The Proof is &#x27;not-so&#x27; Trivial - Physics Edition

Original post by lerjj

Can you give me a hint? I got this somehow:

Spoiler

I get the impression the second part is going to be harder...

What's c?

Reply 61

Original post by Phichi

What's c?

for some reason LaTeX decided \dfrac{v^2}{g} should read "cv²g". I have no idea why. (annoying because that's the only term that's I've gotten correct)

Reply 62

Original post by lerjj

for some reason LaTeX decided \dfrac{v^2}{g} should read "cv²g". I have no idea why. (annoying because that's the only term that's I've gotten correct)

Post your working on a scrap piece of paper, in a spoiler.

Reply 63

Original post by Phichi

Post your working on a scrap piece of paper, in a spoiler.

just noticed a (small) mistake. Probably isn't the main one but I need to check what answer that gives first. :frown:

EDIT: got part a) ! Can't believe I did something so stupid as to forget to add a D/2. Will try part b) tomorrow.

(edited 9 years ago)

Reply 64

spinning muddy wheel.png7.0KB

Original post by WishingChaff

Problem 11**

A wheel of diameter D is rolling along a muddy road at speed v. Particles of mud attached to the wheel are continuously thrown off from all points.

(a) If

v^2 > \frac{Dg}{2}

show that the maximum height above the road attained by flying mud,

H_{max}

, is given by:

H_{max} = \frac{D}{2} + \frac{v^2}{2g} + \frac{D^2 g}{8 v^2}

(b) Show that the range of mud released at this angle is given by:

Unparseable latex formula:

x = -\frac{D}{2} \bigg{[} 1 + \frac{Dg}{2 v^2} + \sqrt{\bigg{(}1-\frac{Dg}{2v^2} \bigg{)} \bigg{(}1+ \frac{Dg}{2v^2} \bigg{)} } \bigg{]}

Solution 11

In the following

\theta

is the angle that the line joining the centre of the wheel to the point where the mud flies off makes with the horizontal diameter of the wheel. Just in case anyone labelled a different angle...

part (a)

Spoiler

Part (b)

Spoiler

I'm pretty sure that took a full hour to write up.

(edited 9 years ago)

Reply 65

[QUOTE="lerjj;53070733"]Solution 11

In the following

\theta

Spoiler

Part (b)

Spoiler

Don't cancel. You've missed a solution in part a. Ignoring the fact it's an invalid one anyway, but still, bad practice and slightly non rigorous.

Posted from TSR Mobile

(edited 9 years ago)

Reply 66

Original post by Phichi

Don't cancel. You've missed a solution in part a. Ignoring the fact it's an invalid one anyway, but still, back practice and slightly non rigorous.

Posted from TSR Mobile

I thought I could cancel because cos(theta) is only zero at 90 (for this scenario) and seeing as that's shooting out sideways that didn't feel like much of a solution. How do I get to the other solution? I need to go proofread my answer, because a lot of the LaTeX has gone funny for some reason...

Reply 67

Original post by Phichi

Don't cancel. You've missed a solution in part a. Ignoring the fact it's an invalid one anyway, but still, back practice and slightly non rigorous.

Posted from TSR Mobile

How should I have found the solutions? I guess that's what the

v^2>\dfrac{Dg}{2}

thing was about? I just checked that my solution made sense with that (which it does because dividing by v² gives 1>sin).

Also, do you know why the answer to part (b) in the question is negative? I don't see how a range can be negative, and unless I've drawn it wrong the mud is being flung in the same direction as v.

Reply 68

The other solution is quite easy to see without any mathematics. Basically, ask yourself the question what would happen if an enormous wheel was moving very slowly on a planet with a large gravitational field? Would the mud be slung very high?

Mathematically the condition in the question comes from the necessity that arcsin is only defined on the domain [-1,1].

Original post by lerjj

How should I have found the solutions? I guess that's what the

v^2>\dfrac{Dg}{2}

(edited 9 years ago)

Reply 69

Original post by WishingChaff

Right, the other solution is that cos(theta)=90, or the particles are 'launched' from the top of the wheel. Which is only the case in the situation you just described.

Is there a reason for defining the range as negative? I attached a drawing of what I thought the question was about, and I would have thought that that would make the range positive.

(edited 9 years ago)

Reply 70

Exactly. I don't see your negative solution you mentioned however, when minus signs appear that don't make physical sense they normally are there because of the initial coordinate system chosen.

Well done with the problem anyway, was a tough projectile problem. I posted this in the STEP paper thread and no one answered.

Original post by lerjj

Right, the other solution is that cos(theta)=90, or the particles are 'launched' from the top of the wheel. Which is only the case in the situation you just described.

Reply 71

Problem 13** (already posted this in different thread but, has many different ways of approaching)

Bob stands at the peak of a very high hill that slopes downward uniformly an angle

\phi

below the horizontal. He fires an arrow with initial speed

v_0

at an angle

\theta

up from the horizontal. Neglect any air resistance, and treat the arrow as a point particle.
(a) At what angle

\theta

should he fire the arrow so that it lands at the maximum distance down the hillside?
(b) For this launch angle, what is the velocity

v

of the arrow just before it strikes the ground?

(Will post my solution in this thread soon - has a nice twist)

(edited 9 years ago)

Reply 72

Problem 14**

A fixed capstan on a ship is used so that a sailor can control a larger load than could otherwise be achieved directly. One side of a long, inextensible rope is attached to the load, which in turn induces a tension

T_0

in the rope. The rope is wrapped in a tight helix around a fixed cylindrical drum of radius

R

, for which the maximum coefficient of static friction between the rope and drum is

\mu

, and the other end of the rope is held by the sailor. If the wrapped rope subtends a total angle

\theta

on the drum (often consisting of multiple turns, so that it may exceed

2 \pi

), with what force

F

must the sailor pull so that the load and rope do not slip? Neglect any gravitational effects.

(edited 9 years ago)

Reply 73

Solution 13 (I had already done this in a previous post - do try the problem before looking at this)

Spoiler

As a note for this problem, I used a slightly unusual method that is not normally taught at M2 (excluding the partial derivative - which I mentioned in a previous post). The point I was trying to make with this method is that there is nothing inherently physical about co-ordinate systems, they are completely up to us. In other words, physics should be independent of what coordinate system we pick. This is in essence a statement of general covariance and is key to the development of General Relativity.

(edited 9 years ago)

Reply 74

Original post by lerjj

How should I have found the solutions? I guess that's what the

v^2>\dfrac{Dg}{2}

Ignoring the above, however Wishing gives a perfect answer. From a mathematical perspective, just factorise out the

cos\theta

and find its unique solution. Then just note that this breaks the conditions given. Obviously you'll be finding the other solution too, which abides by the conditions.

Reply 75

Original post by WishingChaff

Solution 12 (I had already done this in a previous post - do try the problem before looking at this)

Spoiler

In your proof you have the following:

cot(\frac{\pi}{2} - \phi) = tan(\phi)

Surely this would mean

cot(\phi - \frac{\pi}{2}) = tan(-\phi)

Which would leave you with

\phi - \dfrac{\pi}{2} = cot^{-1}(tan(-\phi))

Linking this to the equation

2(\phi + \theta) = cot^{-1}(tan(-\phi)) \, \, \, \Rightarrow \, \, 2(\phi + \theta) = \phi - \dfrac{\pi}{2} + n\pi, \, \, n \in \mathbb{Z}

Seeing as this would yield a negative angle, taking the initial condition that

n = 1

would result in your equation, and the same answer. Really a pedantic thing though.

The co-ordinate system picked is beautiful for the problem however.

Edit: When you subbed your t into your

v_y

equation, how did you get that expression?

I get

v_y = -v_0(sin\theta + 2cos\theta tan\phi )

And just from inspection for your equation for

v_y

, certain values roughly above

\phi = \dfrac{1}{2}

and

\phi = \dfrac{3}{2}

would yield a positive solution, which seeing as you took upwards as positive, would be illogical. A value such as

\phi = \dfrac{\pi}{6}

would yield

v_y = 0

(edited 9 years ago)

Reply 76

langlitz

Original post by WishingChaff

Problem 12** (already posted this in different thread but, has many different ways of approaching)

Bob stands at the peak of a very high hill that slopes downward uniformly an angle

\phi

below the horizontal. He fires an arrow with initial speed

v_0

at an angle

\theta

up from the horizontal. Neglect any air resistance, and treat the arrow as a point particle.
(a) At what angle

\theta

should he fire the arrow so that it lands at the maximum distance down the hillside?
(b) For this launch angle, what is the velocity

v

of the arrow just before it strikes the ground?

(Will post my solution in this thread soon - has a nice twist)

I already posted a "problem 12" on pg 3

Reply 77

Thanks for pointing these out. They were indeed typos in the solution and have now been corrected (I hope).

Original post by Phichi

In your proof you have the following:

cot(\frac{\pi}{2} - \phi) = tan(\phi)

Surely this would mean

cot(\phi - \frac{\pi}{2}) = tan(-\phi)

Which would leave you with

\phi - \dfrac{\pi}{2} = cot^{-1}(tan(-\phi))

Linking this to the equation

2(\phi + \theta) = cot^{-1}(tan(-\phi)) \, \, \, \Rightarrow \, \, 2(\phi + \theta) = \phi - \dfrac{\pi}{2} + n\pi, \, \, n \in \mathbb{Z}

Seeing as this would yield a negative angle, taking the initial condition that

n = 1

would result in your equation, and the same answer. Really a pedantic thing though.

The co-ordinate system picked is beautiful for the problem however.

Edit: When you subbed your t into your

v_y

equation, how did you get that expression?

I get

v_y = -v_0(sin\theta + 2cos\theta tan\phi )

And just from inspection for your equation for

v_y

, certain values roughly above

\phi = \dfrac{1}{2}

and

\phi = \dfrac{3}{2}

would yield a positive solution, which seeing as you took upwards as positive, would be illogical. A value such as

\phi = \dfrac{\pi}{6}

would yield

v_y = 0

Reply 78