The Proof is Trivial!

Guys, can't you do the question by considering

v^2=0

ie. by considering the discriminant of a quadratic in

x

. In fact I tried and I got something very similar but a factor of

g^2

off.

Reply 362

Original post by bananarama2

I would, but I still have a stray minus :tongue:

Care to assist?

Spoiler

In line 6 you have inexplicably multiplied by -1!

Oops, I misread your limits!

(edited 11 years ago)

Reply 363

OK guys, here was how I did the question before posting it:

Spoiler

Reply 364

Original post by DJMayes

OK guys, here was how I did the question before posting it:

Spoiler

You've got less than and I have equals... uh oh... :lol:

Reply 365

Original post by Star-girl

You've got less than and I have equals... uh oh... :lol:

The question has to by nature be an inequality, as there will be more than one value for v which is sufficiently slow to prevent it from falling.

Reply 366

Original post by DJMayes

Why is your PI negative?

Reply 367

Blazy

Problem 55*

The problem is concerned with constructing lines of length root

n

, where

n

is not a perfect square. Suppose you only have a ruler which can only measure integer lengths and assume you can draw perfect right-angled triangles. Show that unless

n

is of the form

\displaystyle 4k-2

, then you only ever need one right-angled triangle to construct a line of length root

n

, e.g. To construct

\sqrt 2

, you create a

\displaystyle \left \{ 1, 1, \sqrt2 \right \}

triangle, to construct

\sqrt3

you create a

\displaystyle \left \{ 1, \sqrt3, 2 \right \}

triangle.

(edited 11 years ago)

Reply 368

Original post by DJMayes

The question has to by nature be an inequality, as there will be more than one value for v which is sufficiently slow to prevent it from falling.

Ah yes, I suppose. So equality denotes the maximum value of v^2, then?

Reply 369

Original post by und

Why is your PI negative?

Could you elaborate? As far as I can see it isn't...

Reply 370

Original post by bananarama2

I would, but I still have a stray minus :tongue:

Care to assist?

Spoiler

I will indeed, I just need to first finish all the modifications on the first part of my solution. :smile:

Reply 371

bananarama2

When I come to the student room Mr M always turns up, like :

[video="youtube;-rPEguSf35c"]http://www.youtube.com/watch?v=-rPEguSf35c[/video]

Reply 372

Solution 50

Using

u

as the initial velocity.

Since the function of

v

is continuous, if

v \to 0^{+}

for some

x

, then at that instant the frictional force must be greater than the weight, thus the rope will remain stationery. It remains to find a range of values of

u

for which

v

will become

0

.

Creating and solving the differential equation, we obtain

v^2=\frac{g(\mu +1)x^2}{l}-2\mu g x+u^2

, so if

v

is to be

0

for some positive

x

, then there must exist a positive root

x=\frac{\mu l \pm \sqrt{(\mu gl)^2-u^{2}gl(\mu +1)}}{g(\mu +1)}

, so a necessary and sufficient condition (assuming all the constant terms are positive) is

u^2 \leq \frac{{\mu}^{2}gl}{(\mu +1)}

as required.

If the rope has mass

m

, then the impulse applied to accelerate the rope to a speed

u

is given by the change of momentum

mu

. Hence

\text{Impulse}^2 \leq \frac{{m^{2}\mu}^{2}gl}{\mu +1}

, giving the maximum impulse as

m\mu \sqrt{\frac{gl}{\mu+1}}

(edited 11 years ago)

Reply 373

Original post by Star-girl

Ah yes, I suppose. So equality denotes the maximum value of v^2, then?

No, I don't think so. If you look at my DE equations solution, equality implies that B = 0, which means that there aren't solutions to x' = 0 as it implies A =0, which is impossible.

Reply 374

Original post by DJMayes

Could you elaborate? As far as I can see it isn't...

Sorry, I meant why is it positive? When you plug

x

into the DE with a positive complimentary function, the constant term has the wrong sign.

Reply 375

Original post by DJMayes

No, I don't think so. If you look at my DE equations solution, equality implies that B = 0, which means that there aren't solutions to x' = 0 as it implies A =0, which is impossible.

Hmmm...

Reply 376

Original post by und

Sorry, I meant why is it positive? When you plug

x

into the DE with a positive complimentary function, the constant term has the wrong sign.

No it doesn't, because the co-efficient of x is negative.

Reply 377