@Bhavay hi :-)

Hello! to u too.

u said if no value of coefficient friction is given we can assume pure rolling motion in pulley question,can this concept extended to objects rolling on an incline or objects moving on horizontal surface?

And yes if no value of friction, it means no friction,right ?

As a general rule if the object is sliding then the question will either tell you the friction coefficient or it will tell you the surface is frictionless.

If it doesn't say anything that usually means the friction is high enough that no sliding takes place.

If you have a particular question in mind you can post it and we can have a look at it.

Yes.

Does Roll means pure rolling ?

@Bhavay I'd guess if a question tells you an object is rolling it normally means pure rolling unless it specifically says otherwise.

@Bhavay that means pure rolling. It doesn't specifically say the object is slipping, and if the object was slipping the queston cannot be answered because α and a would not be related.

Question like these will become much simpler if they clearly states conditions.

@Bhavay I agree, sadly questions are often not as clear as they could be.

Okay sir.

There's one more thing i would like to say , At this point i am self preparing JEE aspirant , and my exam is in next month so i will be be asking a lot of questions sir.

I hope you don't mind that.

I'm always happy to help JEE students. It's such hard work that I'm always ready to help if I can.

Sir for next question , i got my correct answer.Although it doesn't ask for reason , i am providing u the reason( that i think is correct) and i wish u too check that what i think is actually correct or not.

OK ... ?

i simply applied conservation of angular momentum (about the pivoted point).

That's the correct way to do it.

You can't use conservation of enetgy because it is an inelastic collison, but angular momentum is conserved.

now when we apply linear p conser. the new combined velcoity v' = v/7

You can't use conservation of linear velocity because the pivot applies an external linear force.

Even if i consider particle and rod a system?

If the particle and rod were on a frictionless surface then you can use conservation of linear momentum. After the collision the combined rod and particle would end up moving to right.

But with a pivot securing the rod to the surface linear momentum is not conserved. Obviously not since after the collision the linear momentum is zero since the rod just rotates about a fixed point.

U mean the hinge apply a normal reaction so that linear velocity of system becomes 0?

Yes

Also, just to clarify only angular momentum about pivot is conserved right?

Yes, because the pivot doesn't apply a torque.

That's because the system rotates about the pivot, so any force the pivot exerts always has a distance of zero from the centre of rotation.

:-)

Now i have some doubts from the SHM question i did from last night.

OK ... ?

Question Incoming.

Hmm, I guess you are meant to assume θ = 30° so the disk is in equilibrium i.e. the moments from the two masses are equal and opposite.

Yes! , but why ?

that's exactly how they started in the solution.

You mean why assume θ = 30°

First of all is this even SHM?

It's only approximately SHM.

What you'll find is that for small displacements the restoring torque is approximately linear in the displacement angle.

Just like a pendulum.

So this is an angular shm?

You could use angle or sideways distance along the plane. I think angle is probably simpler.

u mean every angular shm can be approximated to linear shm?

Hmm, maybe distance would be simpler. What method does the solution use?

@JohnRennie Angular one. If u want i can post the solution too.

@Bhavay all SHM is related to angular motion. Even with a simple pendulum you can use either sideways distance moved by the bob or angle of the string rotated from the vertical. Both give the same answer.

@Bhavay yes, let's have a look at the solution.

Yes, you end up with the torque approximately proportional to the angle of rotation. That's using the small angle aproximations cos(θ) = 1 and sin(θ) = θ

Sir ...?

Yes?

i don't understand how did they calculated I.

They are calculating the MOI about the point of contact with the ground. The first part is just the parallel axis theorem. The MOI of a disk about its centre is ½MR² so moving a distance R gives you the MOI about a point on the edge as ½MR² + MR². OK so far?

U mean the dashed line, in the diagram i dont see the ground.

They haven't drawn the ground in the figure in the question, but it says the disk is rolling on a horizontal plane.

@JohnRennie Okay so far.

Wait is disc rotating abt center of rotation?

The instantaneous centre of rotation is the red dot, so we are calculating the MOI about the red odt.

Why are we taking iaor?

Will it somehow make our calculation easier?

The disk is effectovely pivoting about the point of contact with the ground. That's why that point is the pivot.

u mean the read dot ,an analogy here- will behave as pendulum with length r?

No, I mean the red dot is the pivot i.e. it is fixed.

then what's doing the shm, the two blue dots?

The disk and the two masses.

The two masses (the blue dots) are fixed to the disk so the disk and the two masses rotate together.

So the total MOI is the MOI of the disk about its edge plus the MOI of the two point masses about the red dot.

If they rotate , how can the red dot be fixed?

For a small angle we can take the red dot as approximately fixed.

You're right it wouldn't be fixed because the disk would roll sideways, but if the rotation is small enough the sideways motion is smll enough to take the red dot as fixed.

Will using Energy conservation be simpler here?

I.e Differentiating total energy of system in a mid position with time.

No, doing it the way the solution does it is the simplest way.

Okay.

How did they calculate the distances of masses from pivoted point?

I got it for $m_a$

That's the calculation they are doing for the torque

I've exaggerated the size of the angle α for clarity

How is Rsin(30+alpha)=2Rsin15?

This is the distance calculation.

ra obviously equals R sqrt(2). Yes?

Yes.

For rb let's zoom in ...

So ½ r_b = R sin(15°)

U just neglected alpha didn't u sir?

otherwise it should be 15+alpha/2.

Yes, I'm assuming alpha is very small and ignoring it. That's what the solution does.

Okay so this problem is finally over.

are u still free?

Yes

I am in for a treat today.

@Bhavay that sounds ominous :-)

That looks straightforward. It's just a complicated way of setting the velocity of the mass on a spring i.e. when the spring extension is zero the velocity of the mass is given by ½mv^2 = mgh i.e. v² = 2gh

It says the mass of the pan is zero, so the velocity of the mass doesn't change when it hits the pan.

Initially shouldn't the spring be extended ?

I think the question means the mass is not initially connected to the spring.

The drawing does make it look as if the mass is connected to the top of the pan then released to fall onto the pan, but I don't think it is.

Yes, Does that mean when the block hits the pan , it will be the mean position as Ke is max there?

No (sorry)

While the mass is falling the spring is at its natural length.

Yes.

But if you consider the mass hanging motionless from the spring there would be an extension given by kx = mg. So the mean position is a distance x = mg/k below the point at which the mass hits the pan.

I can draw a diagram if it would help.

Sure,

What difference does falling mass have on the SHM, when we compare it when the block sits motionless and apply mg?

This is with the mass motionless.

So the equilibrium length of the spring with the mass attached is a distance x = mg/k below the natural length.

Duh,Sir.

Now consider the mass falling and attaching to the bottom of the spring.

Considered.

At the instant the mass attaches to the spring it has fallen a distance h so the velocity is v² =2gh, but it is still above the equilibrium point. It has to travel down a farther distance x = mg/k to reach the equilibrium point.

I think it will extend the spring >mg/k due to that extra velocity.

obviously , mean position will still be at mg/k as there net force =0.

Yes. At the moment the mass attaches to the spring the KE of the mass is mgh. But the spring is displaced a distance x = mg/k from the equilibrium point, so the spring still has a PE ½kx² and the total energy is the sum of these two.

So the total energy is mgh + ½m²g²/k

So the amplitude of the vibration about the new equilbrium point is found by equating this to the PE:

i think u should say spring displaced from it's natural length rather than equilibrium.

½ kx ² = mgh + ½m²g²/k

Oh wait, no, have I missed a term ...

is mgh -ve?

The mhg term is the KE of the block after it has fallen a distance h, so it's a positive number.

@JohnRennie Then haven't u added an extra term here?

If we do this rigorously we would say h is a negative number because the height has decreased (assuming upwards is positive) but g is also negative because g points downwards. So the product gh is positive.

but this will mean . that as we are going down our PE and KE both are increasing.

I'm just doing the calculation. Give me a few minutes ...

Hmm, I must be making a mistake with the calculation somewhere as this isn't working out nicely.

I'm afraid I have to go now. I'll have another look at the calculation when I get a chance.

NP sir, u have been a tremendous help.Bye.