Is w1/w2 = √r2/r1 the Correct Derivation for Circular Central Force?

AI Thread Summary
The discussion centers on the derivation of the relationship ω1/ω2 = √(r2/r1) for two objects spinning on a plate. Participants clarify the meanings of the symbols, where ω represents angular velocity and r denotes the distance from the center of the plate. The conversation highlights the importance of understanding the conditions under which the objects will fall off the plate, specifically relating to centripetal acceleration and friction. After some algebraic exploration and corrections, the correct derivation is confirmed, affirming the relationship between angular velocities and radii. The participants conclude with a sense of resolution and understanding of the concept.
SwedishFred
Messages
28
Reaction score
0
Member warned about posting without the template and with no effort
Hi
we are going through cirkular centralforce and I am complete stuck...
I can't find the derivation to why w1/w2=√r2/r1 is correct
w1/w2=√r2/r1sorry I am lost...
best regards
Fred
 
Physics news on Phys.org
SwedishFred said:
Hi
we are going through cirkular centralforce and I am complete stuck...
I can't find the derivation to why w1/w2=√r2/r1 is correct
w1/w2=√r2/r1sorry I am lost...
best regards
Fred

Maybe you could explain what those symbols mean?
 
yes of course , sorry
There are two objects that are spinning on a plate and they fall off, and the physical reasoning is that the angular velocity (ω) has to meet the condition
ω1/ω2=√r2/r1

ω1= angular velocity objekt 1
ω2= angular velocity objekt 2
r2= objet 2 position to the center of the plate
r1= objet 1 position to the center of the plate

Best regards
 
SwedishFred said:
yes of course , sorry
There are two objects that are spinning on a plate and they fall off, and the physical reasoning is that the angular velocity (ω) has to meet the condition
ω1/ω2=√r2/r1

ω1= angular velocity objekt 1
ω2= angular velocity objekt 2
r2= objet 2 position to the center of the plate
r1= objet 1 position to the center of the plate

Best regards

You should probably the objects will fall of when the acceleration exceeds that which can be produced by the frictional force holding them on. Do you know that the acceleration is given by the expression ##v^2/r## where ##r## is distance from the center and ##v## is rotational speed? What's the relation between ##v## and angular velocity?
 
I Think that the relation is V=ωr correct? (and the friktionskoefficient is the same for both the objects)
 
That is correct. Now we need an equation to express 'staying on the plate' versus 'flying off' a bit more in physics terms using the variables in our exercise.
The template had an item for that; unfortunately it has disappeared (how ?, strange !). So here is a copy:

Homework Statement



Homework Equations



The Attempt at a Solution

 

Homework Statement



There are two objects that are spinning on a plate and they fall off, and the physical reasoning is that the angular velocity (ω) has to meet the condition
ω1/ω2=√r2/r1

ω1= angular velocity objekt 1
ω2= angular velocity objekt 2
r2= objet 2 position to the center of the plate
r1= objet 1 position to the center of the plate

Homework Equations


ω1/ω2=√r2/r1
V=ωr

The Attempt at a Solution



√r2/r1 * ω1/ω2=0

√((ω1r2)/(ω2r1))=0

hmm this doenst feel right..
 
SwedishFred said:

Homework Statement



There are two objects that are spinning on a plate and they fall off, and the physical reasoning is that the angular velocity (ω) has to meet the condition
ω1/ω2=√r2/r1

ω1= angular velocity objekt 1
ω2= angular velocity objekt 2
r2= objet 2 position to the center of the plate
r1= objet 1 position to the center of the plate

Homework Equations


ω1/ω2=√r2/r1
V=ωr

The Attempt at a Solution



√r2/r1 * ω1/ω2=0

√((ω1r2)/(ω2r1))=0

hmm this doenst feel right..

You haven't really done anything except for make an algebra mistake. Start from the physics. What condition will make an object fall off the plate?
 

Homework Statement


[/B]
Object 1 lies on a spinning plate at a distance ##r_1## from the axis. It flies off at angular speed ##\omega_1## (at lower speeed it is held in orbit by friction)
Object 2 lies on the same plate at a distance ##r_2## from the axis. It flies off at angular speed ##\omega_2##.
Friction coefficients are the same for both objects.
Masses of objects may or may not be the same.

Show that ##\omega_1/\omega_2=\sqrt{r_2/r_1}##


Homework Equations



Friction force required to stay in circular orbit F = ...
Maximum centripetal acceleration friction force can provide aat fly-off speed =


The Attempt at a Solution



...
 
  • #10

Homework Equations



Friction force required to stay in circular orbit F = m·v2/r
Maximum centripetal acceleration friction force can provide a at fly-off speed ac =v2/r

on the right track?
 
  • #11
SwedishFred said:

Homework Equations



Friction force required to stay in circular orbit F = m·v2/r
Maximum centripetal acceleration friction force can provide a at fly-off speed ac =v2/r

on the right track?

Right track. Now replace the ##v## with an expression involving angular velocity.
 
  • #12

The Attempt at a Solution


F = m·v2/r = (m* (ωr)2)/r
ac =v2/r =(ωr)2/r
 
  • #13
SwedishFred said:

The Attempt at a Solution


F = m·v2/r = (m* (ωr)2)/r
ac =v2/r =(ωr)2/r

Ok, now simplify that expression. ##a_c## doesn't depend on the mass, agree? So for any two objects ##a_c## is the same.
 
  • #14
agree
(ωr)2/r = 2ω3r2
 
  • #15
SwedishFred said:
agree
(ωr)2/r = 2ω3r2

I don't recognize the algebra you did you get that. Can you explain?
 
  • #16
Ohh sorry calculation fault
I don't know the name in English (kvadreringsregler) (a+b)2 = a2+2ab+b2

this one is wrong! se next one..
 
Last edited:
  • #17
or the potens rule axbx=(ab)x
 
  • #18

The Attempt at a Solution



I am overseeing the weight i F.
ω²r=ω²r
ω²/ω²=r/r
ω/ω=√r/r

am I right ??
[/B]
 
  • #19
SwedishFred said:

The Attempt at a Solution



I am overseeing the weight i F.
ω²r=ω²r
ω²/ω²=r/r
ω/ω=√r/r

am I right ??[/B]

Right idea. It would look much better if you'd distinguish the two values of ##\omega## and ##r##. Start from ##\omega_1^2 r_1 = \omega_2^2 r_2##.
 
  • #20

The Attempt at a Solution



ω21r122r2
ω2122=r2/r1
2122)=(r2/r1)
ω12=(r2/r1)

I Think i got it right ?!
 
  • #21
SwedishFred said:

The Attempt at a Solution



ω21r122r2
ω2122=r2/r1
2122)=(r2/r1)
ω12=(r2/r1)

I Think i got it right ?!

Yes, you've got it!
 
  • Like
Likes SwedishFred
  • #22
Thanks a lot sir!
we will meet again ;-)
best regards
Fredrik
 
Back
Top