Problem Solving Strategies

Prateek Mourya

5:40 AM

Hello sir

John Rennie

@PrateekMourya hi :-)

Prateek Mourya

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

John Rennie

Any rigid body can be thought of as being made up from multiple pointlike masses.

Prateek Mourya

Yes

John Rennie

And as you say it's obvious that τ = I α for a point mass.

Prateek Mourya

5:51 AM

Yes

John Rennie

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)

Prateek Mourya

Yes

But why dont then velocities add the same way

John Rennie

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.

Prateek Mourya

Can the velocity of system and velocity of cemtre of mass be different things?

John Rennie

The phrase "velocity of the system" generally means "velocity of the centre of mass of the system"

Prateek Mourya

5:58 AM

Ok so in terms of velocity and forces we have different intuition for their additions

For system

John Rennie

You aren't adding forces. You are adding torques.

Prateek Mourya

Sorry for silly doubt i forget for a moment what does velocity of system mean:)

John Rennie

Torque = F x r so the torque is normal to both the F vector and r vector.

Prateek Mourya

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?

John Rennie

Yes

Prateek Mourya

6:01 AM

Just like i accepted it Newton's laws of motion

Because of newtons experiments

John Rennie

Yes

Prateek Mourya

Thanks sir

Now i will be able to study this chapter peacefully

John Rennie

:-)

Ashish Ahuja

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:

John Rennie

@AshishAhuja hi :-)

Ashish Ahuja

Hi!

John Rennie

How do you want to do this. Do you want to say what you've done so far?

Ashish Ahuja

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 $\theta = 2 \times arctan(\frac{a}{g})$

John Rennie

Yes, that's what I would say.

Ashish Ahuja

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

John Rennie

Yes, it's a simple mistake. Give me a moment while I finish off answering a question in another room and I'll explain.

Ashish Ahuja

sure, np

till then I'll just write down the last few steps (from $tan(\theta)$)

$tan(\theta) = \frac{h}{w}$

thus

$\theta = arctan(\frac{a}{2g})$

John Rennie

Ashish Ahuja

yeah, that depicts the variables I've used correctly

John Rennie

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).

Ashish Ahuja

6:46 AM

ok

John Rennie

Moving up a distance $h$ increases the PE by $mgh$. Yes?

Ashish Ahuja

yes

John Rennie

But moving a distance $w$ horizontally decreases the PE by $-maw$.

So the total PE change is $\Delta U = mgh - maw$

Ashish Ahuja

hmm ok

John Rennie

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?

Ashish Ahuja

6:49 AM

yup

John Rennie

Giving us:

$$ g \cos\theta = a \sin\theta $$

Ashish Ahuja

yes

John Rennie

Hmm, that gives $\tan\theta = g/a$, how did I get the reciprocal of the actual answer?

Ashish Ahuja

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

John Rennie

That angle is the total angle moved relative to the starting position.

Ashish Ahuja

6:56 AM

hmm, starting position as in the vertical position correct?

John Rennie

Yes

Ashish Ahuja

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 $1 - cos(\theta)$

John Rennie

Oops, yes.

Sorry, brain fade.

Ashish Ahuja

yeah np, that happens to me very often

but it looks like we're still not getting the answer with that change?

continuing from

$mgh = maw$

John Rennie

Hang on, let me grab a pen and try this.

Ashish Ahuja

7:04 AM

sure np

John Rennie

7:22 AM

@AshishAhuja aha, got it :-)

Ashish Ahuja

cool

John Rennie

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?

Ashish Ahuja

yup

John Rennie

So $w = \sin2\theta$ and $h = 1 - \cos2\theta$

Ashish Ahuja

correct

John Rennie

7:27 AM

As before we have $mgh = maw$ so $a \sin2\theta = g(1 - \cos2\theta)$

Ashish Ahuja

yes

John Rennie

And $\sin2\theta = 2\sin\theta\cos\theta$ and $1 - \cos2\theta = 2\sin^2\theta$

Ashish Ahuja

ah ok

John Rennie

And the rest is algebra ...

Ashish Ahuja

yup, give me a second, let me just verify whether I get the same answer..

ok, I am :)

John Rennie

7:30 AM

It was just a matter of being careful with the algebra :-)

Ashish Ahuja

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.

John Rennie

OK :-)

Ashish Ahuja

Thank you :-)

@JohnRennie For how much longer will you be around?

John Rennie

Two or three hours

Ashish Ahuja

ah ok; I gtg now, will try it out a bit later. Bye.

John Rennie