Problem Solving Strategies

@Nobodyrecognizeable my brother lives in London. I'm going to stay with him for the weekend then drive over to my mother's house for the rest of the week. She does indeed live in a tiny village in the country.

yes, the slope has three sections:
1. a straight slope
2. a concave semicircle
3. a convex semicircle

And it's obviously the last section where you might leave the track.

Nobody recognizeable

@JohnRennie pardon me you were showing us your home village whose name I forgot. Anyway I'll talk about this after the problem.

@JohnRennie yep.

John Rennie

In the last section the track travels in a circle, so if you stay on the track there is a centripetal acceleration. Do you know the formula for centripetal acceleration?

Nobody recognizeable

@JohnRennie $\frac{ mv^2}{R}$

John Rennie

Yes. And if the track makes some angle $\theta$ with the horizontal the vertical component of this centripetal acceleration is $v^2/R \cos\theta$. Yes?

Nobody recognizeable

5:16 AM

@JohnRennie obviously.

John Rennie

And that vertical acceleration is being provided by the gravitational force.

Nobody recognizeable

@JohnRennie yep so these are equal but there is a reaction force although.

John Rennie

So the largest value of the vertical acceleration is the gravitational acceleration, $g$. If the vertical component of the centripetal accleration is larger than $g$ you will leave the track.

So the point at which you leave the track is given by:

$$ \frac{v^2}{R}\cos\theta = g $$

Nobody recognizeable

@JohnRennie yep. But what about the reaction force?

John Rennie

The reaction force is just the difference between $g$ and the centripetal force. It's when the reaction force goes to zero that you leave the track.

Nobody recognizeable

5:22 AM

@JohnRennie yep . Ok. But how to get $\theta$

John Rennie

There is a relationship between $v$ and $\theta$. If we call $v_0$ the value of $v$ when you first start the upwards semicircle then the value of $v_0$ you get from the potential energy change.

Nobody recognizeable

@JohnRennie yep.

Can i use $ h=R \theta$?

@JohnRennie $v_0 = \sqrt{2gh}$ but is this gonna be constant throughout the root?

John Rennie

The first step is to get the value of $v)$, and that depends on $h$

Nobody recognizeable

2 mins ago, by Nobody recognizeable

@JohnRennie $v_0 = \sqrt{2gh}$ but is this gonna be constant throughout the root?

John Rennie

At the point A your height is $R\sin30 = R/2$, so your change in height is $h-R/2$. So your velocity is $v_0 = \sqrt{2g(h-R/2)}$

As you move round the semicircle your height increases so the velocity decreases. Your height at an angle $\theta$ is $R\sin\theta$, so your velocity at an angle $\theta$ will be $v = \sqrt{2g(h-R\sin\theta)}$

So now you can write the centripetal acceleration in terms of $\theta$, and then calculate the vertical component of that acceleration.

Nobody recognizeable

5:41 AM

@JohnRennie $ 2g (h-Rsin30°)cos30° =g$

John Rennie

The centriptal acceleration is:

Abcd

@JohnRennie Are you there?

Nobody recognizeable

2g$(h-Rsin\theta) cos\theta $ @JohnRennie

John Rennie

$$ \frac{v^2}{R} = 2g(\frac{h}{R} - \sin\theta) $$

And you get the vertical component by multiplying by $\sin\theta$, so:

$$ a_v = 2g\sin\theta(\frac{h}{R} - \sin\theta) $$

Nobody recognizeable

@JohnRennie yep

John Rennie

5:45 AM

So set that equal to $g$ and solve for $\theta$

@Abcd hi

Abcd

@JohnRennie Morn. Isn't the component of a along b given by $\dfrac{\vec a . \vec b}{|b|}$

John Rennie

Yes i.e. $a\cos\theta$

Nobody recognizeable

@JohnRennie $\theta =\ arcsin{h/R}$

Abcd

@JohnRennie Answer given is $\dfrac{(\vec a \cdot \vec b)\vec b}{|b|^2}$

John Rennie

@Nobodyrecognizeable Use \arcsin for the inverse sine

@Abcd Ah, OK. $\frac{\vec{a}\cdot\vec{b}}{|b|}$ gives the modulus of the component along $\vec{b}$

So multiply that by $\hat{b}$ to get the vector.

Abcd

5:50 AM

@JohnRennie Oh right got it thanks.

John Rennie

@Abcd Trick question! Sneaky! :-(

Nobody recognizeable

@JohnRennie $\arc sin{h/R}$

John Rennie

@Nobodyrecognizeable is it as simple as that? You are going to get a quadratic in $\sin\theta$.

$$ g = a_v = 2g\sin\theta(\frac{h}{R} - \sin\theta) $$

$$ 2\sin^2\theta - 2\sin\theta\frac{h}{R} + 1 = 0 $$

Hmm, that doesn't have an especially simple solution. I get:

$$ \sin\theta = \frac{h}{r} \pm \sqrt{\frac{h^2}{r^2} - 2} $$

Nobody recognizeable

@JohnRennie yep.

@JohnRennie they are giving 3R/4 as answer.

John Rennie

But the minimum value of $h$ is easy. That's just the condition for the discriminant to be real.

Nobody recognizeable

6:01 AM

2 mins ago, by Nobody recognizeable

@JohnRennie they are giving 3R/4 as answer.

John Rennie

3R/4 as the minimum value of $h$?

Nobody recognizeable

@JohnRennie yep.

John Rennie

Hmm, I guess there must have been an error in the calculation along the way. I think the method is fine. However I need to work now for about an hour.

Nobody recognizeable

@JohnRennie ok ill come back later . Thanks anyway for helping. Goodbye. Have a nice day.

harambe

7:02 AM

@JohnRennie good morning

John Rennie

@harambe morning :-)

harambe

Are you free for sometime?

John Rennie

I'm only here for a short time before I have to jump in my car and set off on a long trip.

harambe

Oh so it's better to relax for now . Have a good day sir!

John Rennie

@harambe I'm around for a bit if you have a quick question ...

harambe

7:08 AM

@JohnRennie okay. I had one so better finish it

@JohnRennie imgur.com/a/4i1v6tD

Q11

It's like the previous question we discussed yesterday but there is no wire here

Would charge distribution be still same?

Or It will be by equal distribution on the surfaces

John Rennie

It says the spheres are touched, so they are electrically connected. That's basically the same as connecting them with a wire. So when the R and 3R spheres are touched we get 3/4Q on the 3R sphere and 1/4Q on the R sphere just like before.

harambe

Okay got it.

@JohnRenniebone last question . If between two charges if I keep a dielectric of some thickness then it's force reduces. Why is it so

Abcd

Prove that if in a tetrahedron if two pairs of opposite edges are
perpendicular then the third pair is also perpendicular.

Method:

$\vec a+ \vec b + \vec c + \vec d +\vec e + \vec f = 0 \tag0$ where the vectors represent the sides.

also, $(\vec a + \vec b). (\vec c+\vec d)= 0 \tag{1}$

$(\vec e+\vec f) .(\vec c+\vec d)= 0 \tag{2}$

From 1 and 2,

$\vec e + \vec f = k(\vec a+ \vec b)$

Substituting in $0$,

$\vec a + \vec b = -\dfrac{\vec c + \vec d}{k+1}$

Now, $(\vec e+ \vec f). (\vec a+\vec b) = -\dfrac 1{k+1}((\vec a + \vec b). (\vec c+\vec d))= 0$

John Rennie

@harambe Are you asking how physically a dielectric manages this trick, or how the equations describe it?

harambe

I am interested in physically.... I have a question based on this. Perhaps would you like to see the question and then show the working?

John Rennie

7:19 AM

OK ...

Abcd

@JohnRennie Hi, can you please see my question above?

John Rennie

@Abcd I'll have to print your post and read through. It's a bit long to take in at a glance. For now I'm just finishing off harambe's electrostatics question.

harambe

imgur.com/a/gqWoHki

Do dielectrics block some electric field lines hence reducing the net field

Abcd

@JohnRennie you mean literally "print" on paper using printer?

harambe

I never understood about dielectrics too detailed

John Rennie

7:23 AM

@harambe A dielectric in effect increases the distance between charges. Suppose we write the force law as:

$$ F = \frac{1}{\epsilon_r} \frac{kQ_1 Q_2}{r^2} $$

harambe

Okay

John Rennie

Then we can write this as:

$$ F = \frac{kQ_1 Q_2}{(\sqrt{\epsilon_r}r)^2} = \frac{kQ_1 Q_2}{R^2}$$

Where $R = r\sqrt{\epsilon_r}$

And since $\epsilon_r > 1$ this means the effective distance $R$ is greater than the actual distance. So it's as though the dielectric increases the distance between the charges.

harambe

Okay.. What's with the sq root

John Rennie

@harambe $R^2 = \epsilon_r r^2$

harambe

Okay

John Rennie

7:30 AM

In the question $\epsilon_r = 4$ so the sqrt is 2.

So the effective thickness of the slab is $2 \times r/2 = r$

harambe

Okay

John Rennie

and there is a distance $r/2$ outside the slab, so the effective distance increases to $r/2 + r = 3r/2$

harambe

Got it

John Rennie

So the answer is ... ?

harambe

4/9

I am seeing this fornulae for first time. I would need to do some research on it. Thanks for the help

John Rennie

7:45 AM

@Abcd I have run out of time I'm afraid

John Rennie

2:12 PM

@Abcd it isn't clear to me which sides are which. Something like this?

Abcd

5:05 PM

@JohnRennie Just a b c d e f vectors taken in order

then c vector joining head of b to tail of d

and so on...

John Rennie

@Abcd that doesn't really help me ...

Abcd

@JohnRennie See the next message

@JohnRennie is the next message of any help??

John Rennie

I'm guessing your plan is to prove a.b = 0 (using my labelling) by expanding a and b as the sum of the other vectors. But I can't work out what your a,b,c etc refer to.

Abcd

@JohnRennie let me explain from beginning.

@JohnRennie Just consider a random vector in space denoted by $\vec a$

John Rennie

OK ...

Abcd

5:08 PM

Now take vector b and join it to head of vector a.

John Rennie

Are we still considering a tetrahedron?

Abcd

Now take c and join it to head of b from its tail

do all this in a manner that creates a tetrahedron

@JohnRennie If its not clear yet I can edit your diagram and show what I mean

John Rennie

Have you got a gmail account? I'll share the document.

Abcd

@JohnRennie i will just use snipping tool its faster

John Rennie

Try this link

You should have write access

Abcd

5:13 PM

@JohnRennie how to change line to arrow

John Rennie

Click on the line to select it

Abcd

selected

John Rennie

Use the menu I've shown to change the line ends

Abcd

@JohnRennie I think I have got my mistake.

I was trying to use that sum of vectors in a closed path is 0

But in a tetrahedron we cant find a closed path unfortunately.

Seeing @JohnRennie?

Like @JohnRennie see I have edited the diagram.

But we wont get a unified closed path for the tetrahedron.

Like if we had a triangle we would have quickly got $\vec a +\vec b +\vec c = 0$

John Rennie

What I would say is that the opposite edges a and a are normal if a.c = 0

And a = b - e, and c = erm hang on ..

Hmm ... the point is you can write a and c as sums of other vectors, do the dot product then use the fact that you're told the other pairs of vectors are normal.

I have to go know. Dinner is ready. I'll be here as usual tomorrow.

Abcd

5:26 PM

Okay. Good night.

harambe

9:33 PM

@sammygerbil good evening.

sammy gerbil

@harambe hello

harambe

Can you help me with a doubt @sammygerbil

sammy gerbil

@harambe I will try

harambe

imgur.com/a/MXM6o1g

For this question, I first tried equated the force on +Q... Got me the distance between the charges -Qo and -4Q0 to be l/3

But How would I check for (a) and (b)

On second note I got (b) option as well. Confused about (a)

sammy gerbil

@harambe Option (a) is true because of Earnshaw's Theorem : en.wikipedia.org/wiki/Earnshaw%27s_theorem

You can have positions of equilibrium, but none of them are stable.

harambe

9:49 PM

Oh kay. Thanks for the info

@sammygerbil imgur.com/a/Iv1GFm8

How come velocity of A and B would be equal but not momentum

Why would they gain unequal kinetic energy

Isn't the system isolated so sll the P. E between them would be converted into K. E

sammy gerbil

@harambe At closest approach the relative velocity is zero - they are neither getting closer nor further apart.

Their masses can be different, so their momenta might not be the same.

harambe

Now I think this is application of COM frame

@sammygerbil I am having somewhat trouble understanding the potential energy here. When we bring a charge near another charge then their potential energy decreass right

So the electrostatic force would be repelling charge A and accelerating charge B.

sammy gerbil

@harambe No if you bring two particles of the same charge together the PE increases because you are doing work to move them together.

The particles can have unequal KE for the same reason they have unequal momenta.

harambe

Different masses okay

sammy gerbil

Correct.

harambe

10:03 PM

Is this question similar to a block moving with some velocity and compressing another spring block system

I find these two strikingly simliar now

sammy gerbil

Yes they are similar. The electrostatic repulsion takes the place of the spring force. Except that the ES force is always repulsive, whereas spring force can be attractive or repulsive.

So which options do you think are correct?

harambe

B and D looks okay to me

A is incorrect because they have used "must"

sammy gerbil

Good. What about C?

harambe

Need to think about that a bit...... Let me try this in COM frame first

sammy gerbil

ok but I don't think that will help.

D is correct because of conservation of momentum. What about conservation of energy?

harambe

10:11 PM

It holds true too

sammy gerbil

So? How does it help you decide if C is correct?

harambe

Well the kinetic energy intially would be spent into potential and kinetic energy final

Wait

I got it

The potential energy is increasing so kinetic energy of is decreasing of both particles

sammy gerbil

BTW in the 1st problem if you didn't know Earnshaw's Theorem you could write an expression for the total potential energy U of the 3 particles, as a function of the distances between paricles 1-2 and 2-3. Then you could differentiate once to find positions of equilibrium (turning points in U) and a 2nd time to find out if the equilibrium is stable, neutral or unstable.

But all that takes time in an exam. Much quicker if you know Earnshaw's Theorem.

harambe

Yea

sammy gerbil

In the 2nd problem B, C and D are all correct. For C, particle A loses KE. Some of this becomes KE of particle B, and some becomes electrostatic PE. So KE gained by B is less than that lost by A.

harambe

10:30 PM

@sammygerbil but the equations give a different result

In the COM, at the point of nearest approach both of their velocity will be equal

So m1v1/m1+m2= common velocity where v1 is some velocity of A

Calculating ∆K.E gives me equal

I thought changes in energy were independent of frame

Sorry... It's depending on mass

But going by your explanation which I recall is standard work energy theorem is also correct

sammy gerbil

@harambe In COM frame total momentum is zero. The 2 momenta are always equal and opposite. At closest approach the 2 momenta are zero, and the 2 KEs are also zero.

harambe

@sammygerbil I get it. I think got this now

sammy gerbil

Good.

harambe

10:49 PM

I totally forgot that COM motion is a frame where there is internal motion mso it would make sense for all kinetic energy to convert into potential energy.

But you are right. Using energy theorem is more cleaner here in the lab frame / ground after we know what is happening in COM frame

Transcript for

Problem Solving Strategies