Problem Solving Strategies

5:46 AM

@JohnRennie, Hi sir. Good Morning :)

If you are free, could you please answer the following question:

20

Q: Why does the weighing balance restore when tilted and released

I'm talking about a Weighing Balance shown in the figure: Press & Hold on onside of the horizontal beam and then release it. It makes some oscillations and comes back to equilibrium like shown in the figure. Both the pans are of equal equal masses. When the horizontal beam is tilted by an ang...

I'm unable to understand why the beam balance restores its horizontal position even though the net torque about the axis is zero even in the tilted position sir.

user8718165

@Intellex I think that's cuz of the position of pivot...above or below the beam :)

@JohnRennie Good Morning sir 😊

Intellex

Hi

@user8718165 Yes. That's what being said there. I'm unable to understand the reason for that.

John Rennie

@user8718165 morning :-)

@Intellex I think I answered a similar question. Let me have a quick look ...

Intellex

@JohnRennie Yes sir I saw that

Why is the deviation proportional to weight?

John Rennie

5

A: Why does the beam in a weighing balance get tilted proportional to the weights added to each pan?

Suppose you have a balance that looks like this: where $m_1$ and $m_2$ are the weights you're comparing and $M$ is some large weight fixed to the balance. For simplicity let's $m_1$ is zero, so one one end of your scales you have some weight $m_2$ and there is nothing else on the other end. Y...

user8718165

6:00 AM

@Intellex may be...because of the principle of moments ;)

John Rennie

@Intellex doesn't that answer your question?

Intellex

@JohnRennie Yes sir. It was about deviation. But the question I linked was a different one. When we have same weights on both sides and push one arm, why does it restore to its horizontal position even though the net torque is zero?

@user8718165 Thanks, that was not my question. I just said, it was the one sir has answered.

user8718165

@Intellex Ok

John Rennie

@Intellex my answer addresses that. The COM of the beam is below the pivot, so if there is no torque from the masses on the beam, i.e. if the masses on the beam are equal or zero, there is a net torque on the beam that returns it to its horizontal position.

yuvraj singh

@Intellex I agree, with rennie sir.

Intellex

6:23 AM

yuvraj singh

Can you repeat what actually you want.

Intellex

@JohnRennie Ok sir. I haven't observed a beam balance so closely. Could you please tell out of A,B,C,D,E and F (in the above image) which can freely rotate about the points indicated? I'll try to find the rotational equilibrium based on this.

John Rennie

Typically rotation is possible at the points A, C and D

Intellex

@JohnRennie Fine sir. At B? This is the one which confuses me.

If it can rotate about C and D, then it doesn't matter whether it rotates about E or F (I think)

John Rennie

So when rotated the beam looks like this.

Intellex

6:31 AM

@JohnRennie Ok sir. So B is a fixed point.

John Rennie

The red circles show the points where rotation is possible.

@Intellex Yes.

Intellex

@JohnRennie Thank you. Now I'll try to apply my rotational mechanics knowledge and ask if I've any further doubts :)

BTW sir, how did you draw so fast? Which software are you using to draw diagrams - powerpoint or paint? @JohnRennie

John Rennie

I use Google Draw.

It's very quick and easy to use for simple diagrams like that one, though you'd struggle to do complicated diagrams with it.

Intellex

@JohnRennie I didn't know it's something available from Google!

@JohnRennie Fine sir. Thank you.

user8718165

@JohnRennie sir...how are you? What are you doing?

John Rennie

6:41 AM

@user8718165 just chilling and drinking coffee right now. No dead servers this morning (so far!) so I'm having a relaxed morning :-)

And I have a new laptop being delivered this morning, so that's something to look forward to :-)

Intellex

@JohnRennie That's great sir. I'll always be eager when some new tech stuff arrives :)

John Rennie

Yes :-)

Intellex

@JohnRennie, Regarding beam balance: The net torque exerted by the two pans turns to be zero again! So, I think the balance returns to equilibrium just because the COM of the beam shifts. So heavier the beam, more precise the balance, right sir?

John Rennie

> So, I think the balance returns to equilibrium just because the COM of the beam shifts

Correct! That's the key point.

> So heavier the beam, more precise the balance, right sir?

It's the opposite.

Intellex

@JohnRennie Thank you sir.

John Rennie

6:50 AM

Suppose you had an infinitely massive beam, then it wouldn't move at all when you put a weight on it because the motion would be dominated by the weight of the beam. So that would be a pretty useless balance.

But if you has a massless beam then it would rotate 90° as soon as you put even the tiniest weight on it.

Intellex

@JohnRennie Ok sir. Now understood. Thank you :)

John Rennie

So you want a beam mass such that you get a reasonable rotation for the range of masses you want to weigh.

@Intellex I remember being puzzled by exactly this problem when I was a student. Although it's a simple explanation it's the sort of thing that's hard to work out by yourself.

I only know it because someone else told me when I was a student :-)

user8718165

@JohnRennie wow sir....great to know....will you show the pic after it's delivered?

John Rennie

@user8718165 it's this one.

Allegedly it will be delivered this morning.

Intellex

@JohnRennie That's surprising to hear from you sir. Thank you :)

John Rennie

6:55 AM

@Intellex I only know all this stuff because someone else told me when I was your age. I'm not a genius, just old enough to have been taught a lot :-)

user8718165

@JohnRennie okay Sir.... Awesome

Aladdin

@JohnRennie hi.

user8718165

@JohnRennie Sir... I'm confused on this... What if the weights shift the COM by an mm or so?

John Rennie

@Aladdin morning :-)

@user8718165 the weights cause a torque in one direction and the displacement of the beam COM causes an opposite torque. The beam stops rotating when the two torques are equal. Yes?

Intellex

@JohnRennie "I'm not a genius" - Big lie sir :)

@JohnRennie I would like to say, the net torque provided by both the weights is zero all the time sir. I calculated it for the case above. The torque of one pan is equal and opposite to that of the other. So net torque is zero.

user8718165

7:06 AM

@JohnRennie yeah Sir.... Sorry for late response

@Intellex absolutely....

John Rennie

@user8718165 and the restoring torque due to the displacement of the beam COM is proportional to the mass of the beam.

So if the mass of the beam is zero the restoring torque is always zero. Conversely if the beam mass is infinite the restoringtorque is infinite.

user8718165

@JohnRennie got it Sir.....

Thank you very much sir :-) @JohnRennie

Intellex

@JohnRennie Some additional doubts sir: If the mass of the beam is zero, then the beam would not become horizontal once disturbed, right sir? If the beam is infinitely heavy, then restoring torque is infinite. This means the beam quickly comes to its mean position. Then isn't this pleasing to have a heavier beam sir?

John Rennie

@Intellex No, because it means the beam would never move i.e. when you put a weight on one end the beam angle would remain horizontal.

Intellex

@JohnRennie Ok sir. Is that something to do with its moment of inertia (I think yes)?

John Rennie

7:21 AM

No. The beam rotates until the torque created by putting a mass on one end is equal to the torque created by the displacement of the beam COM. Yes?

(I'm not considering the kinetics here, just the equilibrium angle)

Intellex

@JohnRennie Then yes sir. I was thinking oscillations about the mean when both sides have equal mass (if it was heavier, then I felt the oscillations will damp out so fast)

John Rennie

Typically the damping torque is proportional to the angular velocity $\tau = -k\dot\theta$ for some sonstant $k$.

And $\tau = I\ddot\theta$, so the angular deceleration due to the damping will be given by:

$$ \ddot\theta = -\frac{k\dot\theta}{I} $$

The more massive the beam the larger the moment of inertia, $I$. So a more massive beam is damped more slowly than a lighter beam.

Intellex

@JohnRennie It's counterintuitive for me. Thank you sir :)

John Rennie

@Intellex look at this way, if you had a rotating feather it would be easy to stop it rotating. If you had a rotating elephant it would be hard to stop it rotating :-)

Intellex

7:36 AM

@JohnRennie Yes. Now it seems familiar. But what if we consider the friction force at the pivot sir?

Friction is higher when the normal force is greater. So more massive = more friction = more damping right sir?

John Rennie

Yes, I guess that's true. Though I doubt the main damping is due to friction at the bearing as I imaging friction in the bearing would reduce the sensitivity.

I would guess the main damping is due to drag as the pans move through the air.

Intellex

@JohnRennie Ok sir. I will gain some ideas and again troubleshoot this :) As of now, this alone seems to be confusing and I've a compulsion to proceed. Thank you :)

John Rennie

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

user8718165

9:01 AM

@JohnRennie hello Sir :)

@JohnRennie are you free now Sir?

John Rennie

@user8718165 yes, I'm free

Sorry it took me a while to reply. For some reason I didn't hear the ping.

user8718165

@JohnRennie hello sir....no sorry...it's totally fine Sir...

@JohnRennie Sir I'm still confused about the beam rotation for 0 and infinite mass...I thought I got it...but I was wrong...can you please tell me just once more Sir?

John Rennie

9:18 AM

Let me grab the diagram from the answer I linked above ...

user8718165

@JohnRennie okay sir...

John Rennie

This is a somewhat idealised balance. I've simplified it to make the calculation simple and obvious.

Suppose we put a mass $m_1$ on one end of the beam and nothing on the other end i.e. $m_2 = 0$.

user8718165

@JohnRennie okay sir...Is that big M used for stabilizing sir?

@JohnRennie okay sir...

John Rennie

$M$ is the mass of the beam and the centre of mass of the beam is a distance $h$ below the pivot point.

user8718165

@JohnRennie okay sir...

John Rennie

9:23 AM

I'm assuming the centre of mass is at the blue circle. It doesn't matter what the beam actually looks like, only that the CM is a distance $h$ below the pivot.

user8718165

@JohnRennie okay sir

John Rennie

Now suppose we rotate the beam some angle $\theta$. We need to calcuate the torques due to the masses $m_1$ and $M$.

user8718165

@JohnRennie okay sir

John Rennie

The torque due to $m_1$ is $m_1 g \ell\cos\theta$. Yes?

user8718165

@JohnRennie I was writing...okay sir...

John Rennie

9:26 AM

And the torque due to the COM is $M g \sin\theta$. OK so far?

user8718165

@JohnRennie yeah sir...

John Rennie

And at the equilibrium value of $\theta$ the two torques are equal.

user8718165

@JohnRennie yeah sir...

John Rennie

$$ m_1 g \ell\cos\theta = M g h \sin\theta $$

user8718165

@JohnRennie sir $ m_1 \ell\cos\theta = Mh \sin\theta $

John Rennie

9:29 AM

$$ \tan\theta = \frac{m_1\ell}{Mh} $$

user8718165

@JohnRennie If we set M to infinity $\theta$ is 0 and if M=0 $\theta$ is $\infty$

John Rennie

Yes. And $\theta$ is the deflection from the horizontal, so if $M \to \infty$ then there is no deflection of the beam when we put a mass on it.

@user8718165 yes

user8718165

@JohnRennie got it sir... just one last qn sir...

John Rennie

@user8718165 yes?

Yes, but remember that the net torque for a distributed mass is the same as the torque from a point mass at the COM.

user8718165

@JohnRennie okay sir...sorry

@JohnRennie Got it sir...I'll think about it for a while now...

@JohnRennie Thank you very much sir...for the help :)

Intellex

10:44 AM

@JohnRennie, Hi sir. Are you free now, I've some simple doubts (but confusing for me) from rotational mechanics?

John Rennie

@Intellex yes

Intellex

@JohnRennie, Ok sir. If the resultant torque of all the forces acting on a body is zero about a point, is it necessary that it will be zero about any other point?

John Rennie

No

Intellex

@JohnRennie Could you please explain why is this so, sir? If it's zero about a point then it means it's in rotational equilibrium. How can the same body be in rotational equilibrium and not be in equilibrium when we consider torques about different points?

John Rennie

Consider a point mass. I apply a force to that mass and that force is by definition applied at the COM because it's a point mass. Then the mass starts accelerating but doesn't start rotating. OK so far?

Intellex

10:54 AM

@JohnRennie Yes sir.

John Rennie

Now suppose I take my reference point a distance $d$ normal to the force i.e. a distance $d$ to the side. The angular acceleration about this point is $\alpha = da = dF/m$, and that isn't zero.

Intellex

@JohnRennie Yes sir.

Even net torque is non zero about this new point. How is this so, then it means the object must rotate about this point, am I right sir?

John Rennie

And the torque about that point is $\tau = d\alpha$, so the torque is non zero.

@Intellex the object is rotating about that point. If the velocity of the object is $v$ then the angular velocity is $\omega = v/d$.

The object isn't rotating about its COM, but it is rotating about the chosen reference point.

Intellex

@JohnRennie Ok sir. Now understood sir. Thank you :) ; Similarly for my next similar doubt - If the angular momentum of a body is found to be zero about a point, is it necessary that it will also be zero about a different point? - The answer is no, right sir?

John Rennie

@Intellex correct

Intellex

11:03 AM

@JohnRennie Thank you sir :)

If possible could you please help me regarding the following question @JohnRennie sir?

0

Q: Rolling race where objects roll with slipping

One of the interesting demonstrations of Moment of Inertia includes the "Rolling Race" where objects of same mass and radii but having different Moments of Inertia, are allowed to roll down an incline without slipping, and seeing which one crosses the finish line first. We know that, the object w...

homework-and-exercises newtonian-mechanics friction rotational-kinematics

user586228

What is the meaning of "sum of principal minors"?

I am asking this in context of the characterisitc equation generated after subtracting the eigen values.

John Rennie

@Intellex you'd have to do the calculation. I must admit I found the answer surprising, but the slipping means the objects will have a lower angular acceleration so less energy will go into the rotation leaving more KE for the object. Presumably this cancels out the large moment of inertia so all have the same linear acceleration ...

Intellex

11:20 AM

@JohnRennie, Thank you sir. I'll approach this mathematically. But could you please explain your reasoning regarding "Presumably this cancels out the large moment of inertia..."?

How can two independent entities cancel the effects of each other?

John Rennie

That's the only way I can think of that the linear accelerations would work out to be the same. You'd have to do the calculation to see if it's correct.

Intellex

@JohnRennie But no values for friction coefficient, mass, angle, etc. have been given sir. So, I think I've to make a qualitative approach instead of a quantitative one.

If that's the case do you feel my approach is correct sir?

John Rennie

11:40 AM

@Intellex hi, are you still there?

Intellex

@JohnRennie Yes sir. I was waiting for your reply.

Are you busy?

I think your laptop has arrived. Right sir?

John Rennie

Consider the no slipping case. The force acting down the slope is $F_d = mg\sin\theta$. We get the force acting up the slope using the equation for rotation $F_u r = I\alpha$, where $r$ is the radius of our object so $F_u r$ is the torque. Yes?

Intellex

@JohnRennie Yes sir.

Then we have to equate the angular acceleration and the linear acceleration of the centre of mass for pure rolling.

John Rennie

And then we use the relationship between the linear and angular acceleration $\alpha = ra$ to get the linear acceleration.

Intellex

@JohnRennie Yes sir. I'm aware of it :)

John Rennie

11:45 AM

So the end result is that the linear acceleration depends on the moment of inertia, as we all know :-)

Intellex

@JohnRennie Yes sir. But what about rolling with slipping? It was my original question :)

John Rennie

Now consider the case where the object is slipping. In this case $F_u$ is the frictional force $F_u = \mu N = \mu m g \cos\theta$. Yes?

Intellex

@JohnRennie No. It's $0$.

John Rennie

Suppose I am pushing an object over a horizontal surface with friction coefficient $\mu$, then what is the force I experience due to friction?

Intellex

@JohnRennie $\mu N$ if $N$ is the normal contact force (assuming it's kinetic friction)

John Rennie

11:48 AM

And $N = mg$ so the frictional force is $F_f = \mu m g$. Yes?

Intellex

@JohnRennie Of course yes sir

John Rennie

Now suppose that plane is at an angle $\theta$ to the horizontal. Now what is the frictional force?

Intellex

But in the case of slipping alone (without rotation) - we wont be having any friction right sir?

If slipping is confusing, the object just slides down.

John Rennie

If it's slipping then there is a frictional force $F_f = \mu N$.

That's the case regardless of whether the object is rotating as well.

Intellex

@JohnRennie I'll remove some discrepancies sir: In my question I considered the other extreme case as - on a frictionless surface a body just slides without rolling.

John Rennie

11:52 AM

OK ...

Intellex

@JohnRennie Sorry if that confused you initially sir

John Rennie

In that case $\mu = 0$ so $F_f = 0 mg = 0$

Intellex

@JohnRennie Yes sir.

John Rennie

So the linear acceleration is $a = mg\sin\theta - F_f = mg\sin\theta$.

Intellex

@JohnRennie Yes sir.

John Rennie

11:54 AM

And if $\mu$ is non-zero then $a = mg\sin\theta - F_f = mg\sin\theta - \mu mg\cos\theta$

Intellex

@JohnRennie Yes sir. And this is the rolling+sliding case. And has the same outcome as of the qualitative one sir.

John Rennie

So the end result is that the linear acceleration is independent of the moment of inertia. It depends only on $\mu$.

Intellex

@JohnRennie Yes sir. No problem in this fact.

John Rennie

And if the linear acceleration is the same for equal masses then equal masses will accelerate to the end of the slope in the same time regardless of their moment of inertia.

Intellex

@JohnRennie But things get a little weirder when rolling also happens sir. I think both results should be amalgamated :)

John Rennie

11:58 AM

No. The rolling has no effect on the linear acceleration if there is slipping. That's because there is no longer the constraint that $a = \alpha/r$.

Intellex

@JohnRennie Ok sir. Not even a little bit effect to change the winner?

John Rennie

No.

Intellex

@JohnRennie Ok sir, and this justifies the answer in the book. I'll think why I arrived at an incorrect solution, and ask about this further if I've any doubts. Thank you for your help :)

John Rennie

OK.

BTW the laptop has arrived and it's great! :-)

Intellex

@JohnRennie Great! sir. Was the OS inbuilt?

John Rennie

12:03 PM

@Intellex it's a Chromebook, so it has ChromeOS installed.

Intellex

@JohnRennie Ok sir :)

Bye.

John Rennie

Bye

JohnyO42

1:59 PM

Hello people, i have a soft question about studying physics. So next year I'll be starting uni as a physics major, and i want to get to the level of someone who coulddo reasonably well on a physics olympiad. My prep-plan is the following:
Learn: University Physics with modern physics by freedman,
Feynman lecctures
An Introduction to Mechanics: Kleppner Kolenkow
Griffiths E&M
Also do exercise problems from the following:

Irodov : Problems in general physics
Krotov
200 Puzzling physics problems
old olympiads

skullpetrol

6:20 PM

@JohnyO42 I would recommend old olympiads as your main focus.

JohnyO42

Oh i have more than enough problems and problem books so i think im good on that front

The theory aspect is whats not already set in stone IMO

Transcript for

Problem Solving Strategies