Problem Solving Strategies

harambe

06:05

@JohnRennie good morning

Jasmine

@JohnRennie good morning!

Are you free now?

John Rennie

@harambe @Jasmine hi, yes I'm around for several hours this morning.

harambe

Awesome

@JohnRennie can you help with the graph of lnx/√x

I tried the calculus method but it's not giving me any idea

John Rennie

@harambe I guess so. What are you trying to do?

Jasmine

@JohnRennie in displacement method to calculate focus how is required displacement =$v-u$

@harambe try desmos

harambe

06:21

I have seen desmos but I want to know how to graph it

John Rennie

harambe

@JohnRennie let me send you my work

@JohnRennie the original question is to find the range of the function

Jasmine

This was simple @JohnRennie I got it leave

harambe

The range is from (-infinity to 2/e)

John Rennie

@Jasmine OK. I'd only got halfway through the video you linked anyway :-)

harambe

06:25

@John Rennie How to explain that the function starts from - infinty

The limit at 0 is not defined I guess

John Rennie

Well as $x \to 0$ the value of $\ln(x) \to -\infty$

And as $x \to 0$ the value of $1/\sqrt{x} \to +infty$

Jasmine

@JohnRennie Optics is a hard chapter I feel but everyone says that it's easy do you feel the same?

John Rennie

So it seems obvious the product goes to $-\infty$

harambe

But the denominator tends to. 0 too

John Rennie

@Jasmine optics can get very messy. I find the sign conventions hard to remember and with multiple lenses and mirrors it's easy to get lost in a maze of different values for u, v and f :-)

@harambe huh?

harambe

06:30

What about work and energy... It can be messy too

Jasmine

@JohnRennie yes..

John Rennie

If we write the funstion as $\ln(x) \times \frac{1}{\sqrt{x}}$ then the magnitude of both terms go to infinity as $x \to zero$

harambe

@Jasmine you started revision for mains?

@JohnRennie yeah... I see now. It was a silly doubt.... I was confused due to 1/x

@JohnRennie is infinity x infinity indeterminant form in limit?

John Rennie

@harambe you need to be very careful with infinity. It's ok to say things like $x \to \infty$ because that just means x increases without limit. So you can say $\ln(x) \times 1/\sqrt{x} \to \infty$ meaning the value of the product increases without limit.

But the statement $\infty \times \infty$ is a meaningless statement because infinity isn't a number so you can't multiply it by anything.

harambe

Okay so it just means the values are increasing to a very high value

How to show 2/e is the maxima or the graph

John Rennie

06:38

@harambe It means there is no upper limit to the value.

@harambe do this the usual way. Differentiate the function and set the derivative to zero to find the maximum.

I wonder if there's a more rigorous way to approach the function.

harambe

I get lnx=2x

John Rennie

No, that's wrong.

Hmm

@harambe is the derivative that simple? Let me find a pen and paper to check it ...

harambe

@JohnRennie okay

John Rennie

$$ \frac{d}{dx}( \frac{ln(x)}{x}) = \frac{2 - ln(x)}{2x^{3/2}} $$

Does that look correct?

So the maximum is at $x = \exp(2)$

harambe

@JohnRennie I got $$ \frac{d}{dx}( \frac{ln(x)}{x}) = \frac{ln(x)-2x}{2x^{3/2}} $$

I think I did mistake in derivatives only

@JohnRennie did you get the maxima

I think it should be the maxima

John Rennie

06:56

Suppose we use the product rule, $f = uv$ where $u = \ln(x)$ and $v = x^{-1/2}$

harambe

Okay

John Rennie

Then $du = 1/x$ and $dv = -1/2x^{3/2}$

harambe

Yes

John Rennie

And $$ df = udv + vdu = \frac{-\ln(x)}{2x^{3/2}} + \frac{1}{x^{3/2}} $$

And that gives $$ df = \frac{2 - \ln(x)}{2x^{3/2}} $$

harambe

Okay

John Rennie

07:01

Hmm, our results are similar. Just the minus sign and you got $2x$ instead of $2$

Anyhow the maximum is at $df=0$ so that's when $\ln(x) = 2$ i.e. $x = e^2$

So the maximum value of $f$ is $$ f = \frac{\ln(e^2)}{\sqrt{e^2}} = \frac{2}{e} $$

harambe

@JohnRennie okay. Got it

harambe

07:20

@JohnRennie imgur.com/a/j67KUPA

I am having trouble in identifying the restoring force

At equilibrium, the force due to line charge which is at radial direction would be equal to mg

John Rennie

Yes

harambe

If I displace it downwards or upwards then accordingly the force due to line charge will increase and decrease respectively for any of this case

John Rennie

The field due to an infinite line charge is $2k\lambda/r$

harambe

That would mean the gravitational force can it cant act as restoring force, same for line charge

@JohnRennie yes

John Rennie

If we assume that the charge is above the line charge then the electrostatic repulsion is pushing it upwards and gravity is pulling it downwads. The net force is:

$$ F(r) = \frac{2k\lambda}{r} - mg $$

harambe

07:25

Yeah

John Rennie

Where I'm taking the upwards direction to be positive and $g$ is the number $9.81$ hence the gravitational acceleration is $-g$.

harambe

Okay

John Rennie

Suppose we displace the charge upwards by a small distance $x$, then the net force on the particle is:

$$ F = \frac{2k\lambda}{r + x} - mg $$

harambe

Ok

John Rennie

This doesn't look much like a simple harmonic force, but we can rewrite the expression for the force using the binomal expansion. Start with:

$$ F = \frac{2k\lambda}{r(1 + x/r)} - mg = \frac{2k\lambda}{r} (1 + x/r)^{-1} - mg$$

OK so far?

harambe

07:30

Yes

John Rennie

And we can expand the $(1 + x/r)^{-1} \approx 1 - x/r $ since $x \ll r$

harambe

Okay

John Rennie

So we get: $$ F = \frac{2k\lambda}{r} - \frac{2k\lambda}{r^2} x - mg $$

At equlibrium the net force is zero, and we already worked out at equilibrium $$F = \frac{2k\lambda}{r} - mg = 0$$

harambe

Yea

John Rennie

So that means $$ F(x) = -\frac{2k\lambda}{r^2} x = -kx $$

harambe

07:35

@JohnRennie I see

John Rennie

where $k = 2k\lambda/r^2$

harambe

Yes

John Rennie

And that's the equation for simple harmonic motion!

harambe

@JohnRennie I am slightly surprised. My friend solved this question without undergoing these steps

I wonder how

@JohnRennie one doubt

John Rennie

Yes?

harambe

07:38

When the particle goes down the equilibrium point then the restoring force will be force of the infinite line charge and when it goes above the equilibrium point then restoring force is mg

That would mean the restoring force changes... Is it allowed in shm?

John Rennie

That's not really true. Both forces are always acting, it's just a matter of which one is greater.

@harambe consider the classic SHM example of a mass hanging from a spring.

harambe

Okay

John Rennie

If we move the mass upwards the gravitational force is greater then the spring force, while if we move the mass down the spring force is greater than the gravitational force. So we have two different types of force acting there.

In this case the EM force takes the place of the spring force.

harambe

Oh yeah...... I remember. Wonder why it gave me problem here then

@JohnRennieblet me try this now then

John Rennie

@harambe whenever you see small used in a question it's always worth considering whether the solution involves a binomial expansion.

If you ever work as a physicist you'll get very used to doing this as we do it all over the place to simply calculations.

harambe

07:43

Got it

John Rennie

That's why I immediately thought of it :-)

harambe

@JohnRennie I think we did the similar things in that shm question where a charge was moving on the perpendicular bisector of the line where charge is kept

John Rennie

Yes!

John Rennie

08:01

I'm guessing the mass of the particle cancels out in the final calculation.

harambe

@JohnRennie just got the solution

Yea I substituted but one thing you missed to multiply by q in the restoring force

I got 2s as answer

@JohnRennie imgur.com/a/VnD6ilE

The direction of accelerattion will be towards tangent to the electric field as it is the direction of force

But what can we say about velocity........

I would think the velocity is towards the tsngent field lines also but there is no such option

John Rennie

Velocity is the integral of acceleration so we can't say anything about the velocity unless we know the initial velocity.

For example (c) says the initial velocity is zero. In that case after an infinitesimal time dt we have v=adt so the velocity is in the same direction as the acceleration.

harambe

@JohnRenniebwhen we move along field lines potential decreases. Potential is negative integral of electric field so it makes sense but how to understand intuitively

John Rennie

08:18

The force on a charged particle is along the field line, just like the force on a particle in a gravitational field is downwards.

If the particle moves some small distance $dx$ then it has work done on it equal to $F.dx$ so it accelerates i.e. its kinetic energy increases.

harambe

So conservative forces decrease potential if they act

@JohnRennie yes

John Rennie

That increase in the kinetic energy has to come from somewhere, and it comes from a decrease in the potential energy.

harambe

Makes sense

John Rennie

I need to work for a bit. Back in half an hour to an hour or so ...

harambe

Sure. I busy myself for a bit

harambe

09:52

@JohnRennie are you free for some time

John Rennie

@harambe hi, yes, I'm here for a bit

harambe

@JohnRennie imgur.com/a/XqMvBaV

Q2

I initially solved it but my answer missed by factor of 1/2 but then I remembered about virial theorem

I have one doubt... The total potential energy in virial theorem is sane as the electric potential energy between the charges in this case?

John Rennie

Did you assume the rotating particle was orbiting the central particle i.e. the electrostatic force is equal to the centripetal force?

harambe

Yes

John Rennie

OK, that would have been my guess though the question doesn't make it clear.

harambe

09:59

I wrote the potential energy of the two different configurations and used mechanical energy conservation

John Rennie

@harambe yes, the virial theorem will apply to this system and the potential energy is equal to the electrostatic potential energy. So KE = -0.5 PE.

Though I don't think you need to invoke the virial theorem to solve this problem.

harambe

@JohnRennie it can solve the 1/2 factor inconsistency imo

Work done by external agent is change in K. E

John Rennie

If you missed a factor of 1/2 I think you need to look at your calculation rather than invoke the virial theorem.

The work done is the change in total energy i.e. it is the increase in PE + the decrease in KE.

So jut calculate the total energy before and after.

harambe

Okay got it

John Rennie

The change in energy will be $\frac{kq^2}{2r_2} - \frac{kq^2}{2r_1}$

harambe

10:13

@JohnRennie not getting

I think I proceeded wrong in my first trial.........

John Rennie

Let's start with the orbiting charge at a distance $r$, so the PE is $-kq^2/r$. Yes?

harambe

Yes

John Rennie

The centripetal force is $m$ times the centripetal acceleration so it is $mv^2/r$, and the EM force is $kq^2/r^2$. For a stable orbit we set these equal:

$$ \frac{mv^2}{r} = \frac{kq^2}{r^2} $$

And rearranging gives:

$$ \frac{mv^2}{2} = \frac{kq^2}{2r} $$

and the left hand side is just the KE so we find that the KE is $kq^2/2r$

The total energy is the sum of the PE and KE:

harambe

Okay

John Rennie

$$ E = -\frac{kq^2}{r} + \frac{kq^2}{2r} = -\frac{kq^2}{2r} $$

harambe

10:18

I got till K. E part

John Rennie

@harambe so where did you lose track of the calculation?

harambe

@JohnRennie nowhere

John Rennie

So we agree on the total energy?

harambe

Yes

John Rennie

OK, that means we start with a total energy of $-kq^2/2r_1$ and we end with a total energy of $-kq^2/2r_2$. Yes?

harambe

10:20

Yes

John Rennie

And the difference in the total energy has to be equal to the work done on the system

harambe

Okay

@JohnRennie work done by external agent doesn't necessarily mean energy is increasing or devreasing right.... It could be any right

John Rennie

If you do work on a system you are adding energy to that system. That energy can't just disappear, it has to increase the total energy of the system.

That is energy after = energy before + work done

harambe

@JohnRennie okay but what if a conservative force does work on the system. Would the same thing apply

John Rennie

Work is a transfer of energy. It doesn't matter where the work is coming from. If you have two things, A and B, and A does work on B then the total energy of A must decrease and the total energy of B must increase. The change is equal to the work done. If that wasn't true conservation of energy would be violated.

harambe

10:34

@JohnRennien consider lifting a mass m. When we lift it, I am doing work on it so it's energy should increase but gravity is doing negative work so it's energy should decrease. So how does this decide whether energy is increasing or decreasing here

John Rennie

The increase in the energy of the mass you lift comes from you. That is, your energy decreases by the same amount that the potential energy of the mass increases.

You need to be careful what you mean by gravity here.

What we mean by the gravitational force is that the mass exerts a downwards force $mg$ on your hand, and accordingly your hand exerts an upward force $mg$ on the mass.

So if you lift the mass by some distance $x$ then the work you do on the mass is $+mgx$ and the work the mass does on you is $-mgx$.

harambe

Okay

John Rennie

So the energy of the mass increases by $mgx$ and the energy of you decreases by $mgx$.

harambe

I see. It's a giant puzzle.I still find Work and Energy very confusing....

John Rennie

There is a possible confusion here because strictly speaking it isn't correct to talk about the PE of a single object. Potential energy is something that applies to a syetm as a whole. In this case we have the mass and the earth, and they attract each other with the gravitational force.

When you increase the separation of the mass and the Earth you do work on the whole mass-Earth system i.e. you increase the PE of the mass-earth system.

harambe

10:44

This statement makes more sense to me

John Rennie

Suppose we replace you by a compressed spring i.e. we put the mass on top of a compressed spring.

So the mass-Earth has a gravitational PE, and also the compressed spring has a PE.

harambe

I changed the configuration of earth and mass which would require some energy.... That energy coming from me

John Rennie

If we let the spring raise the mass then the PE of the mass-Earth system increases (becomes less negative), while the PE of the spring decreases.

harambe

Okay

John Rennie

The decrease of the PE stored in the spring is equal to the work done by the spring, and that is equal to the increase in the PE of the mass-Earth (plus any KE the mass gets as it's accelerated by the spring)

So overall the total energy of the spring/mass/Earth combined remains constant.

harambe

10:50

This strikes more intuitive to me

John Rennie

In the typical sorts of exam problems we usually ignore the Earth and just talk about the PE of the mass.

Mr. Xcoder

/s/typical sorts of/all/ :P

John Rennie

The calculation still works, though I can see why if you look too closely at it you can get confused about where exactly the energy is coming from and going to.

@Mr.Xcoder :-)

Part of dealing with exams is understanding what answer the examiner is expecting without worrying too much about the rigorous justification for it!

harambe

Yea.... Maybe after exams I can do more analysis on this... Just solving questions now

Mr. Xcoder

In my (otherwise quite short) contest experience, the claim above is 100% true!

@harambe When are you having your exams?

harambe

10:56

First exam is on January, then it is on April and then if I get selected it is on May

The May exam is final and main exam

Mr. Xcoder

Lol my contests are on December, March and April (exactly one month each before yours)

Good luck and hope you get selected! :-)

harambe

Nice XD. Thanks, You too (^_^)

Mr. Xcoder

Thanks. The first 3 weeks of highschool have been so busy that I only solved about 10 problems during this period... I haven't had time for Physics at all! :-(

And just about anything else, for that matter...

harambe

It's no big deal. If you enjoy physics then you will find a way to deal with that. Before you know, you would be solving physics question during lunch period XD. That's how high school is tho I just used to have fun in my school. Study less, fun more XD

@JohnRennie can you help me with some equilibrium questions

John Rennie

11:12

@harambe I can try :-)

harambe

@JohnRennie imgur.com/a/XqMvBaV

Here why is the total pressure still 1atm...shouldnt it change

John Rennie

I think you uploaded the wrong picture ...

harambe

@JohnRennie sorry for the mistake

imgur.com/a/c2rulk9

John Rennie

@harambe I don't think the answer means the pressure stays at one atm. It clearly doesn't because the pressure increases as the N2O4 dissociates.

No, wait, it is assuming the pressure stays at 1 atm ... at least I think so.

harambe

The total pressure is 1atm at equilibrium.... That's how they have solved it

@JohnRennie I thought of converting kp into Kc

I calculated the answer from it but the answer is coming too big

Or too complex to calculate

John Rennie

11:23

The question doesn't make sense. They say the N2O4 is sealed in a tube, which implies the volume is constant. But in that case the pressure must change because the number of moles of gas is changing.

harambe

Yeah

John Rennie

The answer only makes sense if the volume is allowed to change so the pressure stays constant.

Nehal Samee

Shouldn't it be $0.1(1-α)$ for N2O4?

John Rennie

@NehalSamee yes, that's what I would say.

harambe

It would make sense to convert Kp into Kc then calculate degree of dissociation

Nehal Samee

11:27

@harambe Then you need volume

harambe

Yeah..And it's not given so even that will fail

Nehal Samee

@harambe So... It would be better to follow what @JohnRennie is saying...

harambe

Yep

@John Rennie imgur.com/a/cK9CEYk

Q56

I can't follow the question. Are they talking about another equilibrium with same Kc value

With concertation of CO2 2M

John Rennie

The question says the container initially contains 3 moles of CO and H2O but it doesn't say if that is the quantity before reaction or after equilibirum.

harambe

@JohnRennie How many moles of CO2 must be added... Does this mean at equilibrium

So I calculate the dissociation then calculate concenteration and then see how much additional concentration I should add to get it to 2M

John Rennie

11:44

I have to go now I'm afraid. I may be around later...

Nobody recognizeable

13:24

@JohnRennie hi.

John Rennie

@Nobodyrecognizeable hi

Nobody recognizeable

@JohnRennie are you free now?

John Rennie

@Nobodyrecognizeable yes, for about 15 minutes.

Nobody recognizeable

@JohnRennie does something which is moving around a celestial body has to accelerate to be in that orbit (I am asking for satellite rotation)

John Rennie

@Nobodyrecognizeable Acceleration is a change in the velocity. Yes?

Nobody recognizeable

13:28

@JohnRennie yep.

John Rennie

And velocity is a vector so change in velocity can mean a change in the magnitude of the velocity, a change in the direction of the velocity, or of course both.

Nobody recognizeable

I meant do we need to apply more force after establishing the object into an orbit. If it is already established there? (Looks like you are going to demonstrate centrifugal acceleration) @JohnRennie

John Rennie

@Nobodyrecognizeable I was about to describe centripetal acceleration :-)

Once an object is in orbit that orbit does not need any extra external force to maintain it.

Nobody recognizeable

@JohnRennie how did Cassini fall into saturn then ? There is no friction to stop it's orbiting and no air drag or so. I found youtube videos very confusing as they say cassini

John Rennie

Indeed, applying any external force will change the orbit.

Nobody recognizeable

13:35

Cassini has run out of fuel.(just completed the above statement) @JohnRennie

John Rennie

Cassini was steered into Saturn. Before it ran out of fuel it was steered into an orbit that intersected the atmosphere of Saturn. Atmospheric drag then caused it to crash into Saturn.

Nobody recognizeable

@JohnRennie thanks for clearing the doubt. Have a nice day professor. Goodbye.

John Rennie

Bye

Abcd

16:48

@sammygerbil Please let me know when you are free.

sammy gerbil

17:13

@Abcd hello. I am available for the next 6 hours.

Abcd

Hi, thats great

@sammygerbil I had discussed this with JR 16 months ago but I have completely forgotten. My questions are:
Do the ball and big mass have same horizontal velocity throughout motion?
Do the ball and big mass have same velocity when ball reaches highest point? Why is that?

sammy gerbil

@Abcd No they don't have the same velocity throughout the motion. But they do have the same velocity when the ball reaches the highest point because then their relative velocity is zero - ie the ball has stopped moving up.

Abcd

@sammygerbil Do they have same horizontal velocity throughout the motion

sammy gerbil

No they do not have the same horizontal velocity throughout. I assumed that is what you meant.

Abcd

How is that possible :( ?

blue_eyed_...

17:22

@sammygerbil, A horizontal rod PQ of mass 10gm and length 0.1 m is placed on a smooth plane incident at 60 degree to the horizontal. A uniform vertical magnetic field of value B is applied in the region of PQ. calculate B if the rod remains stationary on the plane when a current of 1.73 A flows in the rod. What is the direction of current in the rod? Please provide some hints

Abcd

Intuition says they should have same horizontal velocity throughout motion.

Because intuition also says "Come on! thats why they are in contact throughout the motion!"

sammy gerbil

@Abcd If they had the same horizontal velocity throughout, the ball would remain in the same position relative to the block. It would not move forward or upward relative to the block.

Abcd

43 secs ago, by Abcd

Because intuition also says "Come on! thats why they are in contact throughout the motion!"

@sammygerbil ^^^

Why do they remain in contact throughout the motion?

@sammygerbil They don't even have same horizontal veloity when ball is moving on the curved surface ?

sammy gerbil

@Abcd Your intuition is different from mine!

Abcd

@sammygerbil can you answer my intuition

Mr. Xcoder

17:25

@Abcd Why would they not? If you apply a force to a ball on very long table, the table is stationary but the ball is moving, but they remain in contact...

Abcd

1 min ago, by Abcd

@sammygerbil They don't even have same horizontal veloity when ball is moving on the curved surface ?

@Mr.Xcoder ^

Like when the ball is heading to top through the vertical region

Mr. Xcoder

My intuition says that unless the ball is going straight vertically upwards (which it doesn't), they don't have the same horizontal velocity

Just at the final point, when the ball no longer moves relative to the curved body, they will have the same horizontal velocity because its vertical component will be 0 and it will remain in the same position relative to the other body

Anyway, I have to go now so I can't continue with my reasoning

sammy gerbil

@Abcd I agree that they remain in contact throughout (until the ball falls off the back of the block). I don't know how to answer your intuition, except by asking you to explain why you think that way. ... They might lose contact if the ball has a large initial velocity and goes up higher than the top of the block.

Mr. Xcoder

@Abcd As a final thought (didn't write this down yet so I may be wrong), is the correct answer (c)?

Abcd

@Mr.Xcoder correct answer is b

Mr. Xcoder

17:33

Ok then I'm not wrong :)

Just checked somethinng.

sammy gerbil

@blue_eyed_... Draw a Free Body Diagram for the rod. There is gravity, normal contact force and the magnetic force. These 3 forces are in equilibrium.

blue_eyed_...

@sammygerbil, I couldn't draw that only

sammy gerbil

@blue_eyed_... The magnetic force on the rod must push it horizontally towards the plane. So the current must be into the page. See hyperphysics.phy-astr.gsu.edu/hbase/magnetic/forwir2.html. ... Why can you not draw the FBD?

Resolve the gravity and magnetic forces along the plane ... they must be equal and opposite.

blue_eyed_...

@sammygerbil, does "a smooth plane incident at 60 degree to the horizontal " means an inclined plane making an angle of 60 degree with the horizontal?

sammy gerbil

@blue_eyed_... I think so. That is how I interpret that sentence. I had to think hard to understand it!

Abcd

17:45

@sammygerbil A uniform ring of radius R is given a back spin of angular velocity $\dfrac{V_o}{2R}$ and thrown on a horizontal rough surface. The velocity of the centre of the ring when it starts pure rolling will be?

@sammygerbil How to do it with conservation of angular momentum.

sammy gerbil

@Abcd Calculate angular momentum about the point of contact with the surface. The COM has angular momentum about this point. Total angular momentum (spin + COM motion) is conserved.

Abcd

@sammygerbil Ya thats what I thought but:

@sammygerbil but there's a problem how can we find angular velocity about point of contact??

Becuase $L= I\omega$

sammy gerbil

@Abcd Is there something missing from the question? It says the ring is "thrown" onto the rough surface. This suggests that it has some initial forward translational motion in addition to the rotational motion.

Abcd

so we would need angular velocity about that point

A uniform ring of radius R is given a back spin of angular velocity Vo/@R and thrown on a horizontal rough surface with velocity of centre to be Vo. The velocity of the centre of the ring when it starts pure rolling will be?

@sammygerbil see now

sammy gerbil

@Abcd Angular velocity of COM is $mvR$ where $v$ is velocity of COM. This is because the point of contact is at rest instantaneously.

Abcd

17:53

@sammygerbil Please see that it has initial rotational motion too!

sammy gerbil

Yes I see. So you will have initial AM of $MV_0 R-I(V_0/2R)$ and final AM of $MVR+I(V/R)$. The final angular velocity is $V/R$ because of the no-slip condition. Note that $I$ is the moment of inertia about the point of contact, not the centre of the ring.

Abcd

@sammygerbil i dot understand IVo/2R term in initial part

sammy gerbil

That is the initial backward angular momentum of the ring. The angular velocity is $V_0/2R$ as given in the question.

It is -ve because the spin is backwards.

Abcd

@sammygerbil but thats not the angular velocity about lowermost point

sammy gerbil

Angular velocity is the same about all points.

Abcd

17:59

@sammygerbil how come

sammy gerbil

Because when an object rotates, all points on it retain the same relative positions. This means that every pair of points rotate about each other with the same angular velocity.

2

Q: How rotating body have same angular velocity and acceleration

Does a body rotating about a fixed axis have to be perfectly rigid for all points on the body to have the same angular velocity and the same angular acceleration

1

Q: Angular velocity of a rigid object about an axis outside of the body

I know that for a rigid body rotating around a fixed axis, the angular velocity of any point with respect to any other point is the same. As a result, the angular velocity is the same for any choice of axis attached to the body as long as that axis is perpendicular to the rotational plane. But ho...

newtonian-mechanics rotational-dynamics reference-frames angular-velocity

Abcd

@sammygerbil hmm I see!

sammy gerbil

For example, you could imagine 3 points on the body A, B, C. If the body rotates such that line AB rotates by angle $\theta$ then the lines AC and BC will also rotate by angle $\theta$ at the same time. So if line AB rotates with angular velocity $\omega=\dot\theta$ then lines AC and BC also rotate with angular velocity $\omega=\dot\theta$. Any line drawn on the rigid body rotates with the same angular velocity.

Point C could be the centre of the ring. Point A could be the point of contact with the ground. Point A rotates about C with the same angular velocity as point C rotates about A.

The rotating body always has the same angular velocity when measured from any inertial (non-rotating) frame of reference. That is because the rotation angle is measured between 2 points in the body relative to a fixed direction in space (ie in the inertial frame of reference).

@Abcd Are you satisfied about the ball (sorry ring) rolling on the sliding block question, or do you still have doubts about the motion?

TheSimpliFire

19:39

Challenge for everyone:

A sample of liquid of mass m1 and temperature T1 is poured into a well-insulated container. A second sample of the same liquid of mass m2 and temperature T2 is added to the container. Let the minimum temperature of the mixture of liquid be T. Given that m2 = T * m1, disregarding units, find T in terms of T1 and T2. You may assume that T1 > T2 and that both T1 and T2 are above 0 Celsius.

sammy gerbil

21:05

@TheSimpliFire I see two problems with your challenge : (1) Why should the mixture reach a minimum temperature? If the container is "well-insulated" the mixture will be at a constant temperature somewhere between T1 and T2, depending on the masses m1 and m2.

(2) If T is a temperature and m1, m2 are masses then how can you write m2 = T * m1 "disregarding units"? The equation is dimensionally incorrect. If you use different units for T (eg Kelvin, Fahrenheit) you will get different answers.

Transcript for

Problem Solving Strategies