Problem Solving Strategies

AbhigyanC

00:01

In the above question, I have a doubt in part C... The answer says that the kinetic energy of the system of the two blocks is never zero and is minimum at maximum compression of the spring (basically, p and r)

What I don't understand is how the kinetic energy of the system can actually change, given that their centre of mass is moving at a constant speed

harambe

00:22

@AbhigyanC The best way to do these two mass system is to go by com frame

In the centre of mass frame, the bodies will be executing simple harmonic motion so linetic energy will be converting into spring energy and vice versa. Basically center of mass frame can be used to see internal motion of the body. That's why it has so much important in rotstion too

At maximum compresdikn, in com frame all the kinetic energy will be converted into spring energy but since the velocity is zero in the center of mass frame, therefore it will ofcourse be minimum in ground frame

AbhigyanC

@harambe All right... Thanks!!

@harambe It makes a lot more sense in the com frame of reference

harambe

Yea. Even I got confused in doing this without com frame so it pays to do it on it xd

Abcd

05:54

@JohnRennie hello

John Rennie

@Abcd morning :-)

Abcd

@JohnRennie Do you know electrochemistry?

John Rennie

I used to though I haven't done it for a long time.

I guess it depends on what exactly you need to know.

Abcd

@JohnRennie electrode potentials

John Rennie

What about them?

John Rennie

06:13

→ 1 message moved to The h Bar

John Rennie

06:27

@Abcd I doubt electrochemistry would be on topic on the PSE. It's normally classified as physical chemistry.

Abcd

Okay

harambe

06:48

@JohnRennie good morning

John Rennie

@harambe morning :-)

I'm working for the next half hour or so ...

harambe

No problem. I usually ask questions from 2.pm IST

John Rennie

07:53

@abcd yes I'm here

Abcd

@JohnRennie Is this OK for sum of digits of a number??

John Rennie

Yes, at first glance that looks fine. I assume you've tested it?

Abcd

Not yet

@JohnRennie Is there any faster way?

John Rennie

@Abcd no, or at least I can't think of a faster way.

Oh, wait, won't sumofdigits(10) return 10 ?

Abcd

It is messing up at the last step.

1369

goes fine till sum = 9+6+3

Then messes up

I think I should put that if thing outside

@JohnRennie No it gives 1

John Rennie

08:05

Let me write the code and test it ...

Abcd

@JohnRennie This is the correct one^

John Rennie

Yes, that's how I'd do it, though you don't need that last if statement since the loop only exits if quotient \lt 10

Abcd

@JohnRennie last if is for the addition of last ones digit of the numbr

@JohnRennie Please see if you are getting right answers without the last if. I dont think you will

John Rennie

Yes, but just put sum += quotient. You don't need the if.

Abcd

Oh that okay.

John Rennie

08:21

If you were implementing sumofdigits as a class method then it can be static since it doesn't use any class variables. If it's a stand alone function then the question doesn't arise.

Abcd

@JohnRennie What's a stand alone function?

John Rennie

@Abcd a function that isn't a member of a class.

Oh wait, if you're using Java there is no such thing.

Abcd

@JohnRennie how can that be

John Rennie

I assumed you were using C

Abcd

No.

John Rennie

08:24

OK, in Java yes you can make the function static.

harambe

@JohnRennie are you free now

John Rennie

@harambe yes ...

harambe

Awesome

Abcd

08:40

@JohnRennie can you see another thing?

?

John Rennie

@Abcd sorry, what was it you wanted to ask?

I was chattering about PCs in the h bar :-)

Abcd

@JohnRennie Is this OK for sum of digits of prime factors for example the output of 23 should be 2+3 = 5 and that of 58 should be 2+9 + 2= 13

int sum_function(int n)
{
int sum = 0;
int quotient = num;
while(true)
{
if (checkprime(quotient == true))
{
sum = sum + sumofdigits(quotient);
break;
}
for(int i = 2; ;i++)
{
if(checkprime(i)== true && quotient%i==0)
{
sum = sum + sumofdigits(i)
quotient = quotient/i;
break;
}
}
}
}

@JohnRennie how to display the code properly in chat?

I have pasted the code with correct indentation but it doesnt get displayed that way

John Rennie

Use the fixed font button

Or just indent everything four spaces

Abcd

 int sum_function(int n)
    {
        int sum = 0;
        int quotient = num;
        while(true)
        {
            if (checkprime(quotient == true))
            {
                sum = sum + sumofdigits(quotient);
                break;
            }
            for(int i = 2; ;i++)
            {
                if(checkprime(i)== true && quotient%i==0)
                {
                    sum = sum + sumofdigits(i)
                    quotient = quotient/i;
                    break;
                }

@JohnRennie In 2nd line quotient = n , not num

John Rennie

It's really hard to just look at code and tell if it works. I'd need to fire up my development system and compile it, and at the moment I don't have Java installed.

Abcd

08:50

Oh okay, leave it.

harambe

@JohnRennie imgur.com/a/uzBaInI

This is giving me a headache

The directions are confusing for me

John Rennie

It seems straightforward enough ...

harambe

@JohnRennie I am confused about angular momentum when objevt undergoes both rotation and translation

Shouldn't the answer just be Mvd

John Rennie

No

harambe

For a body undergoing complete rotation angular momentum is Iw

Like we calculated yesterday

John Rennie

09:00

You have to treat the object as made up infinitesimal masses, $m_i$, at positions $r_i$, and for each of those the angular momentum is $m r_i \times v_i$.

You sum up all the infinitesimal angular momenta to get the total angular momentum.

harambe

@JohnRennie okay

John Rennie

@harambe when we did the wheel yesterday that was a special case because our reference point was the axis of rotation.

In the general case the reference point isn't the axis of rotation.

harambe

OhkY. It makes sense now

@JohnRennie imgur.com/a/uzBaInI

Can you tell me why the expression I have marked is zero

John Rennie

@harambe you haven't marked any expressions ... ?

harambe

09:16

The blue underlined part I have done. Sorry

John Rennie

You mean the line that starts Now or the line that starts Similarly ?

harambe

imgur.com/a/zh1Ckp7

I posted the wrong pic......

John Rennie

In that first expression $r_{i,cm}$ is the distance from the centre of mass to the element $i$.

The expression $\sum m_i r_i$ is the expression for the centre of mass. That is:

$$ M \mathbf R = \sum m_i \mathbf r_i $$

Where $M$ is the total mass and $\mathbf R$ is the position of the centre of mass. OK so far?

harambe

I didn't understand your second line

How does itrelate to centre of mass

John Rennie

Center of mass

In physics, the center of mass of a distribution of mass in space is the unique point where the weighted relative position of the distributed mass sums to zero, or the point where if a force is applied it moves in the direction of the force without rotating. The distribution of mass is balanced around the center of mass and the average of the weighted position coordinates of the distributed mass defines its coordinates. Calculations in mechanics are often simplified when formulated with respect to the center of mass. It is a hypothetical point where entire mass of an object may be assumed to be...

My equation is the standard one for determining the position of the centre of mass.

harambe

09:25

I am seeing this equation for first time. It's good to learn something new

Okay I got it

John Rennie

OK, so $\sum m_i r_i$ gives you the position of the centre of mass.

But the vectors $r_{i,cm}$ are the positions measured in the centre of mass frame. And in the centre of mass frame the position of the centre of mass is at the origin because that's the way the COM frame is defined. And that means $\mathbf R = 0$

harambe

Oh got it

John Rennie

So if $M \mathbf R = \sum m_i \mathbf r_{i,cm}$ that must mean $\sum m_i \mathbf r_{i,cm} = 0$

harambe

Similiarly differentiating the above will give the second expression

John Rennie

Yes

harambe

09:30

Got it

@JohnRennie got the answer

John Rennie

Cool :-)

harambe

@JohnRennie I can even do the disc problem with this derivation

harambe

09:55

@JohnRennie is disc axis of rotation and axle of a disc same?

John Rennie

Normally yes

harambe

@JohnRennie imgur.com/a/UXQl66F

I assumed the force is passing through the the axis

So I got the angular velocity

But I don't know how to get work done

John Rennie

Work done is $\int F(r).dr$ where $F(r)$ is the centripetal force $mr\omega^2$

So you need to get an expression for $\omega$ as a function of the position $r$ of the mass, which is straightforward.

Then just integrate this force $dr$ from $r = R$ to $r=0$

harambe

I am slightly confused. The mass will be experiencing centripetal force from the axis

John Rennie

Yes ... ?

harambe

10:08

And the force by the thread should be different from the centripetal force

John Rennie

@harambe Why?

harambe

@JohnRennie the mass is kept on the frictionless floor so it won't have friction.in order to have centripetal acceleration it should have a force which is F here

Is this the explanation

John Rennie

The mass is confined in a radial channel in the rotating disk, so it can only move in the radial direction. And the question says there is no fiction.

So the only force on the mass is the tension in the thread.

harambe

Okay

John Rennie

So the tension in the thread is equal to the centripetal force.

harambe

10:13

I admit I am not good at circular motion. Maybe I need more practice..........

Yea I got it now

@JohnRennie wouldn't integeration be hard because angular velocity is also not constant

I was thinking of using work energy

Maybe change in K. E will be work done?

John Rennie

The angular velocity does depend on $r$, but that's easy to calculate using conservation of angular momentum. You end up with an expression for $\omega(r)$ that you can integrate.

I need to work for a little while. Back later.

Hema

12:37

A question I'm doing involves eight equal point charges at the corners of a cube at rest. If keeping all others fixed one of the charges is released. What will be its speed at infinity?

How do I approach this problem? I feel 1/8th of the potential energy will be converted toA's KE,but that isnt leading to the right answer. Is there a way of doing this without meticulously subtracting each potential energy pair due to A?

harambe

@Hema from what I can guess the potential energy intially will sll be converted into kinetic energy

Isn't that giving the answer

Hema

13:14

@harambe only one of the charges is released, I actually want to know if there is an easy way to find how much of the initial potential energy is used up when one of the charge is taken to infinity

harambe

Don't know about the easy method. The best bet would be conservation of energy but then again I am not pro at physics. JR or sammy gerbil sir will definitely be able to help you tho

Hema

@harambe ohhh ok

harambe

My guess would be all the potential energy is used because at infinity potential energy will be zero and there is no external force so conservation of mechanical energy will take place

Hema

@harambe but there is still the potential energy of the remaining seven

Hema

13:35

@harambe I got it thanks, I used conservation of energy, thanks for the help.

Never mind I didn't, my answer is wrong by a factor of 2

harambe

@Hema let me try an attempt

Hema

13:53

@harambe ohhh ok

harambe

The answer by me is coming weird. 2(3+3/√2+1/√2) kq^2/ML

Where M is the mass and L is the edge of cube

Mass of the charge

Hema

@harambe that's the answer I'm supposed to get!

Except its 1/ root 3

How did you do it?

What fraction of initial PE?

@harambe Initial PE = 4(3+3/√2+1/√3) kq^2/ML, how much of it did you take as used up by the charge?

harambe

14:17

@Hema the initial potential energy in my case is (3+3/√2+1/√3) kq^2/ML

I am not sure how you got 4 in the expression

Hema

@harambe just a minute

There is no m

4(3+3/√2+1/√3) kq^2/L

harambe

Sorry. Ignore the M

Hema

@harambe this was how I solved it

I took each charge individually and each had interation energies with 3 others at distance L, 3 others at disance √2L and one other at distance √3L

And multiplied it by 8 since 8 charges

And divided the whole by 2 since pairs of charges were counted twice

I think its the right method since the same answer is given in my book

harambe

Wait

@Hema I just calculated potential energy of a single charge q

Hema

@harambe ohhh ok

harambe

14:29

And put it equal to the kinetic energy

Because well we are seeing for just one charge not the system.... My method is like this

Ask JR or sammy sir l. I am still not sure if this is right

Hema

@harambe ohhh ok

Actually I wanted to know if there was a way to get PE of one charge in terms of initial system's PE

Because its a rather lengthy business again calculating for a single charge

Yes I will

sammy gerbil

19:54

@Hema Finding the potential energy of a single charge, is the easiest method, I think. Harambe is correct : its PE when stationary in the cube is converted entirely to KE at infinity. The PE is (3+3/√2+1/√3) kq^2/L. This equals (1/2) mv^2.

If you are asked to find the speed v then there will be an m in the answer, and the whole thing will be inside a square root.

sammy gerbil

20:18

@Hema Your calculation gives you the total electrical potential energy of the system of charges. It is 4x the PE of the single charge which is released. The PE you need to calculate is that of the LAST charge to be placed.

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Problem Solving Strategies