42 Replies
,rotate
@Gyro Gearloose
Note for OP
+solved @user1 @user2... to close the thread when your doubt is solved. Mention the users who helped you solve the doubt. This will be added to their stats.I applied the condition that velocity along common normal should be the same
I got something like 0.5uā34 or something
It wasn't a clean answer
Oh hey I remember this problem...
Yeah, but idk what went wrong here
I'm very sorry for the handwriting but this was as far as I could figure out.
Just a moment, what's v here now?
Component taken along the incline.
Component of resultant?
Wait a min, I gotta write it out better...
Sorry bout that
@Opt
Wait how did we get v
Since projection of of velocity on given vector must be 2u
I'm really sry I didn't understand š
The 2 other vel we got since along common normal they have same velocities
Didn't understand where the v velocity came
I remember this problem from cengage
I just took some component perpendicular to the u/2 component as v.
The main problem in the question was we don't know perpendicular components which would make the problem much easier.
Ik, there was some simpler solution to this but I can't remember ;_;
Component of what
It was something related to relative velocity ig
Velocity.
Bro š
How am I tripping
Why would there be a velocity perpendicular to u/2 is what I'm asking basically
Assume for a moment there is no such component. Then the sphere remains sort of "stuck" on the left cylinder.
Yeah the net resultant velocity would be in that direction won't it
Well, assuming doesn't do any harm tho right? š„²
'Cuz even if there is no velocity in that direction, v would come out to be zero for given constraints.
Yeah but what I'm saying is
If you assume smth that's there in the resultant
Wait I think I get you
@SirLancelotDuLac i kind of get it now thx
I mean, what I said was a bit crass, I apologize. If there is no component of velocity along the incline, meaning the velocity would just be u/2 in the perpendicular direction. Now look at the other wedge and the cylinder. They must always be in contact but our assumption of no perpendicular velocity tells us the block is only moving away from the incline.
I understand this method now, but I'm prolly dumb to ask this why can't we just take u/2 components in x and y and get velocity directly??
u/2 thing is itself a component and component ke components we can't take.
Oh yes nvm I'm dumb af š
I knew I was missing something integral
Lmao sry thx
what?
pretty sure we can
whats wrong wit that
Nah man, I was doing the same error for 30 continous minutes š
the wedges aren't fixed right? if they're moving in opposite directions then how would the cylinder remain in contact with both of them?
this is probably a stupid doubt i'm having, so i'll apologize in advance.
edit: understood that they will eventually separate, and it's asking us for the velocity of the cylinder before they do.
We can't.
i solved a some questions using it and got right answers
whats the reason u say we cant
when we take a component that in itself is a new vector we can again take its component as we please
if that will be of use is another matter
Here we are missing one of the component of the original vector so taking the new vectors components does not give us the original vector tho.
i am just saying that we can take component of a component
Ah, what I meant was "incomplete vector", my bad.
yes