Rotation - Pivoted Rod

  • #1
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Homework Statement


Hi all :biggrin:

attachment.php?attachmentid=33350&stc=1&d=1300734462.jpg



The Attempt at a Solution


This was a solution to my problem i found on internet ...
attachment.php?attachmentid=33349&stc=1&d=1300734462.jpg


I found w2 correctly like in solution ... but then when i thought that that net force will be force due to rod (F) - mg cos37 ... along rod ...

and mg sin37 perpendicular to rod ...

F - mg cos37 = (dm)w2L = .9 (dm) g

mgsin37 = (dm) g (3/5)

But i'm wrong !!! :cry:

WHY??? :confused:

And please someone explain me what is written in solution after finding w2 ... i cant understand it :tongue2:


EDIT:


I got what they have written in answer ... they used τ = αI to find α

they used to find (dm) (dv/dt) = (lα)(dm) ie tangential force ... now i need to know .. why i am wrong in my way ??? .... :confused:
 

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Answers and Replies

  • #2
tiny-tim
Science Advisor
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hi cupid.callin! :smile:
I found w2 correctly like in solution ... but then when i thought that that net force will be force due to rod (F) - mg cos37 ... along rod ...

and mg sin37 perpendicular to rod ...

EDIT:


I got what they have written in answer ... they used τ = αI to find α

they used to find (dm) (dv/dt) = (lα)(dm) ie tangential force ... now i need to know .. why i am wrong in my way ??? .... :confused:
your mg sin37° and mgcos37° are only the weight

you've ignored the tension in the rod …

this method finds the total ("net") force, which is the LHS of F = ma, by just getting the RHS :wink:
 
  • #3
The equation after finding w2 is:
ml2/3 (alpha) = mgl/2 sin37.
It is just an equation connecting torque, moment of inertia and angular acceleration
 
  • #4
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how have i ignored tension?

I found F(tension) - mgcos37 ... i.e. along rod
and mg sin37 ... i.e. perpendicular to rod!!!

That includes tension, Rioght ???
 
  • #5
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The equation after finding w2 is:
ml2/3 (alpha) = mgl/2 sin37.
It is just an equation connecting torque, moment of inertia and angular acceleration
Yes i wrote that in edited part!!!

:)
 
  • #6
tiny-tim
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how have i ignored tension?

I found F(tension) - mgcos37 ... i.e. along rod
ah, no

the tension force is only mgcos37° on a flat slope …

that's because the acceleration perpendicular to a flat slope is zero, and so you can take components perpendicular to the slope: T = mgcos37° …

in this case, the acceleration perpendicular to the slope is not zero, so that doesn't work! :smile:
and mg sin37 ... i.e. perpendicular to rod!!!
that's certainly not tension :confused:
 
  • #7
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ah, no

the tension force is only mgcos37° on a flat slope …

that's because the acceleration perpendicular to a flat slope is zero, and so you can take components perpendicular to the slope: T = mgcos37° …

in this case, the acceleration perpendicular to the slope is not zero, so that doesn't work! :smile:
I guess you are not understanding what i have done

i haven't taken mgcos37 = T
i have used T - mgcos37 = mv2/r = mw2r

And as i have taken X,Y axis as perpendicular to rod, along rod respectively so the only force in X axis is mg sin37 (and no tension of rod)

so shouldn't that be right?
 
  • #8
tiny-tim
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I guess you are not understanding what i have done

i haven't taken mgcos37 = T
i have used T - mgcos37 = mv2/r = mw2r
but the question asks for the total force …

where does that appear in your solution? :confused:
… the only force in X axis is mg sin37 (and no tension of rod)
this is a rigid rod, not a rope or chain …

it does have "tension" perpendicular to its length
 
  • #9
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EDIT:

Not even a string has any tension perpendicular to its length

And i know it has no tension perpendicular to length
............

and i have calculated the net of tension and mg in the first eqn, isnt that right?
 
  • #10
tiny-tim
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a rod also has tension along its lenght ... just its not same in all parts like in string
true :smile:

but a cable (ie a string with mass) also has has tension only along (and not perpendicular to) its length, and yet the tension is not the same all the way along

and a rigid rod also has tension perpendicular to its length
 

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