A point mass m is connected by a massless rod of length L to a fixed point of support, such that the mass is free to rotate about the support in any direction under the influence of
gravity.
(a) Construct a Lagrangian in terms of suitable generalised coordinates, and hence write down the equations of motion.
(b) Consider the special case in which the mass moves on a circular path, the rod making a constant angle £c0 with the vertical. Find the angular velocity of motion around the circle.
test
(a) To construct the Lagrangian, let's consider the rotational motion of the mass. Let θ be the angle the rod makes with the vertical and φ be the angle the mass makes with the rod.
The position vector of the mass can be written as:
r = L(sinθ cosφ, sinθ sinφ, -cosθ)
The velocity of the mass can be calculated by taking the time derivative of r:
v = L(dθ/dt cosθ cosφ - θ̇ sinθ sinφ, dθ/dt cosθ sinφ + θ̇ sinθ cosφ, θ̇ sinθ)
The kinetic energy T of the system can be calculated as the sum of the translational and rotational kinetic energies:
T = (1/2)m|v|^2 + (1/2)Iθ̇^2
Where m is the mass of the point mass, I is the moment of inertia of the point mass about the axis of rotation (which is L^2/3 for a mass m at distance L from the axis), θ̇ is the time derivative of θ.
The potential energy V of the system can be determined by considering the height of the mass above the fixed point of support:
V = mgh = -mgLcosθ
The Lagrangian L is defined as L = T - V:
L = (1/2)m|v|^2 + (1/2)Iθ̇^2 + mgLcosθ
Now, let's calculate the equations of motion using the Euler-Lagrange equations:
d/dt(dL/dθ̇) - dL/dθ = 0
d/dt(dL/dφ̇) - dL/dφ = 0
For the first equation:
d/dt(dL/dθ̇) = d/dt(Iθ̇) = Iθ̈
dL/dθ = -mgLsinθ
So, the first equation becomes:
Iθ̈ + mgLsinθ = 0
For the second equation:
d/dt(dL/dφ̇) = d/dt(0) = 0
dL/dφ = 0
So, the second equation becomes:
0 = 0
(b) In the special case where the mass moves on a circular path, sinθ = sinθ₀ = constant (θ₀ is the angle between the rod and the vertical). Therefore, θ = θ₀ and θ̇ = 0.
Plugging these values into the equation of motion for θ:
Iθ̈ + mgLsinθ = 0
Iθ̈ + mgLsinθ₀ = 0
Since θ̇ = 0, the angular acceleration θ̈ = 0. Therefore, the equation simplifies to:
mgLsinθ₀ = 0
Since sinθ₀ ≠ 0, we can cancel it out:
mgL = 0
This equation implies that the gravitational force acting on the mass in the perpendicular direction to the circular path is zero. Therefore, the mass moves with a constant angular velocity, which is given by the equation:
θ̇ = 0
Therefore, the angular velocity of motion around the circle is zero.
To answer this question, we will follow a step-by-step approach. Firstly, let's define some variables:
- m: mass of the point mass
- L: length of the massless rod
- g: acceleration due to gravity
- θ: angle between the rod and the vertical axis (generalized coordinate)
- θ₀: constant angle of the rod with the vertical in the circular motion case
(a) Constructing the Lagrangian:
The Lagrangian, denoted by L, is defined as the difference between the kinetic energy and potential energy of the system.
1. Kinetic Energy (T):
The point mass has rotational kinetic energy, which is given by the equation:
T = (1/2) * I * ω²
where I is the moment of inertia and ω is the angular velocity.
For a point mass rotating freely at a distance L from the fixed support, the moment of inertia is given by:
I = m * L²
Substituting this value into the equation for kinetic energy:
T = (1/2) * m * L² * ω²
2. Potential Energy (V):
The potential energy of the system is due to the gravitational potential energy of the mass at height h, which is given by:
V = m * g * h
In the case of the point mass connected to the rod, the height h can be expressed as:
h = L * cos(θ)
Therefore, the potential energy can be written as:
V = m * g * L * cos(θ)
3. Lagrangian (L):
The Lagrangian is the difference between the kinetic and potential energy:
L = T - V
Substituting the equations for kinetic and potential energy into the Lagrangian:
L = (1/2) * m * L² * ω² - m * g * L * cos(θ)
(b) Equations of Motion:
To find the equations of motion, we need to use the Euler-Lagrange equations, which state that the derivative of the Lagrangian with respect to each generalized coordinate θ gives the equation of motion for that coordinate.
1. Partial derivative:
∂L/∂θ = m * g * L * sin(θ)
2. Time derivative:
d/dt (∂L/∂(dθ/dt)) = d/dt (m * L² * dθ/dt) = m * L² * d²θ/dt²
Applying the Euler-Lagrange equation:
d/dt (∂L/∂(dθ/dt)) = ∂L/∂θ
m * L² * d²θ/dt² = m * g * L * sin(θ)
Dividing throughout by m * L²:
d²θ/dt² = (g/L) * sin(θ)
These are the equations of motion for the system.
(c) Special Case: Circular Motion:
In the special case where the mass moves on a circular path, the rod makes a constant angle (θ₀) with the vertical.
In circular motion, the angular velocity (ω) is constant. Therefore, d²θ/dt² becomes 0.
Using the equation of motion derived earlier:
0 = (g/L) * sin(θ)
Since sin(θ₀) is constant and nonzero (0 != sin(θ₀)), we can deduce that θ = θ₀.
This means that in circular motion, the angle remains constant at θ₀.
Now, let's calculate the angular velocity (ω) in terms of the given variables:
Using the equation of motion and substituting θ = θ₀:
0 = (g/L) * sin(θ₀)
Since sin(θ₀) is nonzero (0 != sin(θ₀)), we can deduce that the only way for the equation to be satisfied is when g/L = 0.
Therefore, the angular velocity (ω) in circular motion is 0.