores the constraints, 3 counts only the constraints, and 7 has no basis.

Q: A particle of mass \(m\) is moving in a two-dimensional space. Its kinetic energy in terms of polar co-ordinates is given by :

The kinetic energy is \(\frac{1}{2}m(\dot{r}^{2} + r^{2}\dot{\theta}^{2})\).With \(x = r\cos\theta\), \(y = r\sin\theta\):\[ \begin{aligned} \dot{x} &= \dot{r}\cos\theta - r\dot{\theta}\sin\theta \\ \dot{y} &= \dot{r}\sin\theta + r\dot{\theta}\cos\theta \\ \dot{x}^{2} + \dot{y}^{2} &= \dot{r}^{2} + r^{2}\dot{\theta}^{2} \end{aligned} \]So \(T = \frac{1}{2}m(\dot{r}^{2} + r^{2}\dot{\theta}^{2})\): the radial velocity is \(\dot{r}\) and the transverse velocity is \(r\dot{\theta}\). The other options either use \(r\) where \(\dot{r}\) belongs or drop a power of \(r\), and are dimensionally inconsistent.

Q: The number of degrees of freedom for an astronaut constrained to move on surface of a spherical spaceship freely moving in space, is :

The astronaut has 5 degrees of freedom.The spaceship moves freely, so the position of its centre needs three coordinates. The astronaut is constrained to the spherical surface, a constraint of fixed distance from the centre, so his position on it needs only two more, such as the angles \(\theta\) and \(\phi\). In all, 3 + 2 = 5 independent coordinates fix the astronaut's position.Six would count the astronaut as a free particle with the ship's three, ignoring the constraint, or would add the ship's rotation, which does not change where on the sphere he can be. Three ignores the motion of the ship, and four miscounts the surface coordinates.

Q: A particle falls vertically under the action of gravity and a frictional force obtainable from dissipation function \(G(v) = \frac{1}{2}kv^{2}\), k is constant. The Lagrange's equation of motion is :

The equation is \(m\ddot{y} + k\dot{y} + mg = 0\), with y measured upward.With a dissipation function G, Lagrange's equation becomes\[ \begin{aligned} &\frac{d}{dt}\frac{\partial L}{\partial \dot{y}} \\ &\quad - \frac{\partial L}{\partial y} + \frac{\partial G}{\partial \dot{y}} = 0 \end{aligned} \]Taking \(L = \frac{1}{2}m\dot{y}^{2} - mgy\):\[ \begin{aligned} \frac{d}{dt}(m\dot{y}) + mg + k\dot{y} &= 0 \\ m\ddot{y} + k\dot{y} + mg &= 0 \end{aligned} \]The friction term \(k\dot{y}\) always opposes the velocity, so it enters with the same sign as \(m\ddot{y}\). Option D omits friction altogether, and options B and C give the friction term the wrong sign, which would make the resisting force speed the particle up.

Q: Which of the following statements is correct ?

The correct statement is that \(n\) particles with \(k\) holonomic constraints have \(3n - k\) degrees of freedom.Each free particle in space needs three coordinates, so \(n\) particles need \(3n\). A holonomic constraint is an equation \(f(\mathbf{r}_{1}, \ldots, \mathbf{r}_{n}, t) = 0\) among the coordinates, and each independent one lets us eliminate one coordinate. So \(3n - k\) independent generalised coordinates remain.A finite system of particles cannot have infinitely many degrees of freedom (that needs a continuum). \(n - k\) forgets the three dimensions of space, and 6 is the number for a single rigid body, not for \(n\) particles in general.

Q: The Lagrangian of a dynamical system is given by :\[ \begin{aligned} L(q_{1}, q_{2}, \dot{q}_{1}, \dot{q}_{2}) &= \frac{m}{2}(\dot{q}_{1}^{2} + \dot{q}_{2}^{2}) \\ &+ a(q_{1}\dot{q}_{2} - q_{2}\dot{q}_{1}) \end{aligned} \]where \(m\) and \(a\) are constants. Then canonical momenta of the system are given by :

The momenta are \(p_{1} = m\dot{q}_{1} - aq_{2}\), \(p_{2} = m\dot{q}_{2} + aq_{1}\).The canonical momentum is \(p_{i} = \partial L / \partial \dot{q}_{i}\). Differentiating the two parts of L:\[ \begin{aligned} p_{1} &= m\dot{q}_{1} + a(-q_{2}) = m\dot{q}_{1} - aq_{2} \\ p_{2} &= m\dot{q}_{2} + a(q_{1}) = m\dot{q}_{2} + aq_{1} \end{aligned} \]The term \(a(q_{1}\dot{q}_{2} - q_{2}\dot{q}_{1})\) contributes \(-aq_{2}\) to \(p_{1}\) and \(+aq_{1}\) to \(p_{2}\), as for a charged particle in a magnetic field. Option A ignores this term, option B has its signs reversed, and option C keeps only the velocity-dependent pieces wrongly.

Q: A Lagrangian of a particle of mass \(m\) is given by\[ \begin{aligned} L &= \frac{1}{2}m\dot{r}^{2} \\ &\quad - \frac{1}{r}\left(1 + \frac{\dot{r}^{2}}{c^{2}}\right), \end{aligned} \]\(c\) is a constant. Then :

The correct statement is H is conserved.The Lagrangian does not depend explicitly on time, so the Hamiltonian (the Jacobi integral) is a constant of the motion: \(dH/dt = -\partial L/\partial t = 0\).H is not the total energy here, because the potential depends on the velocity. From the previous computation,\[ \begin{aligned} H &= \frac{1}{2}m\dot{r}^{2} - \frac{\dot{r}^{2}}{rc^{2}} + \frac{1}{r} \\ E &= T + U \\ &= \frac{1}{2}m\dot{r}^{2} + \frac{\dot{r}^{2}}{rc^{2}} + \frac{1}{r} \end{aligned} \]They differ by \(2\dot{r}^{2}/(rc^{2})\), which changes along the motion. So 'H represents the total energy' is false, E is not conserved, and 'both H and E are conserved' fails with it. \(H = T + U\) needs a potential independent of velocity and a time-independent constraint.

11.

The Lagrangian of a particle of mass \(m\) in spherical polar co-ordinates is given by :

\[ \begin{aligned} L &= \tfrac{1}{2}m(\dot{r}^{2} + r^{2}\dot{\theta}^{2} + r^{2}\sin^{2}\theta\,\dot{\phi}^{2}) \\ &\quad - mgl\cos\theta \end{aligned} \]

The quantity that is conserved is :

A.\(\dfrac{\partial L}{\partial \dot{r}}\)
B.\(\dfrac{\partial L}{\partial \dot{\theta}}\)
C.\(\dfrac{\partial L}{\partial \dot{\phi}}\)
D.\(\dfrac{\partial L}{\partial r}\)

Exam Year(s)

MH SET - 2019, SET Exam - 2019,

Answer: C

The conserved quantity is \(\partial L/\partial\dot{\phi}\).

\(\phi\) does not appear in L, only \(\dot{\phi}\) does, so \(\phi\) is a cyclic coordinate. Lagrange's equation for it gives

\[ \begin{aligned} \frac{d}{dt}\frac{\partial L}{\partial\dot{\phi}} &= \frac{\partial L}{\partial\phi} \\ &= 0 \end{aligned} \]

so \(p_{\phi} = mr^{2}\sin^{2}\theta\,\dot{\phi}\), the angular momentum about the vertical axis, is constant.

\(r\) and \(\theta\) both appear explicitly in L (through \(r^{2}\dot{\theta}^{2}\), \(\sin^{2}\theta\) and \(\cos\theta\)), so \(\partial L/\partial\dot{r}\) and \(\partial L/\partial\dot{\theta}\) are not conserved, and \(\partial L/\partial r\) is a generalised force, not a momentum.

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12.

For a Lagrangian \(L(q, \dot{q}, \ddot{q}, t)\), the equation motion is of the form :

A.\(\frac{d^{2}}{dt^{2}}\left(\frac{\partial L}{\partial \ddot{q}}\right)\) - \(\frac{\partial L}{\partial q}\) = 0
B.\(\frac{d^{2}}{dt^{2}}\left(\frac{\partial L}{\partial \ddot{q}}\right)\) - \(\frac{d}{dt}\left(\frac{\partial L}{\partial \dot{q}}\right)\) + \(\frac{\partial L}{\partial q}\) = 0
C.\(\frac{d^{2}}{dt^{2}}\left(\frac{\partial L}{\partial \ddot{q}}\right)\) + \(\frac{d}{dt}\left(\frac{\partial L}{\partial \dot{q}}\right)\) + \(\frac{\partial L}{\partial q}\) = 0
D.\(\frac{d^{2}}{dt^{2}}\left(\frac{\partial L}{\partial \ddot{q}}\right)\) - \(\frac{d}{dt}\left(\frac{\partial L}{\partial \dot{q}}\right)\) - \(\frac{\partial L}{\partial q}\) = 0

Exam Year(s)

MH SET - 2019, SET Exam - 2019,

Answer: B

The equation of motion is the \(\ddot{q}\) term, minus the \(\dot{q}\) term, plus \(\partial L/\partial q\), equal to zero: the Euler-Lagrange (Ostrogradsky) equation for a Lagrangian that depends on \(\ddot{q}\).

Varying the action with \(L(q, \dot{q}, \ddot{q}, t)\) and integrating by parts once for the \(\dot{q}\) term and twice for the \(\ddot{q}\) term gives the Ostrogradsky equation:

\[ \begin{aligned} &\frac{\partial L}{\partial q} - \frac{d}{dt}\frac{\partial L}{\partial \dot{q}} \\ &\quad + \frac{d^{2}}{dt^{2}}\frac{\partial L}{\partial \ddot{q}} = 0 \end{aligned} \]

The signs alternate because each integration by parts brings a minus sign. Rearranged with the second-derivative term first, this is option B. Option A drops the \(\dot{q}\) term, and options C and D get the alternating signs wrong.

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13.

A dynamical system consists of three particles \(m_{1}, m_{2}, m_{3}\) in motion in space. The distance between \(m_{1}\) and \(m_{2}\), \(m_{2}\) and \(m_{3}\), \(m_{1}\) and \(m_{3}\) remain constant throughout the motion. The motion will be governed by \(n\) Euler Lagrange equations. Then :

A.\(n = 7\)
B.\(n = 6\)
C.\(n = 9\)
D.\(n = 3\)

Exam Year(s)

MH SET - 2020, SET Exam - 2020,

Answer: B

The number of equations is \(n = 6\).

Three free particles in space have \(3 \times 3 = 9\) coordinates. The three fixed distances \(m_{1}m_{2}\), \(m_{2}m_{3}\) and \(m_{1}m_{3}\) are three holonomic constraints, leaving

\[ 9 - 3 = 6 \]

degrees of freedom. The system is a rigid body (three non-collinear points fix it), which indeed has 3 translational and 3 rotational degrees of freedom. There is one Euler-Lagrange equation per generalised coordinate, so 6 equations. 9 ignores the constraints, 3 counts only the constraints, and 7 has no basis.

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14.

A particle of mass \(m\) is moving in a two-dimensional space. Its kinetic energy in terms of polar co-ordinates is given by :

A.\(\frac{1}{2}m(r^{2} + r^{2}\dot{\theta}^{2})\)
B.\(\frac{1}{2}m(\dot{r}^{2} + r\dot{\theta}^{2})\)
C.\(\frac{1}{2}m(\dot{r}^{2} + r^{2}\dot{\theta}^{2})\)
D.\(\frac{1}{2}m(r^{2} + r\dot{\theta}^{2})\)

Exam Year(s)

MH SET - 2020, SET Exam - 2020,

Answer: C

The kinetic energy is \(\frac{1}{2}m(\dot{r}^{2} + r^{2}\dot{\theta}^{2})\).

With \(x = r\cos\theta\), \(y = r\sin\theta\):

\[ \begin{aligned} \dot{x} &= \dot{r}\cos\theta - r\dot{\theta}\sin\theta \\ \dot{y} &= \dot{r}\sin\theta + r\dot{\theta}\cos\theta \\ \dot{x}^{2} + \dot{y}^{2} &= \dot{r}^{2} + r^{2}\dot{\theta}^{2} \end{aligned} \]

So \(T = \frac{1}{2}m(\dot{r}^{2} + r^{2}\dot{\theta}^{2})\): the radial velocity is \(\dot{r}\) and the transverse velocity is \(r\dot{\theta}\). The other options either use \(r\) where \(\dot{r}\) belongs or drop a power of \(r\), and are dimensionally inconsistent.

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15.

The number of degrees of freedom for an astronaut constrained to move on surface of a spherical spaceship freely moving in space, is :

Exam Year(s)

MH SET - 2020, SET Exam - 2020,

Answer: A

The astronaut has 5 degrees of freedom.

The spaceship moves freely, so the position of its centre needs three coordinates. The astronaut is constrained to the spherical surface, a constraint of fixed distance from the centre, so his position on it needs only two more, such as the angles \(\theta\) and \(\phi\). In all, 3 + 2 = 5 independent coordinates fix the astronaut's position.

Six would count the astronaut as a free particle with the ship's three, ignoring the constraint, or would add the ship's rotation, which does not change where on the sphere he can be. Three ignores the motion of the ship, and four miscounts the surface coordinates.

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16.

A particle falls vertically under the action of gravity and a frictional force obtainable from dissipation function \(G(v) = \frac{1}{2}kv^{2}\), k is constant. The Lagrange's equation of motion is :

A.\(m\ddot{y} + k\dot{y} + mg = 0\)
B.\(m\ddot{y} - k\dot{y} - mg = 0\)
C.\(m\ddot{y} - k\dot{y} + mg = 0\)
D.\(m\ddot{y} + mg = 0\)

Exam Year(s)

MH SET - 2020, SET Exam - 2020,

Answer: A

The equation is \(m\ddot{y} + k\dot{y} + mg = 0\), with y measured upward.

With a dissipation function G, Lagrange's equation becomes

\[ \begin{aligned} &\frac{d}{dt}\frac{\partial L}{\partial \dot{y}} \\ &\quad - \frac{\partial L}{\partial y} + \frac{\partial G}{\partial \dot{y}} = 0 \end{aligned} \]

Taking \(L = \frac{1}{2}m\dot{y}^{2} - mgy\):

\[ \begin{aligned} \frac{d}{dt}(m\dot{y}) + mg + k\dot{y} &= 0 \\ m\ddot{y} + k\dot{y} + mg &= 0 \end{aligned} \]

The friction term \(k\dot{y}\) always opposes the velocity, so it enters with the same sign as \(m\ddot{y}\). Option D omits friction altogether, and options B and C give the friction term the wrong sign, which would make the resisting force speed the particle up.

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17.

Which of the following statements is correct ?

A.A system of \(n\) particles can have infinite degrees of freedom
B.A system of \(n\) particles with \(k\) holonomic constraints has \(3n - k\) degrees of freedom
C.A system of \(n\) particles with \(k\) constraints has \(n - k\) degrees of freedom
D.A system of \(n\) particles has 6 degrees of freedom

Exam Year(s)

MH SET - 2021, SET Exam - 2021,

Answer: B

The correct statement is that \(n\) particles with \(k\) holonomic constraints have \(3n - k\) degrees of freedom.

Each free particle in space needs three coordinates, so \(n\) particles need \(3n\). A holonomic constraint is an equation \(f(\mathbf{r}_{1}, \ldots, \mathbf{r}_{n}, t) = 0\) among the coordinates, and each independent one lets us eliminate one coordinate. So \(3n - k\) independent generalised coordinates remain.

A finite system of particles cannot have infinitely many degrees of freedom (that needs a continuum). \(n - k\) forgets the three dimensions of space, and 6 is the number for a single rigid body, not for \(n\) particles in general.

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18.

The Lagrangian of a dynamical system is given by :

\[ \begin{aligned} L(q_{1}, q_{2}, \dot{q}_{1}, \dot{q}_{2}) &= \frac{m}{2}(\dot{q}_{1}^{2} + \dot{q}_{2}^{2}) \\ &+ a(q_{1}\dot{q}_{2} - q_{2}\dot{q}_{1}) \end{aligned} \]

where \(m\) and \(a\) are constants. Then canonical momenta of the system are given by :

A.\(p_{1} = m\dot{q}_{1},\ p_{2} = m\dot{q}_{2}\)
B.\(p_{1} = m\dot{q}_{1} + aq_{2},\ p_{2} = m\dot{q}_{2} - aq_{1}\)
C.\(p_{1} = a\dot{q}_{2},\ p_{2} = -a\dot{q}_{1}\)
D.\(p_{1} = m\dot{q}_{1} - aq_{2},\ p_{2} = m\dot{q}_{2} + aq_{1}\)

Exam Year(s)

MH SET - 2021, SET Exam - 2021,

Answer: D

The momenta are \(p_{1} = m\dot{q}_{1} - aq_{2}\), \(p_{2} = m\dot{q}_{2} + aq_{1}\).

The canonical momentum is \(p_{i} = \partial L / \partial \dot{q}_{i}\). Differentiating the two parts of L:

\[ \begin{aligned} p_{1} &= m\dot{q}_{1} + a(-q_{2}) = m\dot{q}_{1} - aq_{2} \\ p_{2} &= m\dot{q}_{2} + a(q_{1}) = m\dot{q}_{2} + aq_{1} \end{aligned} \]

The term \(a(q_{1}\dot{q}_{2} - q_{2}\dot{q}_{1})\) contributes \(-aq_{2}\) to \(p_{1}\) and \(+aq_{1}\) to \(p_{2}\), as for a charged particle in a magnetic field. Option A ignores this term, option B has its signs reversed, and option C keeps only the velocity-dependent pieces wrongly.

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19.

For a particle the kinetic energy and the potential energy is given by

\[ \begin{aligned} T &= \frac{1}{2}m\dot{r}^{2}, \\ U &= \frac{1}{r}\left(1 + \frac{\dot{r}^{2}}{c^{2}}\right), \end{aligned} \]

where \(c\) is a constant. Then the Hamiltonian H of the particle is :

A.\(H = T + U\)
B.\(H = P_{r}\dot{r}^{2} - \frac{1}{2}m\dot{r}^{2}\) \(+ \frac{1}{r}\left(1 + \frac{\dot{r}^{2}}{c^{2}}\right)\)
C.\(H = \frac{1}{2}\left(\frac{p_{r}^{2} rc^{2}}{(mrc^{2}-2)} - \frac{1}{r}\right)\)
D.\(H = \frac{1}{2}\frac{p_{r}^{2} r c^{2}}{(mrc^{2} - 2)} + \frac{1}{r}\)

Exam Year(s)

MH SET - 2023, SET Exam - 2023,

Answer: D

The Hamiltonian is \(H = \dfrac{1}{2}\dfrac{p_{r}^{2}\, r c^{2}}{(mrc^{2} - 2)} + \dfrac{1}{r}\).

The Lagrangian is \(L = T - U = \frac{1}{2}m\dot{r}^{2} - \frac{1}{r} - \frac{\dot{r}^{2}}{rc^{2}}\). The momentum is

\[ \begin{aligned} p_{r} &= \frac{\partial L}{\partial \dot{r}} = \dot{r}\left(m - \frac{2}{rc^{2}}\right) \\ &= \dot{r}\, \frac{mrc^{2} - 2}{rc^{2}} \end{aligned} \]

Then

\[ \begin{aligned} H &= p_{r}\dot{r} - L \\ &= \frac{1}{2}\dot{r}^{2}\left(m - \frac{2}{rc^{2}}\right) + \frac{1}{r} \end{aligned} \]

and substituting \(\dot{r} = p_{r} rc^{2}/(mrc^{2} - 2)\):

\[ \begin{aligned} H &= \frac{1}{2}\, \frac{p_{r}^{2}\, rc^{2}}{mrc^{2} - 2} \\ &\quad + \frac{1}{r} \end{aligned} \]

\(H = T + U\) fails because the potential depends on velocity. Option B writes \(p_{r}\dot{r}^{2}\) instead of \(p_{r}\dot{r}\) and is not expressed in \(p_{r}\), and option C has the wrong sign on \(1/r\).

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20.

A Lagrangian of a particle of mass \(m\) is given by

\[ \begin{aligned} L &= \frac{1}{2}m\dot{r}^{2} \\ &\quad - \frac{1}{r}\left(1 + \frac{\dot{r}^{2}}{c^{2}}\right), \end{aligned} \]

\(c\) is a constant. Then :

A.the Hamiltonian H represents the total energy E
B.H is conserved
C.E is conserved
D.Both H and E are conserved

Exam Year(s)

MH SET - 2023, SET Exam - 2023,

Answer: B

The correct statement is H is conserved.

The Lagrangian does not depend explicitly on time, so the Hamiltonian (the Jacobi integral) is a constant of the motion: \(dH/dt = -\partial L/\partial t = 0\).

H is not the total energy here, because the potential depends on the velocity. From the previous computation,

\[ \begin{aligned} H &= \frac{1}{2}m\dot{r}^{2} - \frac{\dot{r}^{2}}{rc^{2}} + \frac{1}{r} \\ E &= T + U \\ &= \frac{1}{2}m\dot{r}^{2} + \frac{\dot{r}^{2}}{rc^{2}} + \frac{1}{r} \end{aligned} \]

They differ by \(2\dot{r}^{2}/(rc^{2})\), which changes along the motion. So 'H represents the total energy' is false, E is not conserved, and 'both H and E are conserved' fails with it. \(H = T + U\) needs a potential independent of velocity and a time-independent constraint.

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