How can I determine the instant the block overcomes static friction?

  • #36
bob012345 said:
What then did you get for the angular velocity? What method?
Final Energy - Initial Energy = Work done by non conservative forces
=> Elastic potential + rotational kinetic energy of s + translational kinetic energy of s - 0 = 0;
; ##\frac 1 2## ##k x_{max}^2## + ##\frac 1 2## ##I_c ω^2## + ##\frac 1 2## ##m_s v_s^2## = 0, I know everything except ##ω## so I can find it.
 
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  • #37
Thermofox said:
I understand why the spring has maximum stretch right before the block moves. I don't get why we use this value. Because when the block starts moving this will move by dx, this causes a compression of dx in the spring. Therefore the deflection when the block moves should be ##x_{max}-dx## and not ##x_{max}## as we said.
Does it? What value of ##dx##?
 
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  • #38
bob012345 said:
Does it? What value of ##dx##?
Well, I don't know the value, but intuitively, if the block moves forward, the spring compresses by that same distance.
 
  • #39
Thermofox said:
Well, I don't know the value, but intuitively, if the block moves forward, the spring compresses by that same distance.
Remember the other end of the system is accelerating. Of course there is a transition from the static friction on the block to the reduced kinetic or dynamic friction when it moves. This is a complex dynamic system which is why it is good we have tools to solve such systems without having to know all the micro details of the motion as you have done.
 
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  • #40
Thermofox said:
Final Energy - Initial Energy = Work done by non conservative forces
=> Elastic potential + rotational kinetic energy of s + translational kinetic energy of s - 0 = 0;
; ##\frac 1 2## ##k x_{max}^2## + ##\frac 1 2## ##I_c ω^2## + ##\frac 1 2## ##m_s v_s^2## = 0, I know everything except ##ω## so I can find it.
Where did the work go to from ##F##? It's missing again.
 
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  • #41
erobz said:
Where did the work go to from ##F##? It's missing again.
Wait, isn't F a conservative force?
 
  • #42
erobz said:
I guess you have to figure out whether or not the force ##F## is large enough to free the block,
The force on the block is not F.
 
  • #43
haruspex said:
The force on the block is not F.
I know, its ##kx##. Still ##F## is still responsible for the action.
 
  • #44
Thermofox said:
Wait, isn't F a conservative force?
No, its an external input of energy into the system. The energy transferred into the system over displacement ##x## is shared between three reservoirs. Elastic potential in the spring, rotational, and translational kinetic energy of the sphere.
 
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  • #45
erobz said:
No, its an external input of energy into the system.
Isn't the work done by ##F## already inside the elastic potential of the spring, since ##F## caused it to stretch?
 
  • #46
Thermofox said:
Isn't the work done by ##F## already inside the elastic potential of the spring, since ##F## caused it to stretch?
The work done by the force ##F## is partially in the spring and partially in the kinetic energies of the sphere. The input of work stretches the spring and gives the sphere kinetic energy. If the input work to the system was nothing (as you claim RHS = 0) , then nothing happens. ##F## is not a conservative force...
 
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  • #47
erobz said:
The work done by the force ##F## is partially in the spring and partially in the kinetic energies of the sphere. The input of work stretches the sparing and gives the sphere kinetic energy. If the input work to the system was nothing, then nothing happens.
##\frac 1 2## ##k x_{max}^2## + ##\frac 1 2## ##I_c ω^2## + ##\frac 1 2## ##m_s v_s^2## = ##W_F## where ##W_F## = ## F x_{max}##
 
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  • #48
Thermofox said:
##\frac 1 2## ##k x_{max}^2## + ##\frac 1 2## ##I_c ω^2## + ##\frac 1 2## ##m_s v_s^2## = ##W_F## where ##W_F## = ## F x_{max}##
Now don't forget to put ##v_s## in terms of ##\omega## and you are on your way.
 
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  • #49
erobz said:
Now don't forget to put ##v_s## in terms of ##\omega## and you are on your way.
Yeah of course, since I don't know ##v_s##
Before I forgot about the work done by ##F## because I didn't consider that it was external to the system and that it was constantly being applied on the system.
 
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  • #50
bob012345 said:
Remember the other end of the system is accelerating. Of course there is a transition from the static friction on the block to the reduced kinetic or dynamic friction when it moves. This is a complex dynamic system which is why it is good we have tools to solve such systems without having to know all the micro details of the motion as you have done.
Now I understand it. I make all of this considerations using ##x_{max}## because if they work for it, then they will surely work for all the other values of ##x\le x_{max}##. It's a fair approximation that lets us study this complicated system. Thanks!
 
  • #51
So finish the problem please. The proof is in the solution.
 
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  • #52
hutchphd said:
So finish the problem please. The proof is in the solution.

##\frac 1 2## ##k x_{max}^2## + ##\frac 1 2## ##I_c ω^2## + ##\frac 1 2## ##m_s ω^2 R^2## = ## F x_{max}##;

; ##ω^2(m_s R^2 + I_c)## + ##k x_{max}^2## = ##2F x_{max}##;

; ##ω^2## = ##\frac {2F x_{max} - k x_{max}^2} {m_s R^2+ \frac 2 5 m_s R^2}## => ##ω## = ##\sqrt {\frac {2F x_{max} - k x_{max}^2} { \frac 7 5 m_s R^2}}## = ##\sqrt {\frac {2~20~0.0491 - 200~(0.0491)^2} { \frac {21} 5 (0.3)^2}}## ##≈ 1,252~rad/s##

Then I was wondering that to find the remaning informations of the sphere (friction force on sphere ##:= f_s## and acceleration of center of mass ##:= a_c##) I could just use the force balance equation of the sphere:
$$\begin{cases}
F + f_s - F_s = m_s a_c \\
N_s = m_s g \\
τ_{fs} = I_c α
\end{cases}$$
=> ##f_s = \frac {I_c α} R## and ## a_c = \frac {F + f_s - F_s} {m_s}## , since I know that ##α = \frac {a_c} R## I have 2 equations and 2 unknowns.

Edit: adjusted moment of inertia of the sphere
 
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  • #53
Thermofox said:
##\frac 1 2 m_s R^2##
It's a sphere, presumably solid, not a cylinder.
Thermofox said:
Then I was wondering that to find the remaning informations of the sphere (friction force on sphere ##:= f_s## and acceleration of center of mass ##:= a_c##) I could just use the force balance equation of the sphere:
$$F + f_s - F_s = m_s a_c $$
Which direction are you taking as positive for ##f_s##?
##F_s## is not constant.
 
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  • #54
haruspex said:
It's a sphere, presumably solid, not a cylinder.

Which direction are you taking as positive for ##f_s##?
##F_s## is not constant.
Yeah I didn't think it through, I've always done problems with cylinders so in my mind the sphere transformed into a cylinder.
##f_s## has the same direction of ##F##, so I'm considering the right direction as positive
I should've said that I'm finding this two values the moment the block starts moving, so that I can use ##F_{s,max}## which is constant
 
  • #55
Thermofox said:
Yeah I didn't think it through, I've always done problems with cylinders so in my mind the sphere transformed into a cylinder.
##f_s## has the same direction of ##F##, so I'm considering the right direction as positive
I should've said that I'm finding this two values the moment the block starts moving, so that I can use ##F_{s,max}## which is constant
Ok. There is another equation relating ##f_s## and ##a_c##.
 
  • #56
haruspex said:
Ok. There is another equation relating ##f_s## and ##a_c##.
But I don't need it, I think.
##\begin{cases}
f_s = \frac {I_c a_c} {R^2} \\
a_c = \frac {F + f_s - F_s} {m_s}
\end{cases}##
If I substitute ##f_s ## or ##a_c## into the other equation don't I get that value?
 
  • #57
Thermofox said:
But I don't need it, I think.
##f_s = \frac {I_c a_c} {R^2} ##
That's the one.
 
  • #58
haruspex said:
That's the one.
Ok, so am I all good?
 
  • #59
Thermofox said:
Ok, so am I all good?
Yes.
 
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  • #60
haruspex said:
Yes.
Thanks for helping!
 
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  • #61
Thanks, well done.
 
  • #62
Thermofox said:
##\sqrt {\frac {2~20~0.0491 - 200~(0.0491)^2} { \frac {21} 5 (0.3)^2}}## ##≈ 1,252~rad/s##
Check the evaluation. (Maybe you already caught this.)
 

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