Angular Speed of a Cubical Block Hitting a Ridge

[Abhishekdas]

Toppling based problem...

Homework Statement

A cubical block of side 'a' is moving with velocity v on a horizontal smooth plane.
It hits a ridge on ita way ahead...
Find the angular speed of the block after hitting the ridge..

Homework Equations

No clue

The Attempt at a Solution

I seriously don't have any idea how to ssolve this question...Will it topple? stop? whatever...

Help me out...

Thank u...

 

[cupid.callin]

mind me asking but what's a ridge according to this question?

 

[Abhishekdas]

A ridge is kind of an obstacle...acc to d diagram which i can post here...

 

[cupid.callin]

If you can post here, then please post it!

 

[tiny-tim]

Welcome to PF!

Hi Abhishekdas! Welcome to PF!

A cubical block of side 'a' is moving with velocity v on a horizontal smooth plane.
It hits a ridge on ita way ahead...
Find the angular speed of the block after hitting the ridge..

(cupid.callin, I think it's fairly obvious what a ridge is )

This is a collision.

Momentum and angular momentum are always conserved in collisions …

so use conservation of angular momentum …

what do you get? ​

 

[Abhishekdas]

Sorry cupid callin i meant to say i can't post it...typing error...

And you Tiny-tim i will try what you said and let you know...
Thanks a lot...

 

[Abhishekdas]

This is a collision.

Momentum and angular momentum are always conserved in collisions …

so use conservation of angular momentum …

[/INDENT]

Hi tiny-tim i considered what you said but still I am confused...The block touches the ridge at bottom point on its right...So there is a torque about its Centre of mass...So how is angular momentum conserved? And since the ridge is fixed its like an infinite mass case ...So we can't take ridge+block as a system...So I am not getting how to conserve Linear momentum too...

So please help...eagerly waiting to hear from you furthur...thanks again...

 

[tiny-tim]

Hi Abhishekdas!

hmm … when I said "Momentum and angular momentum are always conserved in collisions"

sorry … I should have said …

Momentum in any direction is always conserved in collisions if there's no external impulsive force in that direction.

Angular momentum about a point is always conserved in collisions if there's no external impulsive torque about that point.​

In this case, there's an external impulsive reaction force whose direction you don't know, so conservation of (linear) momentum isn't available.

But the only external impulsive torque is zero about the pivot point (the ridge), so if you calculate angular momentum about that point, it will be conserved …

what do you get? ​

btw, the gravitational force isn't impulsive, so during the collision it has zero impulsive torque, so you can ignore it in finding the angular speed immediately after hitting the ridge, as the question asks you.

But from then on, we're in the ordinary, non-impulsive, situation, and so it must be included, so angular momentum is not conserved, and we have to use τ = Iα instead (though conservation of energy would be easier: conservation of energy is only unavailable during the collision ).

 

[Abhishekdas]

Heyyyy...thanks a lot tiny-tim... finally got it...

So that was the thing ...conserving angular momentum about the RIDGE...hmmm...

So initial Angular momentum = mva/2
MI of a cube about an edge(since edge of cube and ridge are almost the same point) = 2(ma^2)/3

So Angular momentum after collision = MI*w...
Equating final and initial Angular momentum i get w=3v/4a...which is d answer...

So I am right with whatever i did right?

And ya...one small doubt we could solve this by taking MI about an edge so simply assuming that the ridge hits the cube at the middle of the right edge isn't it...had it not encountered the ridge at the middle i guess things would have got complicated isn't it...Then also could we have got the same answer? Because apparently L is still conserved about the ridge...Or would w have been different?

Waiting for an answer...thanks again...

 

[tiny-tim]

Hi Abhishekdas!

Yes, that's the correct method.

And ya...one small doubt we could solve this by taking MI about an edge so simply assuming that the ridge hits the cube at the middle of the right edge isn't it...had it not encountered the ridge at the middle i guess things would have got complicated isn't it...Then also could we have got the same answer? Because apparently L is still conserved about the ridge...Or would w have been different?

The ridge in the question is transverse.

If it was longitudinal, and if the cube hit it off-centre, then there would be both a horizontal and a vertical component of the angular momentum vector (both before and after, of course).

 

[Abhishekdas]

Hi...tiny-tim...!

Ok...so in this case the ridge is assumed to be tranverse...ok...
And if it was longitudinal and impact was off centre then there would be horizontal and verical components of L right...Hmmmm...and ya...still these components would be individually conserved right? I mean about the ridge of course...?

And ya...one more case...here the ridge is small...ryt...so we conserve angular momentum about the bottom...now had the ridge been of finite size(smaller than block) then the point about which L is conserved is the point (or line in the transverse case) of contact right?

Just asking you all the possibilities ...lol...hoping to get all these concepts clear into my mind...

Neway thanx...man...

 

[tiny-tim]

these components would be individually conserved right? I mean about the ridge of course...?

Yes. Angular momentum is a vector.

When it is conserved, each component is conserved.

… had the ridge been of finite size(smaller than block) then the point about which L is conserved is the point (or line in the transverse case) of contact right?

Yes, the point of contact …

that's where the impulsive reaction force is, so that's the point about which the impulsive torque will be zero.

 

[Abhishekdas]

Got it...thanx...tiny-tim...