Physics Problems & Solutions: Lagrangian

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Classical Mechanics - Lagrangian

A bead is constrained to slide on a frictionless rod that is fixed at an angle θ with a vertical axis and is rotating with angular frequency ω about the axis, as shown above. Taking the distance s along the rod as the variable, the Lagrangian for the bead is equal to

A. ½ mṡ ² − mgs cos θ
B. ½ mṡ ² + ½ m(ωs)² − mgs
C. ½ mṡ ² + ½ m(ωs cos θ)² + mgs cos θ
D. ½ m(ṡ sin θ)² − mgs cos θ

E. ½ mṡ ² + ½ m(ωs sin θ)² − mgs cos θ

(GR9677 #68)

Solution:

Lagrangian: L = T − U

Potential energy: U = mgh = mgs cos θ
Kinetic Energy: T_kin = ½ mṡ ²
Rotational kinetic energy: T_rot= ½ Iω²
with moment inertia: I = mr² = m(s sin θ)²
→ T_rot= ½ m(ωs sin θ)²

L = T_kin+ T_rot− U = ½ mṡ ² + ½ m(ωs sin θ)² − mgs cos θ

Answer: E

Classical Mechanics - Lagrangian

If ∂L/∂q_n = 0, where L is the Lagrangian for a conservative system without constraints and q_n is a generalized coordinate, then the generalized momentum p_n is

A. An ignorable coordinate
B. Constant
C. Undefined
D. Equal to $\frac{d}{dt} \left (\frac{\partial L}{\partial q_n} \right )$
E. Equal to the Hamiltonian for the system

(GR9277 #43)

Solution:

Lagrangian Equation of Motion for the generalized coordinate q:

$\frac{\partial L}{\partial q}=\frac{\partial}{\partial t}\frac{\partial L}{\partial \dot q}$

Momentum,

$\begin{aligned} p&=mv=m\dot{x} \\L&=T-U=\frac{1}{2}m\dot{x}^2-mgy\\&\rightarrow \frac{\partial L}{\partial \dot{x}}=m\dot{x}=p \end{aligned}$

For the generalized coordinate q:

If $\frac{\partial L}{\partial q_n}=0 \rightarrow \frac{\partial L}{\partial \dot{q_n}}=p_n= \text{constant}$

Answer: B

Classical Mechanics - Lagrangian

A particle of mass m on the Earth’s surface is confined to move on the parabolic curve y = ax², where y is up. Which of the following is a Lagrangian for the particle?

A. $L = \frac{1}{2} m\dot{y} ^2 \left (1+\frac{1}{4ay} \right )-mgy$

B. $L = \frac{1}{2} m\dot{y} ^2 \left (1-\frac{1}{4ay} \right )-mgy$

C. $L = \frac{1}{2} m\dot{x} ^2 \left (1+\frac{1}{4ay} \right )-mgx$

D. $L = \frac{1}{2} m\dot{x} ^2 \left (1+4a^2x^2 \right )+mgx$

E. $L = \frac{1}{2} m\dot{x} ^2 +\frac{1}{2} m\dot{y} ^2+mgy$

(GR9277 #44)

Solution:

Kinetic Energy, $T = \frac{1}{2} m(\dot{x} ^2 +\dot{y} ^2)$

Potential Energy,

Lagrangian, $L=T-U=\frac{1}{2} m(\dot{x} ^2 +\dot{y} ^2)-mgy$

Given the curve y = ax²,

$\begin{aligned} \dot{y}&=\frac{dy}{dt}=\frac{dy}{dx}\frac{dx}{dt}=2ax\dot{x}\\\rightarrow \dot{x}&=\frac{\dot{y}}{2ax} \\\dot{x}^2&=\frac{\dot{y}^2}{4a^2x^2}=\frac{\dot{y}^2}{4ay} \end{aligned}$

$\begin{aligned} L&=\frac{1}{2} m\left (\frac{\dot{y} ^2}{4ay}+\dot{y}^2 \right ) -mgy \\&=\frac{1}{2} m\dot{y} ^2\left (\frac{1}{4ay}+1 \right ) -mgy \end{aligned}$

Answer: A

Classical Mechanics - Hamiltonian

A. (p²/2m) + kx⁴
B. (p²/2m) − kx⁴
C. kx⁴
D. ½mv² − kx⁴
E. ½mv²

(GR8677 #35)

Solution:

Hamiltonian: H = T + U
U = kx⁴
T = ½mv² = p²/2m
H = (p²/2m) + kx⁴

Answer: A

Classical Mechanics - Lagrangian and Hamiltonian

Question 34-36

The potential energy of a body constrained to move on a straight line is kx⁴ where k is a constant. The position of the body is x, its speed v, its linear momentum p, and its mass m.

The body moves from x₁ at time t₁ to x₂ at time t₂. Which of the following quantities is an extremum for the x-t curve corresponding to this motion, if end points are fixed?

A. $\int_{t_1}^{t_2}\bigg(\frac{1}{2}mv^2-kx^4\bigg)dt$
B. $\int_{t_1}^{t_2}\frac{1}{2}mv^2dt$
C. $\int_{t_1}^{t_2}mxv$ $dt$
D. $\int_{x_1}^{x_2}\bigg(\frac{1}{2}mv^2+kx^4\bigg)dx$
E. $\int_{x_1}^{x_2}mv$ $dx$

(GR8677 #36)

Solution:

$\\$Lagrangian: $L=T-U \\$Hamiltonian: $ H=\int_{t_1}^{t_2} L dt \\U=kx^4 \\\\T=\frac{1}{2} mv^2 \\\rightarrow H=\int_{t_1}^{t_2}\bigg(\frac{1}{2} mv^2-kx^4\bigg)dt$

Answer: A

Classical Mechanics - Lagrangian

The Lagrangian for a mechanical system is $L=a\dot q^2+bq^4$ where q is a generalized coordinate and a and b are constants. The equation of motion for this system is

$\\\\\\$A.$ \hspace{5 mm} \dot q=\sqrt{\frac{b}{a}}q^2\\\\$B.$ \hspace{5 mm} \dot q=\frac{2b}{a} q^3 \\\\$C.$ \hspace{5 mm}\ddot q=-\frac{2b}{a} q^3 \\\\$D.$ \hspace{5 mm} \ddot q=+\frac{2b}{a} q^3 \\\\$E.$ \hspace{5 mm} \ddot q=\frac{b}{a} q^3$

(GR0177 #74)

Solution:

The Lagrangian equation of motion for the generalized coordinate q:

$\frac{\partial L}{\partial q}=\frac{\partial}{\partial t}\frac{\partial L}{\partial \dot q}$

$\\\frac{\partial L}{\partial q}=\frac{\partial}{\partial q} (a\dot q^2+bq^4 )=0+\frac{\partial}{\partial q} bq^4=4bq^3\\\\\frac{\partial L}{\partial \dot q}=\frac{\partial}{\partial \dot q} (a\dot q^2+bq^4 )=\frac{\partial}{\partial \dot q} a\dot q^2+0=2a\dot q \\\\\\\frac{\partial}{\partial t}\frac{\partial L}{\partial \dot q}=\frac{\partial}{\partial t} 2a\dot q=2a \ddot q \\\\ \frac{\partial L}{\partial q}=\frac{\partial}{\partial t}\frac{\partial L}{\partial \dot q} \rightarrow 4bq^3=2a\ddot q \rightarrow \ddot q=\frac{2b}{a} q^3$

Answer: D