# A few question on torques and periodic motion.

1. May 10, 2007

### MathematicalPhysicist

i have two questions from kleppner's book, here it goes (iv'e attached a file which have the sketchs of the two problems):
6.17
A rod of length l and mass m, pivoted at one end, is held by a spring at its midpoint and a spring at its far end, both pulling in opposite directions.
The springs have spring constant k, and at equilibrium their pull is perpendicular to the rod. find the frequency of small oscillations about the equilibrium position.

6.18
Find the period of pendulum consisiting of a disk of mass M and radius R fixed to the end of a rod of length l and mass m. how does the period change if the disk is mounted to the rod by a frictionless bearing so that it is perfectly free to spin.

my attempt at solving the problems
6.17
i got to the next equations which as i see contradict each other:
$$(1)=N=cos(\theta)$$
$$(2)=m\frac{d^2x}{dt^2}=2kx-Nsin(\theta)$$
$$(3)=x=\frac{l\theta}{2}$$
the fourth equation is the equation of the torques around the point of contact of the rod with pivot:
$$(4)=\frac{mglsin(\theta)}{2}-2kxsin(\frac{\pi}{2}+\theta)=\frac{ml^2\frac{d^2\theta}{dt^2}}{3}$$
but obviously (4) contradicts 1+2, so where did i go wrong here?

6.18
without friction between the bearing and the disk we would have the equation of torques this way:
$$-(\frac{mglsin(\theta)}{2}+Mglsin(\theta))=I_0 \alpha$$
where:
$$I_0=MR^2/2+ml^2/12$$
and with friction, we would have include f*l.
the torques are calculated around the poivot of the rod with the cieling.

anyway, is something from what i wrote is correct? thwe problems is that for those questions there isn't even an answer clue.

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Last edited: May 10, 2007
2. May 10, 2007

### Staff: Mentor

With respect to 6.17, at first glance, 2 and 3 would appear to be incorrect. The springs will not have the same displacement, except when both springs have zero deflection, if that is possible.

Please refer to the statement "a spring at its midpoint and a spring at its far end, both pulling in opposite directions." The springs are pulling in opposite direction, so one must be in compression while the other is in tension, assuming both have zero deflection at equilibrium. The other part of the statement is "a spring at its [rod] midpoint (l/2) and a spring at its far end (l). So the springs will have different deflections, and x1 and x2, and if x2 is the bottom spring, then x2 = 2*x1.

3. May 10, 2007

### MathematicalPhysicist

ithink you meant the other way around, i.e x1=2x2, and i understand how did you get to that, thanks.
btw, what's wrong with (3)?

4. May 10, 2007

### denverdoc

I think using tan theta=theta for small angles is ok, and Astronuc's objection was again that the two displacements are not the same.

But my own conjecture is that you only need to treat this entirely as a rotational problem, eg

the spring at the top exerts a torque about the pivot=-k*(l*theta)x l

and the moment of inertia is used for both displacements, eg, something along lines of

d"(theta)/dt"=-k*L^2*tan(theta)*sin(pi/2+theta)/Iend-[similar expression for midpoint spring], and finally torque from gravity,

and noting tan=sin/cos and
sin(pi/2+theta)=cos(theta), the problem simplifies nicely.

Last edited: May 10, 2007
5. May 10, 2007

### Staff: Mentor

I hadn't seen the attached picture, so I assumed the pivot at the top and the spring at l at the bottom.

The point is that the for small displacement (small angle about the pivot), sin (theta) ~ theta.

if x = l/2 (theta), at the middle spring, then at l, that spring has twice the displacement x, but it is twice x in the opposite direction, or x2 = l*theta.

Equation 3 is only correct for the spring at l/2.

6. May 11, 2007

### MathematicalPhysicist

ok, what about the second question?

p.s
thanks for the help.