# Dynamics smooth circular bar Problem

1. Feb 8, 2005

### supersix2

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

Thanks!

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2. Feb 8, 2005

### Andrew Mason

The tension, T, in the string provides both the horizontal (centripetal) and vertical (mg) forces.

So:
(1)$$Tcos\beta = mg$$
(2)$$Tsin\beta = m\omega^2r$$

Dividing (2) by (1):

$$tan\beta = \omega^2r/g$$

$$\omega = \sqrt{gtan\beta/r}$$

But $r = Rsin\beta$, so:

$$\omega = \sqrt{g/Rcos\beta}$$

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

AM

Last edited: Feb 8, 2005
3. Feb 8, 2005

### supersix2

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. Feb 9, 2005

5. Feb 9, 2005

### Andrew Mason

Perhaps you can explain how the mass is held there.

AM

6. Feb 9, 2005

### supersix2

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.

Last edited: Feb 9, 2005
7. Feb 9, 2005

### Andrew Mason

That doesn't tell me how it is held there. Is the mass a ring that surrounds the circular bar?

AM

8. Feb 9, 2005

### supersix2

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. Feb 9, 2005

### arildno

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 $$\vec{j},\vec{i}$$ be unit vectors in the vertical and horizontal directions, respectively.
We then have that the outwards normal vector is given by:
$$\vec{n}=\sin\beta\vec{i}-\cos\beta\vec{j}$$
with associated tangent vector:
$$\vec{t}=\cos\beta\vec{i}+\sin\beta\vec{j}$$
or:
$$\vec{i}=\cos\beta\vec{t}+\sin\beta\vec{n}$$
$$\vec{j}=\sin\beta\vec{t}-\cos\beta\vec{n}$$
The centrifugal force is:
$$\vec{f}_{c}=m\omega^{2}R\sin\beta\vec{i}=m\omega^{2}R\sin\beta(\cos\beta\vec{t}+\sin\beta\vec{n})$$
the weight:
$$\vec{w}=-mg\vec{j}=mg(\cos\beta\vec{n}-\sin\beta\vec{t})$$
whereas the normal force is assumed parallell the inwards normal:
$$\vec{N}=-F\vec{n}$$
Hence, equilibrium of forces in the tangential direction requires:
$$m\omega^{2}R\sin\beta\cos\beta-mg\sin\beta=0\to\omega^{2}=\frac{g}{R\cos\beta}$$
Equilibrium in the normal direction:
$$-F+mg\cos\beta+m\omega^{2}R\sin\beta\sin\beta=0\to{F}=\frac{mg}{\cos\beta}$$

Note that when $$\beta=0$$ (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 $$\beta=\frac{\pi}{2}$$ is impossible to achieve.

Last edited: Feb 9, 2005