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(: