Motion of a particle A along x axis is given by x = 3 - 2 cos 2πt and of particle B along y axis is given by y = 4 - 2 cos 2πt. My book says that minimum distance between themis 3 cm and maximum 7 cm.How do I calculate this?

@Hema The squared distance is $S=x^2+y^2$. Write an expression for $S$ and rearrange into the form $S=a+(b-dcos(2\pi t))^2$. You can then see the max and min values of $S$ by inspection.

But it looks simpler to draw a diagram. (The answers given don't look right. Perhaps that is why you are asking?)

@sammygerbil ohhh ok, I'll trydoing it and tell you, I was not verysure how to handle the mathematical aspect actually

@Hema Maybe there is a mistake in the question. It is not as difficult as it appears.

@sammygerbil ohhh ok

@sammygerbil I actually made a mistake in the question, y is given by 4- 2 sin 2πt

@sammygerbil S = 29 - 12 cos 2πt - 16 sin 2πt. I think this will be max if 2πt = 3π/2 therefore S = 29 - 16(-1) = 45. This gives max distance as 3 root 5.

@Hema How can you tell what the maximum will be?

@sammygerbil it seemed to be the highest value the expression of S could have..

I think you need to combine the sine and cosine terms using a trig identity such as sin(A+B)=sinAcosB+cosAsinB.

@sammygerbil ohhh ok

@sammygerbil got it! For a trig expression a sin x + b cos x max and min values are plus and minus √(a^2+b^2). These are plus and minus 20 for S. Thank you!

Yes that gives you the answers you were looking for.

A mass 2m is at rest on top of a vertical spring and in equilibrium a particle of mass m is released from a height h = 4.5mg/k from 2m. Starting from the time the particle sticks to the plate to when spring is in max compression for first time is 2π√(m/ak) then find a. How do I solve this problem? I tried applying energy conservation to these two instants to get 1/2 mv^2 + 1/2kx^2 = 1/2 kA^2,putting v = √2gh and x= mg/k, but I'm not sure what to do next. The answer I'm supposed to get is 3.

@JohnRennie are you there

@Abcd morning :-)

@JohnRennie Hi. $U(x)= \alpha x^4$ for small x. Find time period of periodic motion. U denotes potential energy.

@JohnRennie In exam, I did this question using dimensional analysis. I checked which of the 4 options has dimensions of time and got the right answer ;).

But I want to know the formal method of doing this.

For quartic potentials the time period depends on the amplitude. You don't get simple harmonic motion.

Its about any periodic motion

Not shm

There isn't a simple way to solve the equations of motion.

I have seen it done, but it involves some scary maths.

@JohnRennie question 9

Odd question ...

Statement 2 is obviously false since a planet in an elliptical orbit moves faster when it's closer to the star.

@JohnRennie Why will speed not be same for both orbits

@JohnRennie Semi major axis length appears to be same so I think it they should have same speed?

In a circular orbit the speed is always constant. Yes?

ya

But in an elliptical orbit the speed is not always constant.

So how can an elliptical and circular orbit have the same speed?

@JohnRennie Okay, got it!

Orbit P looks very odd. Is the dot supposed to be the star? If so orbit P is impossible as the star would have to be in the centre of the circle.

No clue...

@JohnRennie question 13

@Abcd f = - du/dx

Direction is negative of force

So the force is restoring only till v_o

After that the slope is 0

If your particle has more than V_o

Them it'll roll away

To infinity and beyond

@AvnishKabaj How did you do 14?

@JohnRennie Are you there???

@Abcd yes?

@JohnRennie What about 14? Is Dimensional Analysis the ONLY way??

Yes. A quartic oscillator can be solved analytically but it's exceedingly complicated and requires maths far above JEE level. (Far above my level too! :-)

@JohnRennie would you please help me with this question? A mass 2m is at rest on top of a vertical spring and in equilibrium a particle of mass m is released from a height h = 4.5mg/k from 2m. Starting from the time the particle sticks to the plate to when spring is in max compression for first time is 2π√(m/ak) then find a. I tried applying energy conservation but I'm stuck after that.

Pls help me with this

@Hema Suppose you gently attach the extra mass $m$ to the existing mass $2m$. The extra mass compresses the spring so the equilibrium position is lower.

@JohnRennie Question D.

Let's do Hema's question next

@JohnRennie but m has a certain velocity when it meets the spring..

@Hema what I'm getting at is that at the moment the falling mass hits the 2m mass the system isn't at its equilibrium position. So the energy of the system is a combination of the KE of the falling mass and the PE in the spring because the spring isn't at its equilbrium length.

On applying conservation at the two instants I got 10 m^2 g^2 = k^2 A^2.. but that doesn't seem to lead anywhere

@JohnRennie Try my question after this

@JohnRennie ohhh ok

The extra mass $m$ compresses the spring a distance $d = mg/k$

@JohnRennie energy will be 1/2 mv^2 + 1/2kx^2 where v = √2gh and x= mg/k right?

@Abcd Then let me add a pls before this statement

@gateprep Yes. That would be better

@JohnRennie ohhh ok

@Hema the PE is $\tfrac{1}{2}kx^2 = \frac{m^2g^2}{2k}$. Yes?

@JohnRennie that gives 14 m^2g^2/k

@JohnRennie yes

@Hema 14?

The small mass falls a distance $h=4.5mg/k$ so it's PE changes by $mgh = 4.5\frac{m^2g^2}{k}$

So the total energy of the system when the two masses meet and stick is $5\frac{m^2g^2}{k}$

@JohnRennie but shouldn't we take 3m instead as it is at the instant of hitting?

I altered my previous message

Should we take 1/2 3m v^2

At the moment of the collision the $2m$ mass is stationary so its KE is zero

The only kinetic energy is the energy of the $m$ mass, and that's equal to the change in its PE $mgh$.

@JohnRennie ohhhh ok ok

How do we find the time then?

This will give 5 m^2 g^2 = k^2 A^2

The time period is just the usual $\tau = 2\pi \sqrt{M/k}$ where $M$ is the total mass.

But you're given the time from the collision to the bottommost point, and you need to work out what fraction of a whole period that time is.

@JohnRennie how do I do that?

If the total energy is $5m^2g^2/k$ then the amplitude is given by $\tfrac{1}{2}kA^2 = 5m^2g^2/k$

@JohnRennie ping me when u r free to ans my question

@JohnRennie ohhh ok sorry I forgot about the half

@gateprep by the way if I'm not wrong the answer is B- contact force is sum of normal force and friction, and you generally can't make normal force zero unless you lose contact or g is zero

But I'm not fully sure so please feel free to check it with John Rennie Sir

SO why greater?

@JohnRennie Pls respond

@JohnRennie bu there are no further details in the qustion about amplitude

@gateprep if normal force can't be made zero then I suppose frictionalforce must be zero accordingto your question, F = N+0 = N> f as f =0

there is no relation between normal and reaction

I don't know why @JohnRennie is not responding to this

@Hema the idea is to work out what fraction of the amplitude the colliding masses are above the equilibrium position

@gateprep because I'm talking to Hema

@JohnRennie ohhh ok ok

Ok but pls make sure to answer mine

@Hema although I get the amplitude as $\sqrt{10} mg/k$ and that factor of $\sqrt{10}$ looks odd - it doesn't look as if it's going to cancel away nicely.

@JohnRennie ohhh ok

@JohnRennie just a minute, 2m will be at rest at collision moment right?

@Hema yes

So is it possible that the time taken from collision to max compression is one fourth the time period?

@JohnRennie never mind it was a silly idea

No. The time from the equilibrium position to the first maximum compression is T/4. But the collision doesn't happen at the equilibrium position.

@JohnRennie Now can you please see question D?

@Hema does your question give the correct answer?

@JohnRennie the answer is a=3, I believe its right as all the other answers are right in that book

@Hema well the period is going to be $T = 2 \pi \sqrt{3m/k}$

@JohnRennie so basically time taken has to be 1/3 of the time period?

If $a=3$ that means the time to first compression is one third of the total period, which seems plausible.

@JohnRennie how do I reach that result?

But that implies the collision happens at a distance $A/2$ above the equilibrium position

@JohnRennie ohhh ok

How did you get A/2 Sir?

@Hema A complete oscillation is 360º

So one third of an oscillation is 120º = 90º + 30º

So the collision must happen 30º above the equilibrium position. The amplitude is $\sin 30º$ and that's 0.5.

@JohnRennie but won't 90 degrees correspond to an extreme?

@Hema yes. From the equilibrium position to the first max compression is 90º

@JohnRennie ohhh ok

And the collision happens above the equilibrium position so the angle from the collision to the first max compression is greater than 90º

There has to be a simple way to do this that I'm missing ...

Anyhow I have to work now for about half an hour. I'll have a think about it while I'm working.

@JohnRennie ohhh ok Sir, thank you

@JohnRennie m question now

pls

@Abcd

Other kids did it using dimensional analysis

But that didn't strike me at the moment while giving the paper

@AvnishKabaj para has hint

¯\_(ツ)_/¯

Ho gaya na

All that ends well is well

@Hema re that problem - I was ignoring the gravitational potential energy and that's why I couldn't get a sensible answer.

@Abcd D could give a straight line or a non-uniform helix

@JohnRennie ohhh ok ok, I'll try it out and let you know about it Sir

@Hema I'm around for a while if you want to discuss it

@JohnRennie are you free for some time

I need to understand COM frame and how to solve questions using it

@harambe yes I'm around for a bit

COM frame is usually pretty easy to understand

Just give me some time for freshing up

Right, I'll go and make a coffee ...

@JohnRennie are you free now sir ?

@harambe yes

I am really having difficulty in understanding C.O.M frame

Why is it called zero momentum frame and how is it different from ground frame

Feels like I went to ground O in mechanics again

We tend to use the COM frame when analysing collisions

As a general rule when you're analysing collisions you need to (1) conserve momentum and (2) conserve energy.

Applying both conservation laws allows you to calculate what happens in the collision.

My teacher said that we use C.OM frame to study internal energy of the system or something .....I can't understand this

But this can get a bit complicated. The advantage of using the COM frame is that not only is the momentum conserved but it's always equal to zero. That often makes problems simpler.

@harambe I'm not sure how useful general comments like that are. We'd have to consider a specific problem to see when and why the COM frame can be useful.

Let me post one of my teacher's problems

@JohnRennie I actually got A = 15mg/k by using energy conservation and also putting v = w√(A^2-x^2) where x = 2mg/k, and also equilibrium position will be 3mg/k and initial position is already 2mg/k, how do I get the time from this?

@Hema The way I did it was to work out the amplitude using conservation of energy, and I got $A = 2mg/k$.

imgur.com/a/u9JftlH

The collision happens when the displacement from equilibrium is half this i.e. $mg/k$.

@JohnRennie but using gravitational PE difference also this result is obtained?

@Hema yes

@Hema let me have a look at harambe's problem then I'll explain how I did it.

We solved this using C.M frame but I did not understand how

@JohnRennie sure,thank you

@harambe yes, that's a good example of where using the COM frame helps.

Unfortunately I suck at C.O.M ......I am just blank in these questions

@JohnRennie Answer is only straight line.

How can I distinguish when To use C.M frame and when to use ground frame

@harambe well look at your example of two springs in te COM frame:

I am all ears

In the COM frame the system as a whole is stationary and the two masses oscillate towards and away from each other

How are you able to say that

Since the masses are $m$ and $2m$ we know the speed of the $2m$ mass is always half the speed of the $m$ mass, and therefore the distance of the $2m$ mass from the centre (the dashed line) is always half the distance of the $m$ mass.

@harambe in the COM frame the total momentum is zero. Yes?

Yes

So $m_1 v_1 + m_2 v_2 = 0$ or put another way $m_1 v_1 = - m_2 v_2$

Okay so their velocities are opposite for both momentum

Yes, velocities with different signs are opposite.

But what about the spring......is it part of C.OM frame

It will be affecting motion in COM frame in the above casev

We normally take the spring to be massless, so its momentum is zero.

Okay

The force from the spring is an internal force. It can't change the momentum of the system as a whole.

@JohnRennie how did you have the relation between distances between masses and com

@harambe well we established that $m_1 v_1 = - m_2 v_2$

That means $v_1/v_2 = m_2/m_1$

In your problem we $m_2 = 2M$ and $m_1 = M$ so we get $v_1 = -2v_2$

(I'm taking $m_2$ to be the heavier mass)

So v1 and v2 are speeds relative to COM?

So when the masses are moving $m_1$ always moves twice as much as $m_2$

@harambe yes

Okay

@harambe remember that we started with the total momentum must be equal to zero, and that's how we got our relationship between the velocities.

Okay .That clears a lot

So these are the velocities in the frame where the total momentum is zero.

So what comes after this

Your diagram shows the situation at time zero (presumably) with the left mass stationary and the right mass heading off at $v_0$. The total momentum obviously isn't zero.

So is this question wrong.....

So is this question wrong.....

But suppose you're in a car driving right at $2v_0/3$. Then as seen from your car the velocity of the left mass is $-2v_0/3$ and the velocity of the right mass is $v_0/3$. Yes?

Since the car and the c.o.m has same velocity ?

Yes, I've chosen the car velocity to be the same as the COM velocity.

Yes

So to do this problem we do three steps:

1. subtract $2v_0/3$ from all the velocities to transform into the COM frame

2. in the COM frame write down the equations for the velocities as a function of time $v_1(t)$ and $v_2(t)$. This is relatively easy because the symmetry in the COM frame makes things simpler to calculate.

Wait

But isn't those velocities already from COM frame

Why substract them

In the ground frame $v_B = 0$ and $v_A = v_0$. Yes?

Yeah

(I'll use your letters for the masses)

If I subtract $2v_0/3$ from all velocities then what do I get for $v_B$ and $v_A$?

Their velocities in COM frame

Correct. That's my step 1 above.

But that's the initial velocities in the COM frame. The masses are attached to a spring so they change with time.

Dang.......I thought you were substacting from COM velocity but lol it was ground frame velocity

Were you perhaps telling me a shortcut

Well the question describes the system in the ground frame. So step 1 is to work out what the system looks like in the COM frame.

Yeah

So now step 2 is to calculate the motion in the COM frame, and that's easier because it's just simple harmonic motion.

the two masses bounce towards and away from each other on the two ends of the spring.

One thing........

The spring force is causing shm in this frame

Yes

And it is an internal force

Yes

So are we studying motion by internal forces

In COM

Yes

Is it what COM is used for

The COM frame makes this simpler to analyse

Because we can the relationships between the masses like $v_B = -2v_A$

and $x_B = 2x_A$

where $x$ is the displacement from the centre of mass.

So the motion of the blocks by this internal force is equal to the motion of an eqyluivalent body in ground frame

To get to the COM frame we subtracted $2v_0/3$ from all the velocities.

I understood that part

To get back to the ground frame we add back $2v_0/3$ to all the velocities we calculated in the COM frame.

So Vmin will be when velocity is minimum in COM frame

Because for going in ground frame,we have to add 2v0/3

In the COM frame the velocity of B is going to be something like $v_B = -2v_0/3 \cos \omega t$ for some angular velocity $\omega$ that depends on the spring constant and the masses.

So in the ground frame we get $v_B = 2v_0/3 - 2v_0/3 \cos \omega t$

Okay

So in the ground frame the velocity of B oscillates between zero and $4v_0/3$

I see....that makes this much cooler..

And you can do the same for $v_A$

I don't know how I would have solved this using ground frame

You could do it in the ground frame, but the calculation gets very messy very quickly.

Yeah ...one more doubt

Yes?

Why is minimum kinetic energy here 1/2 (m1+m2) (Vcm)^2

Does this energy remain fixed

In both frames total energy is conserved so the sum of KE and PE must be constant.

When the spring is compressed that's transferring energy into the PE of the spring so the total KE has to decrease.

Okay

For example, in the ground frame we start with $v_B=0$ and $v_A=v_0$ so the KE is $\tfrac{1}{2} 2m v_0^2$

(just the KE of the 2m mass)

@harambe OK so far?

Yea

In the COM frame the two masses are stationary twice per oscillation, at their closest approach and farthest distance. So wehn we add $2v_0/3$ to get back to the ground frame we find in the ground frame both masses are moving at $2v_0/3$. Yes?

Yes

So now the KE is $\tfrac{1}{2} (m + 2m) (2v_0/3)^2 $

That's your expression 1/2 (m1+m2) (Vcm)^2

Okay...I think I am getting this

COM is quite frustrating but meh It's do conceptual

I will pondering over this and doing some research.

Working in the COM frame is something you need to get the hang of, but once you've working it out you'll find it's very useful

So I will see you later

OK, bye.

@Abcd straight line is obvious because that's just the magnetic and electric fields balancing out. If the fields don't balance and the particle is moving along the electric field lines then it will accelerate while moving in a circle so you get a non-uniform helix. Does the question say what direction the charge is moving relative to the field lines?

@JohnRennie One doubt

@harambe yes?

imgur.com/a/5qbyVrV

The particles are kept on a smooth surface with 2m given velocity v0

I am told to find the tension in the string

Again we had solved this using com frame

But I can't understand why

In the COM frame you just have two masses rotating in a circle around each other.

So the tension in the string is just the centripetal force.

Let me post the picture of motion in com frame

@JohnRennie is the Centre of mass always located towards heavier mass or lighter mass

That's my version of motion in the COM frame

The dotted line shows the position of the centre of mass

imgur.com/a/VSqgfgX

d=l/3

So I can calculate angular velocity

It is coming to be v0/L

For both the masses

Yes, we've both drawn the same diagram.

I can see that momentum is conserved here at every point

So tension will be providing centripetal acceleration here

Yes

BC the particles have to move in circle to conserve momentum

Yes

So the circle came from symmetry or something courtesy of com

Symmetrical movement around COM

Yes, the two particles both have to move in circles around the centre of mass at the same angular velocity otherwise the centre of mass will move.

This looks awfully close to binary star system

Yes indeed, though in a binary star the distance between the stars can change while here the distance is fixed by the length of the string.

Yeah

@JohnRennie why have they defined U = mu.B corresponding to non conservative magnetic field

So I get the tensions to be 2mvo^2/3L

@Abcd context?

Same tension in both circular motion

@harambe yes, I agree. The tension has to be the same on both because the tension is the same everywhere ina string.

Okay

If the tension on the masses were different that would violate Newton's third law!

Again I found in COM frame ,the motion is due to internal force

Yes

In the COM frame there cannot be any external forces, because external forces would change the momentum

It's called a zero momentum frame that's why...must be it's property

Yes

In the ground frame,is mass m accelerated

Because 2m moves with velocity v0?

Why couldn't this hold true in ground frame

T=2m(vo)^2/L

In the COM frame the speeds of the masses are constant but the direction is continually changing.

So when we write the velocity of the masses as a vector that vector has constant magnitude but continuall changing direction.

I see

To get back to the ground frame we add a constant vector with magnitude $2v_0/3$ to the right.

And the net velocity in the ground frame is the sum of the two vectors. But because the angle between the vectors is continually changing the magnitude and direction of the net velocity also continually changes.

COM helps us to avoid this

Yes, using the COM frame makes the calculation much simpler

I need to head off now. Back tomorrow.

Okay.Have a good day sir

@gateprep chat.stackexchange.com/transcript/message/46068988#46068988 : The contact force F is the vector sum of friction f and normal reaction N : $F^2=f^2+N^2$. Also the friction force is limited by the coefficient of static friction : $f \le \mu N$. Usually $0\le \mu \le 1$ but for some materials $\mu \gt 1$ is possible.

This is quite a confusing question, but since it says "there are 3 forces" then I assume that F is an action force which is applied to the object whereas f and N are both reaction forces which are applied by the object to whatever agent supplies force F. Because of Newton's 3rd Law F equals the vector sum of f and N - ie action = reaction.

(i) If F=0 then f=N=0. In this case F=N=0 and F=f=0 so neither of (a) nor (b) are true. For all values of N (c) implies that f<0. But that is not true in this case either.

(ii) If f=0 then F=N. Now (a) is false but F>f=0 is possible so (b) could be true. (c) is false.

(iii) If N=0 then f=0 also because of the law of friction. F must also be 0. This is the same as case (i).

@gateprep None of (a) (b) (c) or (d) is true for all cases (F=0, f=0 or N=0). So based on my understanding of the question none of the possible answers is correct. Perhaps I have not understood the question. Probably it is not worthwhile to continue trying to solve this.

@Hema You have another problem?

@sammygerbil yes actually, I thought the image had an error so I deleted it,and now its not getting uploaded at all.

@sammygerbil i.sstatic.net/aGFsP.png is the image

@Hema What do you have to do here? Find the oscillation frequency? Do the pulleys have mass?

K1= 2K and K2 = K. According to my book the time period is 2 pi root (3m/k) which I took as meaning that effective spring constant is k/3

I actually am not sure how to get this time period

The safest method is to draw Free Body Diagrams for each object. You also need to write down any equations of constraint - eg fixed length of string if it is inextensible.

@sammygerbil at equilibrium k2x = kx = mg, and T = kx, and also kx = 2kx=2T

But that gives net force on block as kx

And somehow effective spring constant has to be k/3 on the block

I think you are assuming that the mass is not accelerating? However if it is oscillating then it is accelerating.

@sammygerbil then I have to assume that the block has moved a distance d fromits equilibirum position

If spring k2 also extends by d then the string won't move at all

I actually am not very thorough with the topic constrained motion

If spring k1 extends by x then the end of the string (attached to spring k2) goes down by 2x.

@sammygerbil yes..

But k2 still won't extend right?

When the extension of k1 is x then the tension in spring k1 is 2Kx. Assuming the pulleys are massless, this force is balanced by the two tension forces in the string wrapped around the lower pulley : 2Kx=2T so T=Kx.

@sammygerbil ohhhh ok

But then net force on m is kx

T is also the tension in spring k2 because this spring is also massless. So for k2 the extension is y such that T=k2y=Ky=Kx. So y=x.

@sammygerbil so k1 and k2 are both extended by x each

When m moves down by x

So when the tension in the string/spring attached to mass m is T, mass m is displaced distance 2x+y=3x below its equilibrium position. So T=Kx=k'(3x) where k' is the effective spring constant experienced by mass m.

@sammygerbil this is when k1 is extended by x right? Why are we starting from k1?

@sammygerbil never mind I got it thanks

So k'=K/3 which is what you were looking for. We could proceed to write the equation of motion for m but this is not necessary : we know that for mass m oscillating on a spring of (effective) spring constant K/2 the angular frequency is $\omega=sqrt{K/3m}$.

@sammygerbil ohhh ok ok

Yes k1 and k2 each extend by x. The end of the string moves 2 times the extension of k1. So in total m moves by 3x.

We start from k1 because we are trying to find a relation between the tension T in the spring/string attached to m and the displacement of m. This gives us Hooke's Law for the combination of springs : T=k'(3x).

We already know T=Kx from spring k1 so we find that the effective spring constant is k'=K/3.

@sammygerbil ohhh ok ok

(Typing error in previous post when I wrote "(effective) spring constant K/2" this should be K/3.)

@sammygerbil I took it as K/3 only so its fine :)

Next question?

@Abcd Did you get a solution? chat.stackexchange.com/transcript/message/46068711#46068711

@sammygerbil there is a mass attached to a horizotal relaxed spring attached to a wall. At t=0 spring is compressed by 2A and released. The spring collides with a wall a distance A from the mass, losing 2/3rd of its KE and returns. From t=0 what is time taken to come to rest again? (taking root (m/k) as 12/pi)

I thought of using the equation 1/2 k(2A)^2 = 1/2 kA^2 + 1/2 mv^2 then 1/2 kA^2 + 1/6 mv^2 = 1/2 ka^2 where k is the new amplitude

Is it right?

I don't understand what this looks like. Is there a spring attached to both sides of the mass? Is there a diagram?

And what should I do next to get the time?