Problem Solving Strategies

Nobody recognizeable

04:58

@JohnRennie hi.

John Rennie

@Nobodyrecognizeable morning :-)

Nobody recognizeable

@JohnRennie I have some questions.

John Rennie

@Nobodyrecognizeable yes?

Nobody recognizeable

Man I'm trying to upload they are not being uploaded. Looks like I have to crop them. I'm coming back.

John Rennie

Which question?

Nobody recognizeable

05:04

@JohnRennie on question 53 I think I have to use Kepler 2nd law.

Both.

John Rennie

@Nobodyrecognizeable in circular orbit the centripetal acceleration is equal to the gravitational acceleration

Nobody recognizeable

@JohnRennie $r=mv^2/g$

John Rennie

That doesn't look quite right ...

Nobody recognizeable

$mv^2/r = Gm/r^2$ @JohnRennie

John Rennie

The centripetal acceleration is $v^2/r$

What you've written is the centripetal force

Likewise the gravitational acceleration is $GM/r^2$, where $M$ is the mass of the Earth. The gravitational force is $GMm/r^2$, where $m$ is the mass of the Moon.

Also, you're given the period of the Moon, so you know its frequency but not its velocity. In that case I would use the expression for the centripetal acceleration $a = \omega^2 r$

Nobody recognizeable

05:12

$r =GM/R^2$ @JohnRennie

John Rennie

Then $\omega = 2 \pi f = 2 \pi/T $

Nobody recognizeable

Ok.

John Rennie

@Nobodyrecognizeable R is the radius of the Earth? If so, where did that come from?

Nobody recognizeable

I couldn't understand it. @JohnRennie

John Rennie

The centripetal acceleration is $\omega^2 r$ where $r$ is the radius of the Moon's orbit (in the question they use D for this radius).

The gravitational force is the usual Newton's law: $F = GMm/r^2$

where $M$ is the mass of the Earth and $m$ is the mass of the Moon.

The acceleration is the force divided by the Moon's mass $m$, so the acceleration is $GM/r^2$.

If we set these equal we get:

$$ \omega^2 r = \frac{GM}{r^2} $$

If we use $\omega = 2\pi / T$ and rearrange the equation we get:

$$ r^3 = \frac{GM}{4\pi^2} T^2 $$

Which is of course Kepler's law $r^3 \propto T^2$

Nobody recognizeable

05:20

@JohnRennie so I need to know the mass of earth to solve it?

John Rennie

@Nobodyrecognizeable You are told that the acceleration at the Earth's surface is 9.8 m/s^2

That means:

$$ 9.8 = \frac{GM}{R^2} $$

Nobody recognizeable

@JohnRennie so I need to multiply R^2 And divide R^2 simultaneously to get it.

John Rennie

You can use this to substitute for $GM$ in our equation for $r$

This gives:

$$ r^3 = \frac{GM}{4\pi^2} T^2 = \frac{9.8 R^2 }{4\pi^2} T^2 $$

You are asked for the value of $r/R$ ($D/R$ in the question). If we take the above equation and divide through by $R^3$ you get:

$$ \frac{r^3}{R^3} = \frac{9.8 }{4\pi^2 R} T^2 $$

And you are told $R$ and $T$. Make sure you use the correct units i.e. $R$ in metres and $T$ in seconds.

Nobody recognizeable

@JohnRennie it's giving 59.5 thanks.

John Rennie

Yes, that's what I get

Nobody recognizeable

05:29

@JohnRennie in next question I think I need to equalize pressure energy and kinetic energy.

John Rennie

Yes, that's the way I would do it.

The force on the water is the pressure times the area of the nozzle. Supose the water velocity is $v$, then in one second we get a stream of water of length $v$.

The work done is force times distance, so it's pressure times are times $v$.

And this work has to be equal to the KE of the stream of water

Nobody recognizeable

@JohnRennie $v =2pA/M$

Are they referring M/A= density ?

John Rennie

I would write the mass of the stream of water as the volume times the density. the voulme is a cylinder of area $A$ and length $v$ so it is just $Av$. Then $M = Av\rho$, where $\rho$ is the density of the water.

If you substitute this for $M$ the area cancels out and you're left with a simple equation for $v$

Nobody recognizeable

@JohnRennie $ v = (2P/\rho)^{1/2}$

John Rennie

Yes, that's what I get.

Nobody recognizeable

05:41

@JohnRennie 17.32 ?

I'm coming back with some more questions.

John Rennie

@Nobodyrecognizeable yes, that's what I get

Nobody recognizeable

Bye for now. I'm getting back after cropping everything. Thanks for the help.

@JohnRennie ^^

John Rennie

OK, how far have you got?

Nobody recognizeable

Note :Whenever I come here I come with lot of questions as I particularly try to solve a paper. So if you have any problems attending me while I'm asking you are always welcome to say it without any hesitation .

@JohnRennie in question 57. In the R1 frame . What does that denote. I could use just the simple length contraction formula . But with it i can't?

Or should i just use l =$l_0 \gamma$

user 2646

06:02

in The h Bar, 1 min ago, by user157860

@JohnRennie, Hi John, I am trying to use the formula that substituted the Lorentz factor to determine the relativistic mass/energy: $$E = \sqrt{p^2c^2 + m^2c^4}=mc\sqrt{v^2 + c^2}= m/c c ( .75c+ 1) =1.32 m_e$$ Can you help me understand hot to use it? should I get the same result I get with Lorentz? I tried with v 0.866c, but instead of m_0c^2 2 I get * 1.32? What's wrong?

John Rennie

In the lab frame the two rods are moving in opposite directions at the same speed $v$. The first step is to calculate the relative velocities of the rods in their own frame i.e. what is the velocity of the rod $R_2$ in the rest frame of $R_1$.

Nobody recognizeable

@JohnRennie $-2v\hat {i}$ ?

John Rennie

@Nobodyrecognizeable for sall velocities, i.e. velocities much less than the speed of light, you can just add the velocities to get the relative velocity. However when the velocities are large you need to use the equation for the relativistic addition of velocities.

Velocity-addition formula

In relativistic physics, a velocity-addition formula is a three-dimensional equation that relates the velocities of objects in different reference frames. Such formulas apply to successive Lorentz transformations, so they also relate different frames. Accompanying velocity addition is a kinematic effect known as Thomas precession, whereby successive non-collinear Lorentz boosts become equivalent to the composition of a rotation of the coordinate system and a boost. Standard applications of velocity-addition formulas include the Doppler shift, Doppler navigation, the aberration of light, and the...

Nobody recognizeable

@JohnRennie but the length of R_2 has become half of its original. The velocity should be close to the speed of light.

John Rennie

@Nobodyrecognizeable yes, so you need to use the relativistic equation for adding velocities.

$$ v_{rel} = \frac{u + v}{1 + uv/c^2} $$

Which in this case is going to give you:

$$ v_{rel} = \frac{2v}{1 + v^2/c^2} $$

No, hang on, that's wrong ...

Aha, yes, it's a plus not a minus in the denominator

Nobody recognizeable

06:14

Sorry I was reading the wiki post. So with this velocity I should use the length contraction formula? @JohnRennie

John Rennie

Yes

@Nobodyrecognizeable suppose the velocities in the lab frame were +0.9c and -0.9c. The relative velocity can't simply be $v_1 - v_2$ because that would give 1.8c and that is faster than light.

That's why the velocity addition formula is more complicated.

Nobody recognizeable

@JohnRennie q no 58.

John Rennie

@Nobodyrecognizeable You just need to use the Lorentz transformations to find out where the two events are in S'

If we assume that the two frames coincide at $t=0$ then the first event $E_1 = (0, 0)$ will be at the origin in both frames.

So you just need to use the Lorentz transformations for $E_2$

Nobody recognizeable

@JohnRennie as I haven't read Lorenz transformation so leave it.

@JohnRennie now q no 56.

Angular momentum = $r\times v$

The position of frame S_2 = 2\pie R hat {i}$

John Rennie

06:32

This is potentially a simple question, but the details have been chosen to make it quite simple.

Nobody recognizeable

The position of the mass would be wrt s2 = $( 2\pi -1)R \hat{i}$

John Rennie

You're told that the position in $S_1$ is $(R\cos\omega t, R\sin\omega t)$

And you're told that $t = 2\pi/\omega$

Nobody recognizeable

51 secs ago, by Nobody recognizeable

The position of the mass would be wrt s2 = $( 2\pi -1)R \hat{i}$

John Rennie

So what is the position in $S_1$ at time $t$?

Nobody recognizeable

Isn't $S_1 $ at rest and$ S_2 $moving? S_1 Should be at the same point. @JohnRennie

John Rennie

06:37

I meant the position of the object in $S_1$ at time $t$. But you've obviously done this part since you got the correct distance to the object in $S_2$ i.e. $R(2\pi - 1)$.

So you just need to calculate the velocity of the object in $S_1$ and that is just $d/dt$ of the position.

Nobody recognizeable

@JohnRennie s_1 should be the centre of the circle aka (0,0) at Cartesian.

John Rennie

I need to work now for about half an hour. Calculate the velocity of the object in $S_1$.

Nobody recognizeable

Ok... goodbye for now . Have a nice day. I'll come back 1 hrs later.

Hema

08:29

My book defines energy stored per unit volume in any electric field as 1/2 ε E^2 where ε is permittivity of medium. It uses this concept to find energy stored inside a solid non-conducting sphere as integral of 1/2 ε0 E^2 dV with limits 0 and R where dV is volume of an elementary shell of thickness dx and radius x.

The final answer is KQ^2/ 10R. My question is,why is ε0 taken and not ε for the medium? Aren't we calculating energy stored within the non conductor which will have a differentpermittiity from ε0?

John Rennie

09:00

@Hema yes I agree. I assume the non-conducting material in question has $\epsilon_r = 1$.

Hema

@JohnRennie ohhh ok thanks

Abcd

10:18

@JohnRennie Are you there

John Rennie

@Abcd morning :-)

Abcd

@JohnRennie hi,could you explain some CS stuff??

John Rennie

Computer Science?

Abcd

@JohnRennie ya

John Rennie

I can try. What's the question?

Abcd

10:20

@JohnRennie All books just explain what is object. What is the difference between objects and classes etc. But nobody explains when to use objects?

@JohnRennie i dont understand at all when to use objects or not

John Rennie

Modern software is frequently huge - millions of lines. And trying to keep it all under control is often very hard. Using objects helps with this.

What objects do is keep the code clear and simple to understand by hiding all the complexity.

Abcd

@JohnRennie How?

John Rennie

Suppose I wanted to write a program to get a web page (in fact I have done exactly this). Transferring data with the HTTP protocol is very complicated and the program would run to hundreds of lines.

But suppose I write a CWebPage object that has all the complicated code within it. Then when I want to write an app to get a web page I could do something like:

mypage = new CWebPage;
mypage.URL = "http://physics.stackexchange.com/"
mypage.GetPage

And I think you'd have to agree that's pretty simple.

Abcd

@JohnRennie If you didnt have object, what would you have done?

John Rennie

@Abcd I'd have to replace the mypage.GetPage line with several hundred lines of code that created a TCP/IP socket, connected to the server, sent various HTTP protocal commands and eventually read back the data.

Abcd

10:28

@JohnRennie Do you have any Java example?

John Rennie

What sort of example?

Abcd

@JohnRennie of usefulness of objects

John Rennie

In Java the code to read a web page and print it out would be something like:

    String urlString = "http://localhost:8080/";

    // create the url
    URL url = new URL(urlString);

    // open the url stream, wrap it an a few "readers"
    BufferedReader reader = new BufferedReader(new InputStreamReader(url.openStream()));

    // write the output to stdout
    String line;
    while ((line = reader.readLine()) != null)
    {
      System.out.println(line);
    }

    // close our reader
    reader.close();

Again, that's pretty simple because Java provides a URL class and Reader classes that do all the hard work for you.

copied and pasted from here

Abcd

@JohnRennie How are objects helping you here

John Rennie

There is a URL class that has all the code for doing HTTP inside it. I create an object using:

URL url = new URL(urlString);

Then I can just use url.openStream to connect to the web server.

Instead of having to write hundreds of lines of code to do it.

Abcd

10:36

@JohnRennie What would you have done w/out objects? Like what code would you have written?

John Rennie

That would be hundreds of lines of code. I can't paste all the code here.

Abcd

Oh okay.

John Rennie

Plus I'd have to go away and read the RFC that explains how the HTTP protocol works.

And that is pretty complicated.

The point is that some clever programmer did all the hard work for me when they created the URL class.

Yes, it's possible to learn programming without worrying at all what hardware your code is going to run on.

In fact the vast majority of code is written that way.

Abcd

Oh thanks! Relief!

@JohnRennie What is the use of Wrapper Classes?

Why convert primitives to objects? (And what does converting primitive to object even mean)

John Rennie

@Abcd opinions may differ, but I'd say that using a wrapper class is mainly cosmetic.

Abcd

10:50

@JohnRennie cosmetic???

John Rennie

Without them you have two different types of variable - objects and primitives. If you use wrapper classes then all your variables are objects.

This seems ... well ... more elegant

Abcd

@JohnRennie why/how

John Rennie

That's where the opinion comes in. Personally I don't have a problem with using primitive types.

Abcd

Ok

@JohnRennie Do you know Boolean Algebra, Karnaughh Maps etc stuff?

John Rennie

I know only basic Boolean algebra and have never heard of Karnaughh Maps

So my ability to help with those topics is limited.

Surely you don't need that stuff for JEE? Karnaugh Maps? That's pretty advanced stuff.

Abcd

10:56

@JohnRennie Wrapper class methods are static so that we dont have to create an object everytime right?

@JohnRennie Its in my school syllabus unfortunately.

John Rennie

@Abcd I don't think wrapper class methods have to be static, though they can be. Any class can have static methods.

Abcd

@JohnRennie What is the use of static methods? When to use them?

John Rennie

With an object typically that object has some data that is specific to the object. For example the URL object I mentioned earlier has a web page address that is specific to that object. I could have other URL objects that would have different addresses associated with them.

And in many cases the object methods won't work without having that data stored in the object.

But there might be some methods associated with a class that don't rely on having an object with some data in it. In that case there's no need to create an object to use that method because you don't need the info an object stores.

So you could make that sort of method static. Then you don't need to create an object to use it.

Abcd

@JohnRennie Could you give a simple example for this?

John Rennie

Hmm ... let's see if I can think of an example ...

Suppose I have a vector object. I could have a method that set the x, y, and z components of the vector and I could have a method called Modulus that returned the length of the vector.

Obviously the Modulus method can't be static because it doesn't make any sense to call it unless we have a vector object with the x, y and z components set. OK so far?

Abcd

11:04

@JohnRennie yes

John Rennie

But then I discover I might need to use both metric and imperial units, so I add a method to my object .ConvertMetresToInches(int Dist)

This method takes one argument, a distance in metres, and returns the distance converted to inches.

Abcd

ok

this should be static, right?

John Rennie

The method doesn't need any of the data in the object, so I could make it a static method, then I wouldn't have to go to thetrouble of instantiating an object before I could use it.

Abcd

Oh okay, got it! Thanks.

John Rennie

This is probably a slightly artificial example, but it should give you the general idea.

Hema

11:30

How did they get the expression for E A? Isn't net field inside a cavity within a conductor zero?

John Rennie

@Hema only if there is no charge in the cavity

Hema

@JohnRennie ohhh

John Rennie

Remember Gauss' theorem!

If there is a charge inside your Gaussian surface there will be a field at the surface.

Hema

@JohnRennie but that is the outer surface right? Why would there be a field experienced at a point within the cavity?

John Rennie

Suppose I use the Gaussian surface I've drawn in red ...

Hema

11:37

Ohhh ok

Hema

11:49

@JohnRennie the flux through that surface will be q/e0 right? But won't the surface S1 have an equal and opposite flux so that net field within conductor is zero?

John Rennie

The nice thing about Gauss' theorem is that all we care about is the charge inside the surface.

Hema

@JohnRennie you mean flux through S1 is q/e0? But then what is the point of having the charges redistribute to get -q on S1?

I'm really sorry for the late reply, my message just got sent actually, I think there was a wifi problem

John Rennie

You don't want to draw your Gaussian surface at S1 because then there is ambiguity about whether the charges are inside the surface or not.

Hema

Ohhh ok

John Rennie

Use these three surfaces.

And the fact that the field inside a conductor is always zero.

I need to go now. I might be around later today.

Hema

12:03

@JohnRennie ohhh ok

@JohnRennie got it thank you!

Abcd

16:48

@IceInkberry $u = eA \sigma T^4$

This law is for thermal radiation emitted by body.

Now suppose T(body)> $T_o$ (room)

@IceInkberry How can we claim that energy absorbed = $eA \sigma{T_o}^4$

I understand that energy emitted $= eA\sigma T^4$

but the law is not for energy emitted man

If Ice is not there @AvnishKabaj @Jasmine

@sammygerbil Please let me know when you are free.

Anonymous

17:25

I just came hereee, reading.

Anonymous

I have really forgotten thermal physics.

Anonymous

At first, I didn't even fathom what you were saying. After skimming through my notes for a while, I guess that the question you are asking is why is the rate of radiation received proportional to T^4?

Abcd

@IceInkberry no

Anonymous

@Abcd Then what is it?

Anonymous

You said Stephen's law is only for emitting and not for absorbing.

Abcd

17:37

@IceInkberry I am asking is Stefan's law only for emission or is it valid for absorption too??

@IceInkberry its Stefan IIRC

Anonymous

@Abcd You don't want to correct how I spell scientists' names. Do you?

Anonymous

Yes, it does if you consider only the surrounding and the blackbody only.

Anonymous

Because the energy radiated by the surrounding will be absorbed by the black body.

Anonymous

And the energy radiated by the surrounding will be according to that law.

Anonymous

(Surrounding here will act as another body since the transfer of energy is happening only between the surrounding and the black body, nothing else.)

Anonymous

17:42

Also, the rate of energy radiated by the surrounding will be equal to the rate of energy absorbed by the black body.

Abcd

@IceInkberry Listen to me once

Temperature of surrournding is $T$

Rate of Energy absorbed by body $\sigma eAT^4 $

Why?

Temperature of body is $T_o$

rate of Energy radiated is $\sigma e A {T_o}^4$. This is Stefan's law

Anonymous

@Abcd Are you sure that it is the Energy absorbed by the body and not Rate of energy absorbed ?

Anonymous

Because Stefan's law deals with the rate.

Anonymous

@Abcd I saw the history carefully ._.

Anonymous

I already did.

Anonymous

17:48

6 mins ago, by Ice Inkberry

Also, the rate of energy radiated by the surrounding will be equal to the rate of energy absorbed by the black body.

Anonymous

9 mins ago, by Ice Inkberry

And the energy radiated by the surrounding will be according to that law.

Abcd

@IceInkberry oh how?

Anonymous

11 mins ago, by Ice Inkberry

Yes, it does if you consider only the surrounding and the blackbody only.

Abcd

@IceInkberry ??

Anonymous

There are only two things: Surrounding and the black body.

Anonymous

17:51

Whatever is emitted from the black body goes into the surrounding and vice versa.

Abcd

Hmm ok3ay thank3s3

33 dis333333333eas3e back

Anonymous

Enjoy

Transcript for

Problem Solving Strategies