Rotation of body in stregth of materials problem

[Chacabucogod]

Homework Statement

A uniform bar of length 2l is rotated in a horizontal plane about its mid-point as center with an angular velocity ω. This case must be reduced to a problem in statics by the application of d'Alembert's principle, the "forces" acting being "centrifugal" forces.

a) Plot the stress distribution and the elastic displacement against the radial distance from the center, and derive formulae for the maximum stress and the maximum displacement.

ANS:$\frac{γω^2}{2g}(l^2-r^2)$

Homework Equations

$F=m*\frac{v^2}{r}=m*ω^2*r$
P=$\frac{F}{A}$

The Attempt at a Solution

I really have no idea where to begin or how to continue

 

[Chestermiller]

Homework Statement

A uniform bar of length 2l is rotated in a horizontal plane about its mid-point as center with an angular velocity ω. This case must be reduced to a problem in statics by the application of d'Alembert's principle, the "forces" acting being "centrifugal" forces.

a) Plot the stress distribution and the elastic displacement against the radial distance from the center, and derive formulae for the maximum stress and the maximum displacement.

ANS:$\frac{γω^2}{2g}(l^2-r^2)$

Homework Equations

$F=m*\frac{v^2}{r}=m*ω^2*r$
P=$\frac{F}{A}$

The Attempt at a Solution

I really have no idea where to begin or how to continue

Let ρ represent the density of the bar. How much mass dm is contained in the region of the bar between r and r + dr? What is the tension in the bar at location r = l ?

Chet

 

[Chacabucogod]

The mass between r and dr will be

$$dm=4πρlrdr$$

The force at l will be

$$F=mω^2l$$

I'll try to see if I can come up with something using what you just told me.

Thank you

 

[Chestermiller]

The mass between r and dr will be

$$dm=4πρlrdr$$

Actually, the differential mass between r and r + dr will be
$$dm=ρAdr$$
where A is the cross sectional area of the rod and Adr is the differential volume of the rod contained between r and r + dr.

The force at l will be
$$F=mω^2l$$

No. Actually, the tension in the rod at r = l will be zero, since the rod is not contacting anything at that location.

Draw a free body diagram of the differential section of rod situated between r and r + dr. Call F(r) the tension in the rod as a function of r. What is the force acting on the surface of the differential element at r + dr, and in which direction does it point? What is the force acting on the surface of the differential element at r, and in which direction does it point? What is the mass times acceleration of the differential element, and in which direction does it point? Write down your force balance equation for the differential element.

Chet

 

[Chacabucogod]

2 Things:

The cross sectional area would be: 2*pi*r*2l no?

If a point mass is rotating about a point a distance l wouldn't the centripetal force be the F that I just mentioned?

Thank you

 

[Chestermiller]

No. In this problem, r is not the distance from the axis of the cylinder. See the problem statement. r is measured along the length of the cylinder from its midpoint.

Regarding the force, your equation would be correct only if all the mass is concentrated at a point. However, in this problem, the mass is distributed along the length, and different parts of the rod are experiencing different accelerations. You need to take this into account.

Chet

 

[Chacabucogod]

So you are telling me that the rod is rotating over a plane and not on its own axis? That's not how I understood it.

Thank you.

NVM now I get it.

 

[Chacabucogod]

So from the side nearest to the midpoint the force is $mω^2r$ and farther there is $mω^2(r+dr)$ right?

 

[Chestermiller]

So from the side nearest to the midpoint the force is $mω^2r$ and farther there is $mω^2(r+dr)$ right?

No. From the side nearest the midpoint, it's F(r), and farther there is F(r+dr). Now, what is the differential mass times acceleration?

Chet

 

[Chacabucogod]

Rho*a*dr*centripetal acceleration right?

 

[Chacabucogod]

Would this be the sum of forces on an element dr?

$$-ρAdr(ω^2r)+ρAdr(ω^2(r+dr))$$

you end up with a Rho*...*dr^2 which I imagine you won't take into account and it will work right?

 

[Chestermiller]

Would this be the sum of forces on an element dr?

$$-ρAdr(ω^2r)+ρAdr(ω^2(r+dr))$$

you end up with a Rho*...*dr^2 which I imagine you won't take into account and it will work right?

Nope. The force balance I get is:

$F(r+dr)-F(r)=-ρAdr(ω^2r)$

or equivalently:

$$\frac{dF}{dr}=-ρAω^2r$$

See what you get when you integrate this between arbitrary r and r = l.

Chet

 

[Chacabucogod]

Alright! Now I understand it! Thank you for your help and patience.