Angular acceleration, velocity, momentum of a door?

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Homework Help Overview

The problem involves a door modeled as a uniform rod, focusing on its rotational motion around a vertical axis. The scenario describes an external force applied by a person to slam the door shut, with various aspects of angular acceleration, velocity, momentum, and energy being explored. Participants are tasked with analyzing the effects of different points of force application and the role of hinges in the system.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation, Assumption checking

Approaches and Questions Raised

  • Participants discuss the calculation of angular acceleration and the moment of inertia, with some questioning the application of torque and the axis of rotation. There are attempts to derive equations for tangential forces and the implications of pushing at different points on the door. Several participants express uncertainty about their calculations and seek verification of their results.

Discussion Status

The discussion is active, with multiple participants providing their calculations and questioning each other's reasoning. Some participants have offered guidance on the correct formulas to use, while others have pointed out potential errors in previous calculations. There is a mix of agreement on certain answers, but also significant debate over the correct approach and assumptions.

Contextual Notes

Participants note potential typos in the problem statement regarding the distance from the hinges and the time duration of the force application. There is also mention of the need to clarify the axis of rotation and the moment of inertia used in calculations.

  • #31
haruspex said:
You're not answering my question. Where is this vertical axis in relation to the door's width? Does it run up the middle of the door?
I got 10 in a)...Inertia formula that i have used is: I=(ML^2)/3...The question says that we should consider the door as a uniform rod, hence the formula that i have chosen to use...Please correct me if I'm wrong
 
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  • #32
StavangerFinest said:
I got 10 in a)...Inertia formula that i have used is: I=(ML^2)/3...The question says that we should consider the door as a uniform rod, hence the formula that i have chosen to use...Please correct me if I'm wrong
That's it.
 
  • #33
haruspex said:
That's it.

I'm a bit stuck with e) though, I get tangential force to be Ft= ((mr^2α)/d)-mrω^2-F (the force from the mother-in-law which is 100N), and it doesn't seem like the correct answer.
Guidance or any tip is appreciated
 
  • #34
haruspex said:
That's it.
 
  • #35
On c.) could the time be 2,53s (also got t= -1,26), used abc formula with α=10 and ω=2? Used the inertia formula that StavangerFinest did: I = (ML^2)/3.
 
  • #36
statii said:
On c.) could the time be 2,53s (also got t= -1,26), used abc formula with α=10 and ω=2? Used the inertia formula that StavangerFinest did: I = (ML^2)/3.
I got t= 0,40s...What did you use as θf and θi?
 
  • #37
statii said:
used abc formula with α=10 and ω=2?
what is the abc formula?
The time to close is less than 2 s but more than StavangerFinest's 0.4s.
 
  • #38
haruspex said:
what is the abc formula?
The time to close is less than 2 s but more than StavangerFinest's 0.4s.
I did it again and i get 0,79s... Assistants told me I shouldn't have used abc formula but a simple t= Δθ/ Δω...
Can you help me with question e) and thereby f) in this exercise though? I got in f) that

d= 2L/3 (two thirds of L)

pretty insecure abt this one
 
  • #39
StavangerFinest said:
I did it again and i get 0,79s... Assistants told me I shouldn't have used abc formula but a simple t= Δθ/ Δω...
Can you help me with question e) and thereby f) in this exercise though? I got in f) that

d= 2L/3 (two thirds of L)

pretty insecure abt this one
Right on both.
But I do not agree with your formula in post #33. There should not be any ##\omega## reference. Did you change your formula?
 
  • #40
I am stuck on e). What I have done:
T = d x F x sin(theta)
T = I x alpha

I x alpha = d x F x sin(theta)

F (d) = (I x alpha) / (d x sin (theta))
F (d) = (1/3 m x d^2 x a) / (d^2 * sin (theta))

But then all my d's disapear... Am I on the right track?
 
Last edited:
  • #41
StavangerFinest said:
I got t= 0,40s...What did you use as θf and θi?
I used tetha final to be pi/2 and tetha initial to be 0, and abc formula with alpha=0, got 0,79s then :)
 
  • #42
haruspex said:
Right on both.
But I do not agree with your formula in post #33. There should not be any ##\omega## reference. Did you change your formula?
Just ignore that one. I started over from this formula:

F- Fh= M* a (a centre of mass)

and I worked my way down from there.
 
  • #43
leifen said:
I am stuck on e). What I have done:
T = d x F x sin(theta)
T = I x alpha

I x alpha = d x F x sin(theta)

F (d) = (I x alpha) / (d x sin (theta))
F (d) = (1/3 m x d^2 x a) / (d^2 * sin (theta))

But then all my d's dissapear... Am I on the right track?
d shouldn't disappear, no.
 
  • #44
StavangerFinest said:
d shouldn't disappear, no.

Badly formulated by me. I am stuck, because the d's disapear, that's my problem. But am I using the right method and formulas?
 
Last edited:
  • #45
haruspex said:
That's it.
Sorry, have been busy. I still have 40 in a . Alpha =( ΣF*L/2)/ (1/3*M*(L/2)^2) is my formula. Since it is a uniform rod with an axis through the end and the push is in the middle... I saw answer 10 somewhere... So now I'm puzzled
 
  • #46
As haruspex asked: "What is the abc formula?"

As for finding the time for the door to close:
After MIL (mother-in-law) releases the door, its angular velocity is constant. The time interval during which she applies force is fixed at 0.2 seconds (we think so anyway - no confirmation from Haveagoodday )​

Finally: It strikes me that this thread is very weird. We have several respondents, mostly new PF, who jump in at various points, mostly to get help, as if they are also tasked with solving this problem.

Do you folks know each other?Also: It would greatly help if posters would use units with their numerical answers, as well as using the "Reply" feature so it's a bit clearer as to who you're answering or asking.
 
  • #47
SammyS said:
As haruspex asked: "What is the abc formula?"

As for finding the time for the door to close:
After MIL (mother-in-law) releases the door, its angular velocity is constant. The time interval during which she applies force is fixed at 0.2 seconds (we think so anyway - no confirmation from Haveagoodday )​

Finally: It strikes me that this thread is very weird. We have several respondents, mostly new PF, who jump in at various points, mostly to get help, as if they are also tasked with solving this problem.

Do you folks know each other?
I bet we are in the same class. Abc formula is what they call formula for solving square equation. X= -b+/- sqrt b^2...
 
  • #48
coffeemanja said:
I bet we are in the same class. Abc formula is what they call formula for solving square equation. X= -b+/- sqrt b^2...
The quadratic formula.

By the way:
Is it true that the time of contact with the door is 0.2 seconds? (or as some write: 0,2 seconds)
 
  • #49
SammyS said:
The quadratic formula.

By the way:
Is it true that the time of contact with the door is 0.2 seconds? (or as some write: 0,2 seconds)
Yes, that is in the statement of the problem:)
 
  • #50
coffeemanja said:
Yes, that is in the statement of the problem:)
Well, it looks like a typo to me in the statement of the problem, as was the l=2 rather than l/2 .
Haveagoodday said:
An angry mother-in-law slams the door shut, by pushing at the middle of

the door (l=2 from the hinges) with a force of F = 100N, lasting a time

t = 0:.00 s.
 
  • #51
SammyS said:
Well, it looks like a typo to me in the statement of the problem, as was the l=2 rather than l/2 .
image.jpg
 
  • #52
statii said:
On c.) could the time be 2,53s (also got t= -1,26), used abc formula with α=10 and ω=2? Used the inertia formula that StavangerFinest did: I = (ML^2)/3.
in c i get 0.395s
coffeemanja said:
Sorry, have been busy. I still have 40 in a . Alpha =( ΣF*L/2)/ (1/3*M*(L/2)^2) is my formula. Since it is a uniform rod with an axis through the end and the push is in the middle... I saw answer 10 somewhere... So now I'm puzzled
you are using the wrong value of l, you have to put in the total width of the door in the equation, it doesn't matter where she pushes the door, it only matters when calculating the torque.
 
  • #53
Haveagoodday said:
in c i get 0.395s

you are using the wrong value of l, you have to put in the total width of the door in the equation, it doesn't matter where she pushes the door, it only matters when calculating the torque.
Yeah, got that already. Thanks anyway!
 
  • #54
Haveagoodday said:
in c i get 0.395s
It's about double that. Please post your working.
 
  • #55
haruspex said:
It's about double that. Please post your working.
yeah, i recalculated and got 0.785s.
 
  • #56
Haveagoodday said:
yeah, i recalculated and got 0.785s.
Excellent.
 
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  • #57
haruspex said:
It's about double that. Please post your working.
Can I post mine?
Θf=Θi+ωi*t+0.5αt^2
90=0+2t+0.5*10t...and the result is far of the one you say is correct...
 
  • #58
coffeemanja said:
Can I post mine?
Θf=Θi+ωi*t+0.5αt^2
90=0+2t+0.5*10t...and the result is far of the one you say is correct...
you have to use this equation t= θf-θi/wf-wi
and you also have to use radians instead of degrees.
 
  • #59
coffeemanja said:
Can I post mine?
Θf=Θi+ωi*t+0.5αt^2
90=0+2t+0.5*10t...and the result is far of the one you say is correct...
What is the are value you putting for alpha here? What units do you have for omega?
 
  • #60
Haveagoodday said:
you have to use this equation t= θf-θi/wf-wi
wf-wi? Did you mean that?
 

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