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5:40 AM
Hello sir
 
@PrateekMourya hi :-)
 
Sir how to prove that for a rigid body the effect of change in acceleration of system (angular) of each force acting on each particle can be calculated by sigmaIalpha =F×R
I mean i was able to make intuition for torque for one particle how to do it for multiple
That torque is additive and moment of inertia is additive too like mass
For a system
 
Any rigid body can be thought of as being made up from multiple pointlike masses.
 
And as you say it's obvious that τ = I α for a point mass.
 
5:51 AM
Yes
 
And torques just add. They are vectors and add like vectors.
So the total torque on the rigid body is the sum of all the torques on the point masses that make it up.
i.e. total torque = Σ τ = Σ I α = a (Σ I)
 
Yes
But why dont then velocities add the same way
 
The velocities are all in different directions because they are tangential to the circle the particles rotating.
But the torque vectors are all in the same direction because they lie along the axis of rotation.
 
Can the velocity of system and velocity of cemtre of mass be different things?
 
The phrase "velocity of the system" generally means "velocity of the centre of mass of the system"
 
5:58 AM
Ok so in terms of velocity and forces we have different intuition for their additions
For system
 
You aren't adding forces. You are adding torques.
 
Sorry for silly doubt i forget for a moment what does velocity of system mean:)
 
Torque = F x r so the torque is normal to both the F vector and r vector.
 
One more way could be that experimentally it is verified that
One single torque produces same effect
As different torques added together
Can i use this?
 
Yes
 
6:01 AM
Just like i accepted it Newton's laws of motion
Because of newtons experiments
 
Yes
 
Thanks sir
Now i will be able to study this chapter peacefully
 
:-)
 
6:26 AM
@JohnRennie hi, could you help me out with the following problem regarding harmonic motion? It seemed rather straightforward at first but I'm getting different answers with different methods.
The problem essentially states:
"A simple pendulum is at rest in a car (in its vertical position). The car now accelerates with a constant acceleration $a$. Find the maximum angle of deflection $\theta$ the pendulum makes with respect to the vertical."
I'm trying to solve this using two methods:
 
@AshishAhuja hi :-)
 
How do you want to do this. Do you want to say what you've done so far?
 
Yeah, that's what I'm typing out; I've got two methods, both of which give different answers. That's my dilemma
My first method is as follows:
We can consider the effective gravity acting on the pendulum to have changed by an angle $\alpha = arctan(\frac{a}{g})$
The new equilibrium position of the pendulum is at this angle $\alpha$ wrt to the vertical
One of the extreme positions of the pendulum is the vertical position itself; hence it makes sense to say that the other extreme will be the double of this angle $\alpha$, i.e, the answer $\theta = 2 \times arctan(\frac{a}{g})$
 
Yes, that's what I would say.
 
6:33 AM
Ah ok; my second method is used energy conservation which is getting me a different answer. Here goes -
Initial KE and final KE (at the extreme position) is both zero
Let us split the displacement of the bob into two components, a horizontal component $w$ and a vertical component $h$
Now since we are in a non-inertial frame of reference a pseudo-force is present
acting on the bob towards the left (car is accelerating towards the right)
Total work done = change in energy
Work done by gravity + work done by pseudo-force = change in PE (since change in KE = 0)
assuming mass of bob to be $m$
$maw - mgh = mgh$
$aw = 2gh$
$\frac{a}{2g} = \frac{h}{w}$
$tan(\theta) = \frac{w}{h}$
thus
$\theta = arctan(\frac{2g}{a})$
so where am I going wrong?
It's probably something silly but I just can't figure out what
sorry uh my bad
I wrote something wrong in the above steps but the answer is still different; lemme repeat the last few lines
 
Yes, it's a simple mistake. Give me a moment while I finish off answering a question in another room and I'll explain.
 
sure, np
till then I'll just write down the last few steps (from $tan(\theta)$)
$tan(\theta) = \frac{h}{w}$
thus
$\theta = arctan(\frac{a}{2g})$
 
yeah, that depicts the variables I've used correctly
 
Suppose we rotate the pendulum an angle $\theta$ then the bob moves horizontally by $\sin\theta$ and vertically by $\cos\theta$ (we'll take the length of the string to be unity fr convenience).
 
6:46 AM
ok
 
Moving up a distance $h$ increases the PE by $mgh$. Yes?
 
But moving a distance $w$ horizontally decreases the PE by $-maw$.
So the total PE change is $\Delta U = mgh - maw$
 
hmm ok
 
And at the two extremes of the motion the PE is the same, so we get $\Delta U = 0$ i.e.
$$ mgh = maw $$
OK so far?
 
6:49 AM
yup
 
Giving us:
$$ g \cos\theta = a \sin\theta $$
 
Hmm, that gives $\tan\theta = g/a$, how did I get the reciprocal of the actual answer?
 
uhh idk let me try and see as well..
also shouldn't there be a $2 \times$ somewhere, since we are measuring $\theta$ from the vertical position
 
That angle is the total angle moved relative to the starting position.
 
6:56 AM
hmm, starting position as in the vertical position correct?
 
Yes
 
so then at the other extreme the angle made will be $2 \times arctan(\frac{a}{g})$, as I had calculated above? But here we aren't getting the $2 \times$
or is there something wrong with my first method?
@JohnRennie I feel like there's an issue with this
shouldn't $h$ be $1 - cos(\theta)$
 
Oops, yes.
Sorry, brain fade.
 
yeah np, that happens to me very often
but it looks like we're still not getting the answer with that change?
continuing from
$mgh = maw$
 
Hang on, let me grab a pen and try this.
 
7:04 AM
sure np
 
7:22 AM
@AshishAhuja aha, got it :-)
 
cool
 
OK. Let's define the angle $\theta$ by $\tan\theta = a/g$ so $\theta$ is the angle between the extreme and the equilibrium position i.e. the total swing is $2\theta$. OK so far?
 
So $w = \sin2\theta$ and $h = 1 - \cos2\theta$
 
correct
 
7:27 AM
As before we have $mgh = maw$ so $a \sin2\theta = g(1 - \cos2\theta)$
 
And $\sin2\theta = 2\sin\theta\cos\theta$ and $1 - \cos2\theta = 2\sin^2\theta$
 
ah ok
 
And the rest is algebra ...
 
yup, give me a second, let me just verify whether I get the same answer..
ok, I am :)
 
7:30 AM
It was just a matter of being careful with the algebra :-)
 
Although, isn't this very similar to my first method? We're using the energy relation to find $tan(\theta) = \frac{a}{g}$, from which we get $\theta = arctan{a}{g}$ and then we multiply it by two. Do you know how we could do the same by taking $\theta$ as the angle between the two extremes?
Actually, I'll probably try this out myself.
Will let you know if I don't get it.
 
OK :-)
 
Thank you :-)
@JohnRennie For how much longer will you be around?
 
Two or three hours
 
ah ok; I gtg now, will try it out a bit later. Bye.
 
7:35 AM
Bye :-)
 
8:34 AM
Hi @JohnRennie
 
@AshishAhuja hi :-)
 
So umm I just tried solving the same problem using energy conservation at the extremes and I'm getting a different answer, yet again :(
Could you point out what's wrong in my approach? I'll type it out
 
OK ... ?
 
2 hours ago, by John Rennie
user image
I'm using this setup of variables
The change in PE between the two extremes is $mgh - maw$ as you had shown before, right?
 
So now you're using θ for the total swing?
 
8:36 AM
Yes
I want to try and arrive at the same answer without considering the equilibrium position
 
That's the same as we have already done except that the algebra is going to get very messy.
I tried it and got in a tangle, which is why I went back and wrote the angle as $2\theta$.
 
That's exactly what I had expected, but I'm getting something different; could you hold on for a moment, I think I might have gotten what is wrong.
@JohnRennie ah ok
 
@AshishAhuja OK ... ?
 
Ok yeah, my bad; looks like I had done something silly and was getting a definite answer which was pretty different. Sorry to trouble you.
Also, what software did you use to draw those diagrams?
 
I use Google Draw. It's fine for simple diagrams and it's free! :-)
I screen grab the diagram and paste it into MS Paint, then crop it as necessary.
 
8:43 AM
ahh, I'll try it out one of these days then. Thanks.
 

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