Dynamics smooth circular bar Problem

In summary, the mass is held in equilibrium by the radial acceleration and the normal force. The angular velocity is not a limiting factor.
  • #1
supersix2
26
0
I think I did this problem correctly however I am not 100% sure. Unfortunatley the Dynamics book that I have only gives answers to even numbered problems and I cannot check me work.

Anyway here is the problem from the book.

The smooth circular bar rotates with constant angular velocity about the vertical axis AB. The radius R = 0.5 m The mass m remains stationary relative to the circular bar at an angle of 40 degrees. Determine the angular velocity.

Here is my work:

https://www.physicsforums.com/attachment.php?attachmentid=2510&stc=1

My answer just in case you can't read it is 3.877 rad/s.

Thanks!
 

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  • #2
supersix2 said:
I think I did this problem correctly however I am not 100% sure. Unfortunatley the Dynamics book that I have only gives answers to even numbered problems and I cannot check me work.
The tension, T, in the string provides both the horizontal (centripetal) and vertical (mg) forces.

So:
(1)[tex]Tcos\beta = mg[/tex]
(2)[tex]Tsin\beta = m\omega^2r[/tex]

Dividing (2) by (1):

[tex]tan\beta = \omega^2r/g[/tex]

[tex]\omega = \sqrt{gtan\beta/r}[/tex]

But [itex]r = Rsin\beta[/itex], so:

[tex]\omega = \sqrt{g/Rcos\beta}[/tex]

[tex]\omega = \sqrt{9.8*/.766*.5} = 5.06 rad sec^{-1}[/tex]


AM
 
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  • #3
Thanks for the help...but there is no string the mass is just held there by the Forces of the rotating bar. I'm not sure if that would change what you told me.
 
  • #4
Bump. I'm still unsure about this problem. Remember there are no strings attached to the mass.
 
  • #5
supersix2 said:
Thanks for the help...but there is no string the mass is just held there by the Forces of the rotating bar. I'm not sure if that would change what you told me.
Perhaps you can explain how the mass is held there.

AM
 
  • #6
The circular bar is rotating causing a radial accelertation which in turn causes a force that counteracts the component of the object's weight in the normal direction.

However, my intution is playing tricks on me because I don't think I can ignore the component of the objects weight in the tangential directionn. However, since the angular velocity is constant there is no tangential acceleration meaning there is no force in the tangential direction.

This reasoning leave me to believe that the method is used (as shown in my attachment) is the right method but I am not 100% sure.

The method you showed does make sense if the mass is also supporeted by strings (which it is not). The lines I drew in my picture are dimension lines that show the angle and radius.
 
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  • #7
supersix2 said:
The circular bar is rotating causing a radial accelertation which in turn causes a force that counteracts the component of the object's weight in the normal direction.
That doesn't tell me how it is held there. Is the mass a ring that surrounds the circular bar?

AM
 
  • #8
Yea, the mass is a ring attached to the circular bar. I suppose its like a washer or something attached to the bar. The bar is spinning and object is help there by the force of the radial acceleration and perhaps, the force of the tangential acceleration. Although I am not entirely sure there is a tangential acceleratoin since the rotataional velcocity is constant.
 
  • #9
Note that you've got TWO unknowns here:
The angular velocity, and the magnitude of the NORMAL force acting upon the mass from the circle.
To set the problem up let [tex]\vec{j},\vec{i}[/tex] be unit vectors in the vertical and horizontal directions, respectively.
We then have that the outwards normal vector is given by:
[tex]\vec{n}=\sin\beta\vec{i}-\cos\beta\vec{j}[/tex]
with associated tangent vector:
[tex]\vec{t}=\cos\beta\vec{i}+\sin\beta\vec{j}[/tex]
or:
[tex]\vec{i}=\cos\beta\vec{t}+\sin\beta\vec{n}[/tex]
[tex]\vec{j}=\sin\beta\vec{t}-\cos\beta\vec{n}[/tex]
The centrifugal force is:
[tex]\vec{f}_{c}=m\omega^{2}R\sin\beta\vec{i}=m\omega^{2}R\sin\beta(\cos\beta\vec{t}+\sin\beta\vec{n})[/tex]
the weight:
[tex]\vec{w}=-mg\vec{j}=mg(\cos\beta\vec{n}-\sin\beta\vec{t})[/tex]
whereas the normal force is assumed parallell the inwards normal:
[tex]\vec{N}=-F\vec{n}[/tex]
Hence, equilibrium of forces in the tangential direction requires:
[tex]m\omega^{2}R\sin\beta\cos\beta-mg\sin\beta=0\to\omega^{2}=\frac{g}{R\cos\beta}[/tex]
Equilibrium in the normal direction:
[tex]-F+mg\cos\beta+m\omega^{2}R\sin\beta\sin\beta=0\to{F}=\frac{mg}{\cos\beta}[/tex]

Note that when [tex]\beta=0[/tex] (i.e, the mass is on the rotation axis), any value of angular velocity is in this analysis acceptable, since there is no induced centrifugal force, whereas [tex]\beta=\frac{\pi}{2}[/tex] is impossible to achieve.
 
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1. What is the Dynamics smooth circular bar Problem?

The Dynamics smooth circular bar Problem is a classic problem in mechanics that involves a uniform circular bar rotating about a fixed axis. The goal is to determine the equations of motion and the forces acting on the bar.

2. What are the assumptions made in the Dynamics smooth circular bar Problem?

The main assumptions made in this problem are that the bar is perfectly uniform and rigid, and that there is no friction or air resistance present. Additionally, the bar is assumed to be rotating at a constant angular velocity.

3. How do you solve the Dynamics smooth circular bar Problem?

To solve this problem, you first need to set up the equations of motion using Newton's second law of motion. This involves considering the forces acting on the bar, such as the weight, tension in the bar, and centrifugal force. Then, you can use the equations of motion to solve for the angular acceleration and angular velocity of the bar at any given time.

4. What are the applications of the Dynamics smooth circular bar Problem?

The Dynamics smooth circular bar Problem has many applications in engineering and physics. It can be used to analyze the motion of objects rotating about a fixed axis, such as a spinning top or a flywheel. It is also applicable to the design and analysis of various mechanical systems, such as gears and turbines.

5. How does the Dynamics smooth circular bar Problem relate to real-world situations?

The Dynamics smooth circular bar Problem is a simplified version of real-world situations involving rotating objects. By understanding the principles and equations involved in this problem, scientists and engineers can better understand and analyze the motion of more complex systems, such as rotating machinery and vehicles.

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