The Proof is Trivial!

Original post by und

This didn't exactly inspire confidence:

\displaystyle x=\left( \frac{\mu l}{2g(\mu +1)}+\frac{v}{2\sqrt{\frac{g}{l}(\mu +1)}} \right) e^{\sqrt{\frac{g}{l}(\mu +1)}t} + \left( \frac{\mu l}{2g(\mu +1)}-\frac{v}{2\sqrt{\frac{g}{l}(\mu +1)}} \right) e^{-\sqrt{\frac{g}{l}(\mu +1)}t}-\frac{\mu l}{g(\mu +1)}

That's similar to what I've got; not quite though.

Reply 322

shamika

Problem 54 *

A triangle has sides of length at most 2, 3 and 4 respectively.
Determine, with proof, the maximum possible area of the triangle.

Spoiler

Reply 323

und

OP

Original post by shamika

Problem 54 *

A triangle has sides of length at most 2, 3 and 4 respectively.
Determine, with proof, the maximum possible area of the triangle.

Spoiler

It's the one question that I didn't even read when I was doing the paper. Someone I know who said he completed it got one mark. :lol:

Original post by DJMayes

That's similar to what I've got; not quite though.

I was quite hoping you'd say it was completely wrong.

Reply 324

ukdragon37

14

Original post by around

In the Cambridge 4th year some people (tried) to run a Topos Theory reading group. I think in the end only about 5 or so people finished the course.

I started it and then gave up after about 6-7 'lectures' because it was just utterly incomprehensible.

Yeah it's on my list of "things to learn as I get older".

Reply 325

Original post by DJMayes

Problem 50:

An inextensible rope of length l and uniform mass per unit length lies on a rough table with one end on an edge. The co-efficient of friction between the table and the rope is μ. The rope receives an impulse which sets it moving off of the edge of the table at speed v. Prove that, if the rope does not fall off the table, then:

v^2<\dfrac{lg\mu^2}{1+\mu }

By considering this as

\mu

varies between zero and 1, find the maximum possible impulse that could potentially be given to the rope without it falling off of the table.

Solution 50:

Let the mass per unit length of the rope be

m

. If the rope does not fall off the table, then the distance traveled by the rope off the table must be less than

l

. Let this distance be

x

.

When the rope has just come to rest, a length

x

of the rope hangs off the table and length

(l-x)

is still on the table. Consider the case where friction here must be limiting in order to stop the rest of the rope from sliding off the table. For the section of the rope that's hanging, it's weight must be equal to the tension in the rope, which in turn is equal to the frictional force in the horizontal section of rope otherwise there would be an acceleration.

\therefore F=T=\mu m(l-x)g = mxg \Rightarrow x= \dfrac{\mu l}{1+\mu }

upon solving for

x .

Now as the rope travels off the table, the frictional force slows it down from a speed

v

to

0

. As the frictional force is a function of

x

, then the work done by friction in stopping the rope,

W

, is:

\displaystyle W= \int^x_0 F \ dx = \int^x_0 \mu m(l-x)g \ dx = \mu mg\int^x_0 (l-x) \ dx

\displaystyle = \mu mg \left[ lx- \frac{x^2}{2} \right]_0^x = \mu mg\left( lx- \frac{x^2}{2} \right)

The rope also falls, thus losing gravitational potential energy. By the principle of conservation of energy, the work done by friction is equal to sum of the loses in gravitational potential and kinetic energies:

\displaystyle \Rightarrow \mu mg\left( lx- \frac{x^2}{2} \right) = \frac{1}{2} mlv^2 + \frac{mgx^2}{2}

\Rightarrow 2\mu glx -\mu gx^2=lv^2 +gx^2

\Rightarrow lv^2 = 2\mu lgx - gx^2(1+\mu )

Substituting in the expression for

x

that was derived earlier:

lv^2 = 2\mu lg\dfrac{\mu l}{1+\mu } - g(1+\mu ) \left( \dfrac{\mu l}{1+\mu } \right)^2

\Rightarrow lv^2 = \dfrac{2\mu ^2 l^2g - \mu ^2l^2g}{1+ \mu }

\therefore v^2 = \dfrac{lg\mu ^2 }{1+\mu }

This however gives the limiting value of the square of the velocity, since I assumed that conditions were limiting. Hence the range of values for which the rope does not fall must be given by

v^2 < \dfrac{lg\mu ^2 }{1+\mu }

as required.

As shown earlier, equality gives the limiting value of

v^2

and therefore of

v

. Maximum impulse implies maximum change in velocity by the impulse-momentum principle, and so the maximum value for the impulse,

I_{max}

, of the rope without it falling off the table will just be

mlv

where

v

is limiting:

\displaystyle \therefore I_{max} = ml\mu \sqrt{ \frac{lg}{1+\mu } } .

(edited 11 years ago)

Reply 326

shamika

Original post by around

In the Cambridge 4th year some people (tried) to run a Topos Theory reading group. I think in the end only about 5 or so people finished the course.

I started it and then gave up after about 6-7 'lectures' because it was just utterly incomprehensible.

Does anyone understand it?

Reply 327

Original post by Star-girl

Since the rope also falls, thus losing gravitational potential energy, the kinetic energy lost is less than the work done by friction:

\displaystyle \therefore \frac{1}{2} mlv^2 - 0 < \mu mg\left( lx- \frac{x^2}{2} \right)

\Rightarrow lv^2 < 2\mu lgx - \mu mx^2g

I think putting in the GPE here (As KE plus GPE gained = Work Done) gets rid of the two and solves your co-efficient issue. I think you then need to sub this in to the other expression for x and work with possible values of x to get an inequality.

I've put my comment up there in bold. Your method isn't the way I was using when I did it but it looks solid.

Reply 328

ukdragon37

14

Original post by shamika

Does anyone understand it?

Peter Johnstone.

Reply 329

Original post by DJMayes

I've put my comment up there in bold. Your method isn't the way I was using when I did it but it looks solid.

Thanks - I thought of putting it in, but I didn't think it was necessary. I'll try that. :cute:

Reply 330

und

OP

Original post by ukdragon37

Peter Johnstone.

Does he do undergraduate supervisions? I'm scared now. :colondollar:

Reply 331

Original post by Star-girl

Thanks - I thought of putting it in, but I didn't think it was necessary. I'll try that. :cute:

I'm making no promises it will work - I did it the same way I think und has approached it which means that I don't necessarily have the best bearing on what you're trying to do. :tongue:

Reply 332

und

OP

Original post by DJMayes

I'm making no promises it will work - I did it the same way I think und has approached it which means that I don't necessarily have the best bearing on what you're trying to do. :tongue:

I don't know why my x is different to what you got but I'll leave it now anyway since it's been solved using a much nicer method.

Reply 333

Original post by DJMayes

I'm making no promises it will work - I did it the same way I think und has approached it which means that I don't necessarily have the best bearing on what you're trying to do. :tongue:

I'll chug on and see if I can solve it. :biggrin:

Reply 334

ukdragon37

14

Original post by und

Does he do undergraduate supervisions? I'm scared now. :colondollar:

I'm not actually sure, but John's is certain to have more than enough supervisors for undergrads to not have to bother him if they don't want to. :laugh:

His Elephant, currently standing at 1278 pages over two volumes, is one of the few books I might own in my life that is able to serve as an excellent murder weapon (along with TAOCP). :tongue:

Reply 335

Original post by und

I don't know why my x is different to what you got but I'll leave it now anyway since it's been solved using a much nicer method.

It's sign differences in a couple of places, but otherwise essentially identical.

(And it hasn't been fully solved yet)

Reply 336

Original post by DJMayes

...

Is

0 \leq \mu \leq 1

?

Reply 337