Lagrangian Formulation Questions and Answers

Answer: A

The Lagrangian is \(\frac{1}{2}ma^{2}(\dot{\theta}^{2} + \omega^{2}\sin^{2}\theta) + mga\cos\theta\).

The particle has two velocity components. Along the circle it moves with speed \(a\dot{\theta}\). The circle's rotation carries it round the vertical axis on a horizontal circle of radius \(a\sin\theta\), with speed \(a\omega\sin\theta\). These are perpendicular, so

\[ \begin{aligned} T &= \tfrac{1}{2}m(a\dot{\theta})^{2} + \tfrac{1}{2}m(a\omega\sin\theta)^{2} \\ &= \tfrac{1}{2}ma^{2}(\dot{\theta}^{2} + \omega^{2}\sin^{2}\theta). \end{aligned} \]

With \(\theta\) measured from the downward vertical, the height below the centre is \(a\cos\theta\), so \(V = -mga\cos\theta\) and \(L = T - V\) is the keyed expression.

The other options have the wrong kinetic energy: the rotational term must be \(\omega^{2}\sin^{2}\theta\), quadratic in \(\omega\), not \(2\omega\sin\theta\), \(2\omega^{2}\cos\theta\) or \(\omega\sin 2\theta\). They also use \(\sin\theta\) in the potential, which belongs to an angle measured from the horizontal.

Answer: A

The equations of motion are \(2\ddot{r} = r\dot{\theta}^{2} - g(1 - \cos\theta)\) and \(\frac{d}{dt}(r^{2}\dot{\theta}) = -g r \sin\theta\).

First simplify: \(L = m\dot{r}^{2} + \frac{1}{2}m r^{2}\dot{\theta}^{2} - m g r (1 - \cos\theta)\). Apply Lagrange's equations \(\frac{d}{dt}\frac{\partial L}{\partial \dot{q}} = \frac{\partial L}{\partial q}\).

\[ \begin{aligned} r:\ \tfrac{d}{dt}(2m\dot{r}) &= m r \dot{\theta}^{2} - m g (1 - \cos\theta) \\ \theta:\ \tfrac{d}{dt}(m r^{2}\dot{\theta}) &= -m g r \sin\theta \end{aligned} \]

Dividing by \(m\) gives the first option. The other options change the sign of the gravity term in one equation or the other: the potential \(m g r(1 - \cos\theta)\) enters \(L\) with a minus sign, so both generalised forces carry a minus sign.

Answer: D

The action is \(-\frac{1}{3}\sqrt{2g}\) J s.

Hamilton's action is \(S = \int (T - V)\,dt\). With \(m = 1\) kg, the particle falls 1 m in \(\tau = \sqrt{2/g}\), with \(v = gt\) and height \(h = 1 - \frac{1}{2}gt^{2}\):

\[ \begin{aligned} L &= \tfrac{1}{2}g^{2}t^{2} - g\left(1 - \tfrac{1}{2}gt^{2}\right) = g^{2}t^{2} - g \\ S &= \frac{g^{2}\tau^{3}}{3} - g\tau \end{aligned} \]

With \(\tau = \sqrt{2/g}\), \(g^{2}\tau^{3} = 2\sqrt{2g}\) and \(g\tau = \sqrt{2g}\), so

\[ S = \frac{2}{3}\sqrt{2g} - \sqrt{2g} = -\frac{1}{3}\sqrt{2g} \]

\(-\frac{2}{3}\sqrt{2g}\) is \(-\int V\,dt\) alone, dropping the kinetic part. The action is negative here because the potential energy outweighs the kinetic energy for most of the fall, which rules out the positive options.

Answer: A

The constant is \(\alpha = +\frac{k}{m}\).

The equations of motion are \(\ddot{x} = -\frac{k}{m}x\) and \(\ddot{y} = -\frac{k}{m}y\). Differentiate the candidate constant:

\[ \begin{aligned} \frac{d}{dt}(\dot{x}\dot{y} + \alpha xy) &= \ddot{x}\dot{y} + \dot{x}\ddot{y} + \alpha(\dot{x}y + x\dot{y}) \\ &= \left(\alpha - \frac{k}{m}\right)(x\dot{y} + \dot{x}y) \end{aligned} \]

This vanishes for every motion only if \(\alpha = k/m\). Physically, the isotropic oscillator conserves the energy of each Cartesian motion separately and also these cross terms, which is why it has more constants of motion than energy and angular momentum alone. \(-\frac{k}{m}\) and \(-\frac{2k}{m}\) have the wrong sign, and 0 leaves the \(-\frac{k}{m}(x\dot{y} + \dot{x}y)\) term.

Answer: D

(i) is not correct and (ii) is correct.

Adding a total time derivative \(\alpha\,df(q)/dt = \alpha f'(q)\,\dot{q}\) changes the canonical momentum:

\[ \begin{aligned} p &= \frac{\partial L}{\partial \dot{q}} \\ &= m\dot{q} + \alpha f'(q) \end{aligned} \]

so statement (i) is false. But the extra term contributes nothing to the Euler-Lagrange equation:

\[ \tfrac{d}{dt}(\alpha f'(q)) - \alpha f''(q)\dot{q} = 0 \]

The equation of motion stays \(m\ddot{q} = -m\omega^{2}q\), so (ii) is true. Lagrangians differing by a total time derivative give the same action up to end-point terms and hence the same dynamics, while the canonical momentum is shifted by a gauge-like term.

Answer: B

The answer is the motion along the z direction is simple harmonic.

Write \(\vec{r} = x\hat{\imath} + y\hat{\jmath} + z\hat{k}\). The constraint (the cylinder's surface, without friction) can only push perpendicular to the surface, that is horizontally, so it has no \(z\) component. Along \(z\) the only force is \(-kz\):

\[ m\ddot{z} = -kz \]

which is simple harmonic motion with \(\omega = \sqrt{k/m}\).

The force \(-k\vec{r}\) is conservative and the frictionless constraint does no work, so total energy is conserved. Linear momentum is not conserved, since a net force acts. The force \(-k\vec{r}\) exerts no torque about O, and the constraint force, which points towards the axis, has no torque about the axis, so the axial component of angular momentum is conserved; the other components only oscillate with \(z\). Nothing makes the angular momentum grow with time.

Answer: B

The Lagrangian is \(\dfrac{m}{2}(\dot{r}^{2} + r^{2}\dot{\theta}^{2}) + \dfrac{m}{r}\).

In plane polar coordinates the kinetic energy is \(T = \tfrac{m}{2}(\dot{r}^{2} + r^{2}\dot{\theta}^{2})\). An attractive force of magnitude \(m/r^{2}\) towards the origin comes from the potential \(V = -m/r\), since \(-dV/dr = -m/r^{2}\), a pull inwards. So

\[ \begin{aligned} L &= T - V \\ &= \frac{m}{2}(\dot{r}^{2} + r^{2}\dot{\theta}^{2}) + \frac{m}{r} \end{aligned} \]

Option A has the sign of \(V\) wrong, which would make the force repulsive. Option C uses \(m/r^{2}\), the force, in place of the potential, and option D has \(r\dot{\theta}^{2}\) for \(r^{2}\dot{\theta}^{2}\).

Answer: C

The equation of motion is \(\ddot{x} = \dfrac{m_{1} - m_{2}}{m_{1} + m_{2}}g\).

Lagrange's equation is \(\tfrac{d}{dt}L_{\dot{x}} - L_{x} = 0\), where subscripts denote partial derivatives. Here \(\partial L/\partial \dot{x} = (m_{1} + m_{2})\dot{x}\) and \(\partial L/\partial x = m_{1}g - m_{2}g\), so

\[ (m_{1} + m_{2})\ddot{x} = (m_{1} - m_{2})g \]

The heavier mass accelerates down with this reduced gravity. An equation of motion gives the acceleration \(\ddot{x}\), so the two options with \(\dot{x}\) are wrong, and inverting the fraction would give an acceleration greater than \(g\), which is impossible.

Answer: B

The Lagrangian is \(\frac{1}{2} m_{1}\dot{x}_{1}^{2} + \frac{1}{2} m_{2}\dot{x}_{2}^{2}\) \(- q_{1}q_{2}/|x_{1} - x_{2}|\).

The Lagrangian is \(L = T - V\). The kinetic energy of the two particles moving along x is \(T = \frac{1}{2}m_{1}\dot{x}_{1}^{2} + \frac{1}{2}m_{2}\dot{x}_{2}^{2}\). In esu (Gaussian) units the Coulomb potential energy of the pair is \(V = q_{1}q_{2}/|x_{1} - x_{2}|\), with no \(4\pi\epsilon_{0}\) factor. So

\[ L = T - \frac{q_{1}q_{2}}{|x_{1} - x_{2}|} \]

The option with \(+q_{1}q_{2}/|x_{1}-x_{2}|\) is the total energy \(T + V\), the Hamiltonian, not the Lagrangian. The two options with negative kinetic terms correspond to \(V - T\) and \(-(T+V)\), neither of which is the Lagrangian.

Answer: C

The keyed answer is \(r \gt 3n - k\).

For \(n\) free particles there are \(3n\) coordinates. A holonomic constraint is an equation among the coordinates, so it can be used to eliminate one of them: with \(k\) holonomic constraints the system needs exactly \(3n - k\) independent generalised coordinates.

A non-holonomic constraint restricts velocities (or is an inequality) and cannot be integrated into a relation among coordinates, so it cannot be used to eliminate one. The number \(r\) of independent coordinates the motion needs therefore stays larger than \(3n - k\); this is the textbook sense in which the paper uses 'degrees of freedom'.

\(r = 3n - k\) is the holonomic count, \(n - k\) forgets the three coordinates per particle, and \(r \lt 3n - k\) gets the direction wrong.