@sammygerbil would you please help me with that question? I still haven't found a solution.

@sammygerbil got your answer. Thanks

@JohnRennie good morning

@harambe morning :-)

Are you free for sometime

Yes, I should be around for several hours

Really...... I wanted some help with rotation

OK ... ?

I am revising rotatikn about fixed axis but I am not able to solve it

Now that I have learned somedhst about COM frame, maybe I can understand what is going on

@JohnRennie when we see rotation in com frame, do we only see objects rotating

I mean just pure rotation

The COM frame just means the total momentum is zero, so objects can still be moving around as long as their total momentum adds up to zero. So motion needn't be purely rotation.

If you're talking about the COM frame of a single object then yes in that frame the object is just rotating.

Okay.

@JohnRennie how should I solve questions where an object is hinged and does rotstion around a fixed axis

Have you got an example?

Let me search

@JohnRennie I don't have one but I have questions

OK ...?

@JohnRennie imgur.com/a/MD8lllR

You can explain to me the one which would be easy and conceptual maybe

I just need a general idea how to approach these questions

hi

@harambe the basic equation for rotational motion is the angular equivalent to Newton's second law:

$$ T = I \frac{d^2\theta}{dt^2} $$

where $T$ is the torque.

Yea. Torque is equivalent to force

So if you look at Q14 you just need to calculate the moment of inertia about the pivot point and also calculate the torque.

@harambe in which class are you studying?

You get the moment of inertia using the parallel axis theorem.

@JohnRennie here the rod will rotate about the pivot right

@harambe yes

I calculste the moment of jnertia. It is coming M$L^2$\12 +Mx2

I forget the moment of inertia of a rod, but adding $Mx^2$ is the correct use of the parallel axis theorem.

@JohnRennie how to add something on bookmark.

@harambe I can't remember how to bookmark in chat. I know it can be done but I've never needed to do it so I've forgoten how.

How do you render mathjax in chat then

@harambe I think it's hard to do on a phone.

You need a laptop really.

Oh

@JohnRennie not able to activate mathjax sigh

Wht is the next step after calculating I axis

@harambe I believe it can be done on a phone, but it's really tricky.

Something about creating a bookmark page... I will try this when I am bored

@harambe calculate the torque as a function of the rotation angle. Obviously at the start it's just $mgx$, but as the rod rotates the toque will change.

@JohnRennie here will the C.O.M move in circular motion

@harambe yes, I think you can assume the rod pivots about the pivot point without sliding. So the pivot to centre of mass distance is always $x$.

Yea. We are told that the pivot point remains fixed in such questions

But why will C. O. M move in circulation motion

I get it... All points move in circular motion along axis, right?

With same angulsr velocity and acceletstion

In circular motion the distance to the pivot is constant. Yes? i.e. the circle has a fixed radius (which is pretty much how you define a circle)

Would the fact all points rotates in circle along axis is the consequece of this for rigid body?

If the rod doesn't slide then the distance from the pivot to the centre of mass is always $x$ i.e. the COM moves at a constant distance from the pivot. So it can't help but move in a circle of radius $x$.

So the motion of com as seen from axis isn't circular always..... Will it depend on the geometry of the object

The centre of mass of an object is a fixed point in that object.

So if the centre of rotation is also fixed then the distance from the rotation point to the COM will be constant i.e. the COM will move ina circle.

Okay. I got it

So should I draw fbd for centre of mass here

Because in my notes we had drawn fbd for centre of mass

And solved it

With respect to fixed axis

@harambe hi

Hi....?

@harambe why with a "?"

@harambe the only force acting at the COM is $mg$ downwards. The torque is $F \times x$ so the torque is $mg x\sin\theta$ where $\theta$ is the angle to the vertical.

I can see that in my fbd now. Okay

@JohnRennie now how should I integerste this or know where the angular acceleration is maximum

@harambe the equation for the acceleration is:

$$ \ddot \theta = T/I = mgx \sin\theta / I $$

And $\theta$ starts out at 90º

Okay.

Let me work on this

The only variable in the equation is $\theta$ - everything else is constant.

And what happens to $\sin\theta$ as $\theta$ reduces from 90º towards zero?

Sin theta becomes zero so angular acceleration becomes zero at that point

Correct, so where is $\ddot\theta$ a maximum?

At theta =0

Got it

A maximum? Isn't that when $\sin\theta$ is a maximum?

I said this for angulsr velocity maybe

Angular acceleration will be maximum where sin theta will be maximum

I made a slight mistake

@harambe yes

@JohnRennie so rest is calculus now

The question doesn't ask you to calculate the equation of motion. It just asks for the equation for $\alpha$.

And that's just $\alpha = T/I$

So we've done that question.

@JohnRennie got the answer

@harambe cool :-)

@JohnRennie can you help me with another example

Yes. I'm using part of my brain for something else at the moment, but I'm happy to devote the remaining brain cells to to rotational motion :-)

Do you like rotational

Well first time I studied this i admit I learnt nothing

Rotational motion problems aren't something you do for fun :-) You do them to learn the principles you need to pass the JEE!

Lol yea. That's the motto for us XD

imgur.com/a/L7VsC7j

Here should I follow the same principle. Observe the trajectory of COM and write torque equation

That's a really hard question!

Oh, hang on, OK they're only asking what happens the instant the string is cut. That's not so hard.

I have this question in my notes too so I was hoping I can at least solve this one with enough honesty now

My teacher have solved this in com frame or something

@JohnRennie should I send in my note solution and then you can explain to me or are you gonna explain to me in your own way

I don't think I can devote enough attention at the moment to do the question justice. Sorry :-(

I'll have to quit for now and maybe look at the problem later.

Oh okay. We can discuss this when we have time

Are you available for other questions

imgur.com/a/IVSNDkC

@JohnRennie when you get time, can you just explain me what to do in the first question

Are they shortening the rod or something or we are calculating a new distance for impulse. In either way, I don't know what to do

if someone prove Goldbach's conjecture then where he can go

ok. is there any online course of number theory

@AdarshKumar you'd be better off asking in the math chat room.

@JohnRennie ok sir

@harambe the entire question isn't in the picture

@AvnishKabaj talking about the first one

@harambe

Find out the rods rotational velocity by conserving angular momentum about the hinge or any other point

Then conserve linear momentum

As no impulse is being delivered by the hinge

You'll obtain a velocity of the com from the rotational velocity

Other from colm

Both the velocities should be the same

You should get the value of x from there

@Hema Oscillation of Spring-Pulley System : This is a mass-on-a-spring type of problem. If we find the equation of motion and put it into the form $\ddot x+\omega^2 x=\text{constant}$ then we can find the period of oscillation from $T=2\pi /\omega$. There are 2 chief methods of finding the equation of motion : (i) apply Newton's 2nd Law to each mass, or (ii) differentiate the energy equation. I shall use method (i).

It looks as though the pulley $m_2$ will rotate, in which case we would have to take into account the rotational acceleration of the pulley as well as its linear acceleration. However, the answer you have been given can only be obtained if we ignore rotational motion. We must assume that the mass of the pulley is concentrated at a point at its centre so that its moment of inertia is zero. Then we can ignore its rotation. (This makes the problem much easier than it would be otherwise.)

Suppose the spring is extended upwards by distance $x$ from its equilibrium length. Then the tension in the spring is $F=kx$. Suppose the tension in the string attached to the pulleys is $T$. Then the resultant force on the pulley equals the upward acceleration of this pulley : $2T-F=m_2\ddot x$.

When $m_2$ moves up by a distance $x$ then $m_1$ moves down by $2x$. So when the acceleration of $m_2$ is $\ddot x$ then the acceleration of $m_1$ is $2\ddot x$. The resultant downward force on $m_1$ is related to its acceleration by $m_1g-T_1=2m_1 \ddot x$. (For $m_1$, down is the direction in which $x$ increases.)

Combine all the equations to eliminate $T, F$. We get $(4m_1+m_2)\ddot x+kx=2m_1g$. From this we can deduce that $\omega=\sqrt{\frac{k}{4m_1+m_2}}$ hence $T=2\pi \sqrt{\frac{4m_1+m_2}{k}}$.

The non-zero constant on the RHS of the equation of motion tells us that the equilibrium position of the oscillations is not at $x=0$ (where the spring is at its relaxed length) but at some other position.

@sammygerbil good evening

Whoops! I missed out the gravitational force on mass $m_2$. However, this only makes a difference to the RHS of the equation of motion, which should be $(2m_1-m_2)g$.

@harambe Good evening.

@harambe Do you have a question to discuss?

I have but I will be free in about a hour. Then we can discuss if you like

ok. I need to take a break also.

@Hema Rotation of pulley $m_2$ : Instead of assuming that the mass of this pulley is concentrated at its centre, we could assume that there is no friction between the pulley and the string. The string then slips around the pulley without making it rotate, so rotational inertia does not enter into the problem.

@harambe I am ready for your question.

imgur.com/a/EWk9WEU

Don't know how to proceeed

What does impulse delivered by pivot becomes zero mean..... Do the normal reactions at the hinges are zero

@sammygerbil

@harambe Isn't this the problem MrXcoder posted above? Have you seen the reply by Avnish?

I have seen it but I wanted to solve this in centre of mass frame and the question doesn't fall in conservation of angular momentum in the given section that's why

@harambe Not all problems are easier to solve in the COM frame.

I know but I wanted to try it as some other questions are solved based on that

@sammygerbil just wanna know what does impulse delivered by pivot becomes zero mean

What will happen when it becomes zero... Will rotation be be without angular aceleration

Impulse is a brief but large force. Imagine the rod floating freely in space, not attached to any hinge. If the rod is hit at the bottom end, two things happen : its COM will move forward in the direction of the impact force, and the rod will rotate about its COM.

This means that the top end of the rod will actually move backwards initially.

Okay

If there is a hinge at that point, the hinge will prevent movement by supplying a force in the forward direction. This motion starts very suddenly in response to the hit at the other end, so the force supplied at the hinge is also very sudden, ie an impulse force.

Okayy so I get what it means now

This force is short duration because it only needs to change the axis of rotation from the COM to the hinge. When this is accomplished, this force is no longer required.

Okay. I will give it a shot now

If you hit the rod at the lower end the impulse force at the hinge is forward. If you hit the rod near the top the impulse force at the hinge is backward because the rod will try to rotate clockwise about the COM but the hinge will prevent this, which requires a backward force.

@sammygerbil so when the impulse by pivot is zero then that means rotstion is smooth

That is there won't be any torque hence angular acceleration

When the impulse at the hinge is zero then there is no motion of the top end of the free rod (where the hinge would be) when hit in the same place. Read and try to understand the comment by Avnish above.

If the rod is hit at the COM the rod does not rotate but it still moves forward, so the top end moves forward also. If the rod is hit a little below the COM then the forward motion of the COM is balanced by the anticlockwise rotation of the rod, so that (initially) the top end of the rod is stationary.

The comment by Avnish explains how to calculate this point.

Okay. I reread @AvnishKabaj comment and you guys are saying the same thing

That's good to know.

One thing that puzzled me is that, after the rod is hit it starts rotating about the hinge, so there is circular motion for which the hinge must be supplying the centripetal force (vertically upwards). But the question says the impulse at the hinge is zero. What about the centripetal force?

Don't know about that but yeah even I can't see a centripetal force

The answer is that the centripetal force is not an impulse force. It exists but it is a constant force and is much smaller than the instantaneous impulse force. The question is only asking about the impulse force, which is very large and instantaneous and horizontal, not the centripetal force which is much smaller but has a long duration and is vertical.

Okay