- #1
Wavefunction
- 99
- 4
Homework Statement
A massless spring of length [itex] b [/itex] and spring constant [itex] k [/itex] connects two particles of masses [itex] m_1 [/itex] and [itex] m_2 [/itex].
The system rests on a smooth table and may oscillate and rotate.
a) Determine the Lagrange's equations of motion.
b) What are the generalized momenta associated with any cyclic coordinates?
Hint: With the right choice of generalized coordinates, 3 of the 4 should be cyclic. If the force
between the masses is a central force, some of the coordinates we used earlier in the course to
describe central-force motion might be useful.
c) Determine Hamilton's equations of motion.
Homework Equations
eq(1) [itex] \frac{∂L}{∂q_j}-\frac{∂}{∂t}\frac{∂L}{∂\dot{q_j}}=0 [/itex]
eq(2) [itex] H=\sum_ip_i\dot{q_i}-L = T+U[/itex]
The Attempt at a Solution
Part A
Step 1) find the Lagrangian
[itex] T = \frac{m_1}{2}[\dot{r_1}^2+r_1^2\dot{θ_1}^2]+\frac{m_2}{2}[\dot{r_2}^2+r_2^2\dot{θ_2}^2] [/itex]
[itex] U = \frac{k}{2}[(r_1-b)^2+(r_2-b)^2] [/itex]
[itex] L= T-U = \frac{m_1}{2}[\dot{r_1}^2+r_1^2\dot{θ_1}^2]+\frac{m_2}{2}[\dot{r_2}^2+r_2^2\dot{θ_2}^2]-\frac{k}{2}[(r_1-b)^2+(r_2-b)^2] [/itex]
Step 2) Use eq(1) to determine Lagrange's eqs of motion:
[itex] \frac{∂L}{∂r_1} = m_1r_1\dot{θ_1}^2-k(r_1-b) [/itex]
[itex] \frac{d}{dt}\frac{∂L}{∂\dot{r_1}} = \frac{d}{dt}[m_1\dot{r_1}] =m_1\ddot{r_1} [/itex]
eq(3) [itex] m_1r_1\dot{θ_1}^2-k(r_1-b)-m_1\ddot{r_1}= 0 [/itex]
[itex] \frac{∂L}{∂θ_1} = 0 [/itex]
[itex] \frac{d}{dt}\frac{∂L}{∂\dot{θ_1}} = \frac{d}{dt}[m_1r_1^2\dot{θ_1}] = m_1[2r_1\dot{r_1}\dot{θ_1}+r_1^2\ddot{θ_1}] [/itex]
eq(4) [itex] -m_1[2r_1\dot{r_1}\dot{θ_1}+r_1^2\ddot{θ_1}]=0 [/itex]
Going through the same steps for [itex] m_2 [/itex]
eq(5) [itex] m_2r_2\dot{θ_2}^2-k(r_2-b)-m_1\ddot{r_2}= 0 [/itex]
eq(6) [itex] -m_2[2r_2\dot{r_2}\dot{θ_2}+r_2^2\ddot{θ_2}]=0 [/itex]
Now I can apply the given constraints: the system is only allowed to rotate on the frictionless table top and oscillate.
eq(a) [itex] θ_1 + π = θ_2 [/itex] (This basically means that spring can only extend/compress along the radial direction)
eq(b) [itex] r_1 = a^2r_2 [/itex] (Since the system can't translate radially this condition along with eq(a) ensures that the particles oscillating.)
From both of these constraints: [itex] \dot{θ_1}=\dot{θ_2} [/itex], [itex] \ddot{θ_1}=\ddot{θ_2} [/itex], [itex] \dot{r_1}=a^2\dot{r_2} [/itex], and [itex] \ddot{r_1}=a^2\ddot{r_2} [/itex].
Part B
They are constants (e.g [itex] \frac{d}{dt}[m_1r_1^2\dot{θ_1}] = 0 [/itex] )
Part C
eq(7) [itex] \dot{q_i} = \frac{∂H}{∂p_i} [/itex]
eq(8) [itex] \dot{p_i} = -\frac{∂H}{∂q_i} [/itex]
The Hamiltonian is:
[itex] H = \frac{m_1}{2}[\dot{r_1}^2+r_1^2\dot{θ_1}^2]+\frac{m_2}{2}[\dot{r_2}^2+r_2^2\dot{θ_2}^2] + \frac{k}{2}[(r_1-b)^2+(r_2-b)^2] [/itex]
For [itex] r_1 [/itex]:
eq(7) yields: [itex] \dot{r_1} = \frac{p_1}{m_1} [/itex]
eq(8) yields: [itex] \dot{p_1} = m_1r_1\dot{θ_1}+k(r_1-b) [/itex]
For [itex] r_2 [/itex]
eq(7) yields: [itex] \dot{r_2} = \frac{p_2}{m_2} [/itex]
eq(8) yields: [itex] \dot{p_2} = m_2r_2\dot{θ_2}+k(r_2-b) [/itex]
For [itex] θ_1 [/itex] (L, in the following parts stands for angular momentum, NOT the Lagrangian)
eq(7) yields: [itex] \dot{θ_1} = \frac{L_1}{m_1r_1^2} [/itex]
eq(8) yields: [itex] \dot{L_1} = 0 [/itex]
For [itex] θ_2 [/itex]
eq(7) yields: [itex] \dot{θ_2} = \frac{L_2}{m_2r_2^2} [/itex]
eq(8) yields: [itex] \dot{L_2} = 0 [/itex]
Thank you in advance for checking my work(: