Jim's maths and physics problems

[Twiglet]

My solution agrees with Ipetrichs:

Block 1.

F=ma

In the plane, F(body a)=mgSin60 - 100=69.74
F(body b)=119.74

a(body a)=3.49m/s^2
a(body b)=5.99m/s^2

s=ut+1/2at^2

we want to find the distance travelled & time when the particles catch each other,

B is released 5 seconds after A, so effectively,
Initial velocity for both particles is 0, as they start at rest

s(a)=s(b)

1/2xa(body a) (t+5)^2=1/2a(body b)t^2
1.745(t^2+10t+25)=2.995t*2
1.25t^2-17.45t-43.63=0

provides solutions for t from the standard quadratic formula:

x= [-b +/-(b^2-4ac)^1/2]/2a

a=1.25
b=-17.45
c=-43.63

so

t= [17.45 +/-22.86 ]/2.5

clearly negative time is unphysical so t must be 16.124s seconds (this is the time particle B has been travelling, a will have been going 5 seconds longer)

The distance travelled by both particles is obviously identical and should yield the same result for s=1/at^2
s(a) =0.5x3.49*(21.124)^2=778.66m
s(b)=0.5x5.99x(16.124)^2=778.65m which is good enough

v=u+at
v(a)=3.49x21.124=73.72m/s
v(b)=5.99x16.124=96.58m/s

[PsychoSerenity]

I keep meaning to get in on these. I'll see if I can find the time tomorrow, to find out if I can actually do any of them. I did physics and maths A-level, but it all seems rather a long time ago.

[Twiglet]

I'll suggest one:

A central force has the form F = (r/r)*f(r)

What possible forms of the force function f(r) give closed noncircular orbits? That is, orbits where r(a=2*pi) = r(a=0), where a is the position angle.

I could use a hint on this. Are you asking for all possible forms f(r) can take, or just the Keplers law/perihelion-apehelion stuff and conditions for escape from orbit?

[JimC]

After re-checking my plane solutions, I found a silly error; when I recalculated, got the same as lpetrich and Twiglet...

I first did a quick sketch of a velocity time graph with 2 lines; at a time t, you will have 2 triangles, whose area represents the displacement of each object. I found an expression for the 2 areas, set them to be equal, and ended with the following quadratic in t:

2.5t2 - 59.9t +149.75 = 0

Of which 1 solution is under 5 seconds, and so is ignored, the other is 21.12 s

The rest is easy... (it did turn out to be a bit of a ski slope, didn't it... )

This level of problem would be at the hard end of secondary physics, but certainly doable...

I like problems that involve simultaneous equations; they appeal to me, for some reason...

[lpetrich]

I'll suggest one:

A central force has the form F = (r/r)*f(r)

What possible forms of the force function f(r) give closed noncircular orbits? That is, orbits where r(a=2*pi) = r(a=0), where a is the position angle.

I could use a hint on this. Are you asking for all possible forms f(r) can take, or just the Keplers law/perihelion-apehelion stuff and conditions for escape from orbit?

All possible forms. I don't need conditions for escape from orbit - just which central-force functions give closed noncircular orbits.

[JimC]

I'll suggest one:

A central force has the form F = (r/r)*f(r)

What possible forms of the force function f(r) give closed noncircular orbits? That is, orbits where r(a=2*pi) = r(a=0), where a is the position angle.

I could use a hint on this. Are you asking for all possible forms f(r) can take, or just the Keplers law/perihelion-apehelion stuff and conditions for escape from orbit?

All possible forms. I don't need conditions for escape from orbit - just which central-force functions give closed noncircular orbits.

(for me, this will be like being an audience member at a serious music recial - someone knowlegeable about the musical forms, but not quite up to actually performing...)

[Twiglet]

I can't do that one either. I'd be interested to see the solution to see if it makes me kick myself for being so silly, or whether it gives me a sort of "ah" sigh of "I used to be able to do that long ago"

I'm guessing, purely intuitively (i.e. stating the blooming obvious) that f(r) is going to take the form a/r^2 :p

[lpetrich]

Hints:

Trigger Warning!!!1! :

Find the equations of motion in polar coordinates around the force's center.
One of them is rather rather easy to integrate.
Since this is for finding closed orbits, we can use the position angle as the independent variable.
Using this independent variable, it's easier to use 1/r than r directly.
Start with a circular orbit and do a perturbation expansion around it.
You will get some constraints out of doing that.

[lpetrich]

Here's the first part of it: the derivation of the equations to solve.

Trigger Warning!!!1! :

We start with Newton's laws of motion, which yield this equation:
dr/dt = (r/r)*f(r)

Let r = r*{cos(a), sin(a)}
where r is the radius and a is the position angle, and where both are functions of t. This yields two equations:
d2r/dt2 - r*(da/dt)2 = f(r)
r*(d2a/dt2) + 2(dr/dt)(da/dt) = 0

The second equation can easily integrated to yield the conservation of angular momentum:
r2(da/dt) = h

where h is a constant. We can substitude this into the first equation to give
d2r/dt2 - h2/r3 = f(r)

One can multiply this one by (dr/dt) and integrate over t to give conservation of energy, but we will not need that here. Change the independent variable from t to a:
d/dt = (da/dt)(d/da) = (h/r2)(d/da)

This yields:
h2*( (1/r2)(d/da)((1/r2)(dr/da)) - 1/r3) = f(r)

Set r = 1/u and do some rearranging:
d2u/da2 + u = - 1/(h2u2)f(1/u)

One can now do a perturbation expansion around a circular orbit (u = constant), starting with this lowest-order noncircular orbit:
u = u0(1 + e*cos(n*a))

What kind of number must n be for an orbit to be closed at this level of approximation?

One can get the closure constraint by finding the contributions to u of the next higher powers of e, and seeing what constraints on n and f one can find.

[Magicziggy]

Problem 5

I resolved the forces acting in the direction of the slope and yielded a resultant force that enabled me to calculate the acceleration for each block.
I set up a spread sheet using a = 19.6SQRT(3) -5 for block 1 and a = 19.6SQRT(3) - 2.5 for block 2
I put t = 0, 1,2 etc in column A and in col B had a.t (velocity) and Col C: 0.5a.t^2 (distance)
Then in Cols D and E i did the same for block 2 but starting at t=5 and subracting 5 from column A.
Where the distances crossed over I zoned in, successively making the increment smaller.

I yielded the same results as already shown but didn't have to solve a quadratic.
Just another way of achieving the same thing.

A minor point is that this method does not yield a solution that has to be discounted.

[JimC]

Problem 5

I resolved the forces acting in the direction of the slope and yielded a resultant force that enabled me to calculate the acceleration for each block.
I set up a spread sheet using a = 19.6SQRT(3) -5 for block 1 and a = 19.6SQRT(3) - 2.5 for block 2
I put t = 0, 1,2 etc in column A and in col B had a.t (velocity) and Col C: 0.5a.t^2 (distance)
Then in Cols D and E i did the same for block 2 but starting at t=5 and subracting 5 from column A.
Where the distances crossed over I zoned in, successively making the increment smaller.

I yielded the same results as already shown but didn't have to solve a quadratic.
Just another way of achieving the same thing.

A minor point is that this method does not yield a solution that has to be discounted.

That is a perfectly valid method, IMO, which I would put in the general category of an iterative method. With the right set up, you could have cells specifying the mass of the blocks, the frictional forces on each, the time delay and the slope, and therefore have a very general way of finding solutions to a broad class of problems.

I love doing stuff with Excel - I once made one that iterated the logistic equation. By fine-tuning a parameter, I could make the iteration edge into chaotic behaviour...

[lpetrich]

Here's the second part: finding the constraints and the solutions:

Trigger Warning!!!1! :

We'd earlier obtained an equation that may be expressed as:
d2u/da2 + u - k*g(u) = 0

where k = 1/h2 and g(u) = - f(u)/u2

We start with a circular orbit: u = u0. That gives k = 1/g(u0)

Let us solve
d2u/da2 + u - g(u)/g(u0) = 0

with angular momentum fixed. To find a noncircular orbit, add a perturbation: u = u0*(1 + e*cos(n*a))

n must be a nonnegative integer, otherwise the orbit will not be closed. It can be made positive without loss of generality. Inserting that expression and and expanding to power 1 in e gives
-n2 + 1 - u0 g'(u0)/g(u0) = 0

which means that the force obeys a power law: g(u) = g0 u1 - n^2 or f(r) = f0 rn^2-3

Expanding g(u) in powers of e yields ek (cos(n*a))k, which implies that ek multiplies terms of the form cos((k-2m)*n*a), where m = 0, 1, ... So we try
u = u0*(1 + e*cos(n*a) + c0*e2 + c1*e2*cos(2*n*a) + c2*e3*cos(3*n*a))

and expand the equation solution up to e3. We don't need a e3*cos(n*a) term, because it can be folded into the e*cos(n*a) term. This has the solution
c0 = (n2 - 1)/4
c1 = - ((n2 - 1)/12
c2 = n2*(n2 - 1)/96

with a leftover value of
- n2*(n2 - 1)*(n2 - 4)/12 * u0

This yields n = 1 or 2, with f(r) = f0/r2 or f0*r -- the inverse-square (Keplerian) and the harmonic-oscillator cases.

For the Keplerian case, g(u) = g0 and u = k*g0(1 + e*cos(a))

For the harmonic-oscillator case, r = x1 * cos(w*t) + x2 * sin(w*t)

for constant vectors x1 and x2.

Interestingly, both sorts of orbits are ellipses, though with the force center at one focus in the Keplerian case, and at the center in the HO case.

[lpetrich]

Here's something interesting that's sort of related to my earlier one. It's working out what would happen with different numbers of dimensions of space and time.

Gravity and electromagnetism both have limiting quasistatic forms that can be expressed in terms of single potentials (gtt, At). These potentials are given, in general, by the equation

D2V = sumi d2V/d(xi)2 = (source terms: mass or charge density or whatever)

In the Newtonian limit, motion under the influence of this potential has the form
(force) = - g*(gradient of V)

where g is mass or charge or whatever. Here's an interesting conundrum: how many space dimensions can there be before orbits (bound states) become unstable?

[lpetrich]

Solution to the first part: the potential and force for n space dimensions

Trigger Warning!!!1! :

Take the source to be a point, described as a delta function. By symmetry, V will only be a function of the distance from that source:
r = sqrt(sumi(xi2))
The solution may have a singularity at the origin, but it's related to the source being a delta function there.

Let us now find it.
dV/dxi = xi/r*V'(r)
d2V/dxidxj = deltaij/r*V'(r) - xixj/r3*V'(r) + xixj/r2*V''(r)
D2V = (n-1)/r*V'(r) + V''(r)

It has solutions of the form V(r) = V0 + V1*r2-n

For n = 2, one gets V(r) = V2*ln(r/r0), which can be obtained from the above expression by having
V1 = - V0 + V2/(2-n) - V2ln(r0)

The force that results from this potential has the form (r/r)*f(r), where f(r) = f0r1-n

One can now plug this force law into the equations of motion to see what happens.

[newolder]

Solution to the first part: the potential and force for n space dimensions

Trigger Warning!!!1! :

Take the source to be a point, described as a delta function. By symmetry, V will only be a function of the distance from that source:
r = sqrt(sumi(xi2))
The solution may have a singularity at the origin, but it's related to the source being a delta function there.

Let us now find it.
dV/dxi = xi/r*V'(r)
d2V/dxidxj = deltaij/r*V'(r) - xixj/r3*V'(r) + xixj/r2*V''(r)
D2V = (n-1)/r*V'(r) + V''(r)

It has solutions of the form V(r) = V0 + V1*r2-n

For n = 2, one gets V(r) = V2*ln(r/r0), which can be obtained from the above expression by having
V1 = - V0 + V2/(2-n) - V2ln(r0)

The force that results from this potential has the form (r/r)*f(r), where f(r) = f0r1-n

One can now plug this force law into the equations of motion to see what happens.

Does a fifth force from the fifth dimension have any connection to this thing? Or am I on a wron trax, agin?

We investigate the dynamics of particles moving in a spacetime augmented by one extra dimension in the context of the induced matter theory of gravity. We examine the appearance of a fifth force as an effect caused by the extra dimension and discuss two different approaches to the fifth force formalism. We then give two simple examples of application of both approaches by considering the case of a Ricci-flat warped-product manifold and a generalized Randall-Sundrum space.