Falling Ruler Problem | Solution from Video at 8:03 | Help Available

[Ahmedemad22]

Can anyone help me on this problem from this video starting from 8:03 , I've been working on it for a week ,and I couldn't find a solution.

 

[BvU]

hello Ahmed,

as per the guidelines : show us in your post what you have so far.

 

[Ahmedemad22]

Net torque =mglsin(x)
Mv^2/l=mgcos(x)-n*sin(x)
N is normal force from the floor.
Are my equations right?

 

[BvU]

$\tau = {mgL\sin x\over 2}$

What does $Mv^2/l=mg\cos x-n\sin x\$ describe ? (M same as m ?)

And: where is the similarity with the igloo example ?

 

[Ahmedemad22]

Yes , the same
There are two problems in that video I 'm trying to solve the second one starting from minute 8:04

 

[BvU]

I know. But there is a link between the two problems: Walter says as much. Think of the trajectory that the center of mass of the rules describes.

 

[Ahmedemad22]

Okay , but I wonder if the normal force is making centripetal force component. that is what confused me

 

[BvU]

You can answer that yourself: which way should a centripetal force point ?

Note also that he needs his timer for some reason -- what reason ?

 

[Ahmedemad22]

It should point to the center of the circle that com moves around , but the normal force here has a component that point outward , so it's decreasing the centrepetal force , so it should be in the equation .
I think he used his timer to apply force to the ruler

 

[BvU]

Yes, I suppose it should in an equation. But yours is describing what, exactly ?

I think he used his timer to apply force to the ruler

Correct. What would happen if he didn't
use it ?

 

[Ahmedemad22]

Describing normal force from the floor.

Maybe, ruler wouldn't slide , correct?

 

[BvU]

Describing normal force from the floor.

$Mv^2/l=mg\cos x-n\sin x\$: in $\ mv^2\over 2\$ I see some energy. In $\ mgl\cos x\over 2\$ also. But the other one ? Perhaps it's good to set up a list of known/unknown variables and a few more equations -- all of these important constituents of the template !

Maybe, ruler wouldn't slide , correct?

Do you have a ruler and a reasonably smooth surface nearby ?

 

[Ahmedemad22]

I can't understand what you mean, can you explain further?

Yes , I have made that experiment twice.

Only thing still confusing me is what makes the centrepetal force to the center of the mass .

 

[BvU]

can you explain further?

What is $nl\sin x\over 2$ ?

Yes , I have made that experiment twice.

and did you observe

Maybe, ruler wouldn't slide , correct?

as correct or as proven wrong ?

what makes the centripetal force to the center of the mass

gravity is one component

 

[Ahmedemad22]

I didn't write nlsin(x)/2 in the equations.

Yes , i did

It wouldn't slide if he didn't put the timer. And it slided when i did it

So Mv^2/l=mgcos(x) ?

 

[BvU]

It wouldn't slide if he didn't put the timer.

How do you now ?
I asked what YOU observed

And it slided when i did it

'it' being 'blocked the foot of the ruler from sliding to the left' (while the ruler is falling to the right) ?

From what you post I can not determine what you observed. Please try again.

It wouldn't slide if he didn't put the timer. And it slided when i did it

No
(I suppose with v you mean the velocity of the center of mass of the ruler ?)

 

[Ahmedemad22]

I pushed the ruler a bit and it started to fall then it's gone away from the block i put behind it , and slided until it lied at the surface.

Yes v is com velocity.

This problem is more trickier for me , I even tried to search about it but i didn't find anything helpful.All I'm seeking for are the equations and F.B.Ds

 

[BvU]

I pushed the ruler a bit and it started to fall then it's gone away from the block i put behind it , and slided until it lied at the surface

Yes. So you reproduced Walters result.
What do you observe when you do not put a block behind it ?

This problem is more trickier for me , I even tried to search about it but i didn't find anything helpful

Yes, Walter said it was more complicated. And no, there does not seem to be a worked out solution within google range. We have to think for ourselves

 

[Ahmedemad22]

It slided also but only when it lied at the ground.

But , can a force which is applied from the pivot point make a radial acceleration?(i.e the normal force from the surface that touches the pivot point , this question might be a littel far from the main one but it'll help me through this problem).

 

[BvU]

It slided also but only when it lied at the ground

In my case the bottom end of the ruler (that is falling to the right) slides to the left, already when theta is about 30 degrees.

can a force which is applied from the pivot point make a radial acceleration?

It certainly can. That's why the block is needed -- for the horizontal component

i.e the normal force from the surface that touches the pivot point

That's the vertical component and the normal force works outwards, but: yes

I may be confusing you somewhat. There is also unfinished business:

So Mv²/l=mg cos(x) ?

No. The ruler is not just translating but also rotating. So more complicated than the girl on the igloo: there is a rotational energy term too.

Remember Walter claiming the outcome would be exactly the same (the same cosine of the angle) ?

The need for the block becomes clearer now: it exerts a sideways force. Not at the center of mass but at the pivot point. If we slide that up to the center of mass we get a horizontal component (to increase v) but we have to add a torque (opposite to the torque from $mg\sin\theta$ ). Fortunately we don't calculate $\theta$ as a function of time. All we keep in mind is that the center of mass follows a circular trajectory -- until ...

 

[Ahmedemad22]

So both normal force from ground and the other one from the block does centrepetal acceleration in addition to gravity?

 

[Ahmedemad22]

And does normal force from the block does torque?I don't understand why we need opposite torque

 

[Ahmedemad22]

So normal force from the block directed radially or in the x direction

 

[BvU]

Yes they do contribute, but with a minus sign: they both work in the wrong direction -- opposing gravity and rotational inertia.

And does normal force from the block does torque?I don't understand why we need opposite torque

 

[Ahmedemad22]

But torque from the normal force from the block should be zero as it acts at pivot point

 

[BvU]

Correct. If you take the bottom end as axis of rotation That's a good way to set up equations for $\theta(t)$. But we were after the angle at which the centripetal force would go below what's needed for a circular trajectory of the c.o.m.

 

[Ahmedemad22]

The angle is 48 as sir walter said , and I'm trying to reach it and get that answer

 

[Ahmedemad22]

and what is function of theta