Rotational Loading Homework: Find Force Pulling Arm away from Pivot Point

[Shidoran]

Homework Statement

I'm not using numbers because I only want to understand the mathematical relationship. This is not a coursework question, but I imagine it stems from coursework level physics. I am trying to design a rotating arm for my own use and I want to calculate bracing requirements between the vertical and horizontal arm.

A mass M is at the base of a vertical post of length R2, which is connected to a horizontal arm of length R1 by a 90 degree angle. The arm of length R1 rotates at some rpm around pivot point P. I want to understand the mathematical method to find the force (a moment at the mass perhaps?) that tries pulls the arm R2 away from R1(ie making the angle larger than 90 degrees).

Homework Equations

F = Ma (Newton's law)
T = radius * mass * gravity = (mass * distance^2) * (change in rotational angle/change in time)
T=2*pi*r/v=2*pi/w

The Attempt at a Solution

I think I'm trying to find the vertically oriented torque at point M caused by the rotation of M and A around point P. I'm a little lost to be honest! I appreciate anyone hinting me in the right direction, I did physics quite a while ago and I am horribly rusty.

 

[Quinzio]

You need to find the centrifugal acceleration, the multiply it by the arm of the force.

This is the resulting torque that the point A will experience
(mass_{M})\ \omega ^2\ r_1 \ r_2

 

[Shidoran]

Hi

Thanks for the reply. So I'm not trying to find a torque at the mass M, but actually a torque at point A? I guess that makes sense.

I imagine you're talking about centripetal acceleration? Is that the same as centrifugal?

So all I'd need to do is find:

A(R1) = V(R1)^2/R1

This is not affected though by the mass at M?

Sorry just an additional question,

If I want to convert the torque to an actual linear force (tangential to R1) I imagine that's mgsin(theta) but if I'm rotating constantly in a full circle (ie 360 degrees) it doesn't quite sit with me to use 360 degrees... am I wrong?

 

[Shidoran]

I'm sitting here at the moment:

T(A) =M*omega(A)^2*r1*r2 (as you said)

where omega(A)=v^2(A)/r1

then T(A)=M((V(r1)^2)/r1)^2*r1*r2 - equation 1

So to convert something like rpm to get a velocity (as V would be the unknown), V=circumference of travel * rpm = pi*d*rpm

So if I find V at point A from pi*2*r1*rpm and substitute into equation 1 I should get the torque I am looking for. If I want the tangential force I can then use:

F(A) = m g sin 360 but I need this concept in its rotational equivalent format?

Am I on the right track?

 

[Quinzio]

Without a drawing it's not easy to follow you.
In you drawing you asked for the force that tries to push M outside.
To counteract this force, you drew a brace.
Do you need to calculate the tension in the brace ? Eg. if the brace was a string ?

 

[Shidoran]

Sorry I may have overshot the mark of my original question. Attached is what I think I am understanding, am I on the right path? This will allow me to test the difference that increases in the Mass M have on the torque applied at point A.

Is it possible for you to show me how you derived the first formula? (ie the T=(mw^2)R1R2)?

Thanks again for your help.

 

[Quinzio]

\omega={V/r_1}
without the square

F = (mass_{M})\ \omega^2\ r_1
is the centrifugal force

the torque is
\tau = F b
b = r_2
\tau = (mass_{M})\ \omega^2\ r_1 \ r_2

 

[Shidoran]

Hey thanks for the help.

Hmm when I'm checking the units (SI) something's not adding up quite right. Is it as simple as substituting in the rpm speed (which is technically a frequency) with the rule omega=2*pi*f?

The rads/sec doesn't seem to cancel out and I'm not left with Nm for the units of torque. As r1 cancels out in the substitution, I end up with this:

Nm = N * (rads/s)^2 * m^2

What did I miss? it should end up with Nm = Nm for it to be correct...