Why Am I Wrong in Calculating Forces on a Rotating Rod?

[cupid.callin]

Homework Statement

Hi all

The Attempt at a Solution

This was a solution to my problem i found on internet ...

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 !

WHY?

And please someone explain me what is written in solution after finding w² ... i can't understand it 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 ? ...

 

[tiny-tim]

hi cupid.callin!

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 ? ...

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 ​

 

[ashishsinghal]

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

 

[cupid.callin]

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 ?

 

[cupid.callin]

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!

:)

 

[tiny-tim]

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!

and mg sin37 ... i.e. perpendicular to rod!

that's certainly not tension

 

[cupid.callin]

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!

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?

 

[tiny-tim]

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?

… 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

 

[cupid.callin]

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, isn't that right?

 

[tiny-tim]

a rod also has tension along its length ... just its not same in all parts like in string

true

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