Physics 410 | First Hour Exam | March 3, 2006 10-10:50 am |

Picture temporarily Omitted

- The figure shows a strange and probably useless vehicle with
an unusual spring arrangement. The wheels and the "payload" are
suspended at their respective centers of mass, and can rotate
without friction. Their masses and moments of inertia are as shown
in the figure (the wheels have negligible moment of inertia). The
spring has zero unstretched length and has
spring constant k, and the connecting rods
are massless, of length
*l*and hinged on frictionless bearings at all connecting points.- Give the number of degrees of freedom of this system

3: for example, move left wheel, move right wheel, rotate payload

- Choose your generalized coordinates and explain what they
are - most simply by drawing appropriate arrows and letters
(thus: ¬ x ®) on the figure. Use
foresight with respect to part (e)
See figure. These are the x- and y- coordinates of the top mass, and the angle φ of tilt of that mass. The foresight anticipates that x and φ will be cyclic.

- Find the system's potential energy in terms of your chosen coordinates.

The P.E. of M is just Mgy, since it is suspended at the center of mass (which is also the center of gravity). For the P.E. of the spring we need to know the amount of its stretch, which is 2(

*l*^{2}- y^{2})^{½}, so that energy is^{1}/_{2}k·4·(*l*^{2}-y^{2}), giving a total energy ofU = Mgy + 2k( *l*^{2}- y^{2}) - Find
the equilibrium of the system. Does the equilibrium condition determine all the generalized coordinates uniquely? Interpret physically.
Since you have U, the best way to do this is by putting ÑU = 0:

Since U does not depend on x or f, its derivatives in those directions is automatically zero, so the whole gradient is indeed zero; but also it does not¶U ¶y= Mg - 4ky = 0 therefore y _{equil}= Mg/4k*determine*x and f.

If you chose a different coordinate, your condition will come out in terms of that coordinate. If you chose the position of the two wheels, call them x_{1}and x_{2}, U will depend only on x_{2}- x_{1}. If you did not notice that, it will still be correct to write ¶U/¶x_{1}= 0*and*¶U/¶x_{2}= 0 - you will get the same condition twice. - Find the equations of motion, with particular
attention to conserved quantities

We already know U, and T

_{M}=^{1}/_{2}M(x^{·2}+ y^{·2}). For the wheels you can either use the fact that the K.E. = K. E. of center of mass + K.E. relative to C.M., the latter determined by y because the distance of each wheel from C.M. (of the*wheels*) is d = (*l*^{2}- y^{2})^{½}, or you work out^{1}/_{2}m[(x-d)^{·2}+ (x+d)^{·2}] =^{1}/_{2}m[2x^{·2}+ 2d^{·2}] and find d^{·}from the chain rule, with the result (leaving out the constant term in U)

Since x and f are cyclic, pL = ( ^{1}/_{2}M + m)×x2

+ æ

è^{1}/_{2}M +my ^{2}*l*^{2}-y^{2}ö

ø×y2

+ ^{1}/_{2}I×f2

- Mgy + 2k y ^{2}_{x}and p_{f}are conserved:

The y-equation isp _{x}= (M + 2m)×x= const, p _{f}= I×f= const d dté

ëæ

èM + 2my ^{2}*l*^{2}-y^{2}ö

ø×yù

û+ Mg - 4ky = 0. - If there is another
conserved quantity that did not appear in (e), give it now

Teacher was obviously thinking of the total energy (Hamiltonian, derived from L by a suitable sign change to get T

**+**U),

if it seems strange to see the spring energy negative, note that around y = 0 the spring indeed acts as if it had a negative spring constant, it pushes M away from y = 0.E = ( ^{1}/_{2}M + m)×x2

+ æ

è^{1}/_{2}M +my ^{2}*l*^{2}-y^{2}ö

ø×y2

+ ^{1}/_{2}I×f2

+ Mgy - 2k y ^{2}

- Give the number of degrees of freedom of this system
- A certain system of one degree of freedom has generalized
coordinate q, which "lives" on the real line, -¥ < q < +¥. (In other words, q is not periodic, q and q + 2p
represent different configurations of the system.) The equation of
motion of the system is

which can also be written as (big hint)××q= gcosq + ×q2

cosq sinq 1+cos^{2}q(*) d dtæ

è×q(1+cos ^{2}q)ö

ø+ ×q2

cosq sinq- gcosq = 0 . (**) - Find a possible Lagrangian from which this equation can be derived.
The item after the d/dt is clearly ¶L/¶q
^{·}, and it easily integrated (almost everyone recognized that this is easy because for this*partial*derivative, the "undotted" q's are regarded as constant - molodtsy! as they say in Russian, that is, way to go guys!). So we have L =^{1}/_{2}q^{·2}(1+cos^{2}q) + f(q) where f(q) is some unknown function of q. Substitute this into (**) to find ¶f/¶q = g cos q hence f = g sin q and

This is not unique. You could add to it dF(q)/dt where F is any function of q without changing the equation of motion.L = ^{1}/_{2}(1+cos^{2}q)×q2

+ g sin q - Describe and/or draw a possible physical system that obeys this equation of motion, and mention what the coordinate q is in your system. One straightforward solution is a particle sliding on a wire bent into a vertical sine curve, with q the particle's horizontal position. Another is a particle moving in a vertical circle coupled to a particle moving horizontally directly below the first particle, with q the angle of the first particle (though this would violate the provision that q should not be periodic).

**Safety net**: If the above seems impossible (even though it is not), then*show*that for*small*q (to linear order in q) the equation of motion (*) reduces to

and solve parts (a) and (b) for that case.××q= g/2 - Find a possible Lagrangian from which this equation can be derived.
The item after the d/dt is clearly ¶L/¶q

File translated from T

On 06 Mar 2006, 21:58.