# Rotation - Pivoted Rod

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 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 ??? ....

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tiny-tim
Homework Helper
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

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

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 ???

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
Homework Helper
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

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
Homework Helper
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

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?

tiny-tim