1. Aug 22, 2009

### redoxes

There is a rod treating as rigid body, the rod which mass is m and lenghth is R rotate about one end of itself with the angular acceleration $$\alpha$$. Apply a tangential external force f on another end to make the rod rotate. Now divide the rod into elements of mass. For each mass element dm with the distance r from the fixed end, it acted by an tangential component of resultant force dF which satisfy:

$$dF=dm\cdot a_{t} = dm\cdot \alpha r =\lambda dr \cdot \alpha r$$($$\lambda$$ is the linear density)

$$\alpha =\frac{M}{I}=\frac{fR}{\frac{1}{3}mR^{2}}$$ (M is the moment of force,I is the moment of interia about the fixed end)

So the tangential component of resultant force F of the rod will be:

$$F = \int_{0}^{R}\lambda dr \cdot \alpha r=\frac{m}{R}\frac{fR}{\frac{1}{3}mR^{2}}\int_{0}^{R}r\cdot dr=\frac{3}{2}f$$

As we see, F is not equal to f, is this a paradox? what is wrong in this argument?

2. Aug 22, 2009

### D H

Staff Emeritus
There is a constraint force at the fixed end of the bar that is keeping the fixed end fixed, and you are ignoring this.

3. Aug 22, 2009

### redoxes

I think ,at the fixed end ,the tangential component of resultant force dF is zero according to $$dF =\lambda dr \cdot \alpha r$$,which have included the constraint force.

4. Aug 22, 2009

### D H

Staff Emeritus
Try again.

If there were no constraint force the bar as a whole would go flying off.

5. Aug 22, 2009

### Mapes

Isn't this http://en.wikipedia.org/wiki/Carroll%27s_paradox" [Broken]? If you sum the moments around the free end, you'll see that there must be a nonzero component of force perpendicular to the bar at its fixed end, as D H says. Otherwise the bar wouldn't begin to rotate.

Last edited by a moderator: May 4, 2017
6. Aug 22, 2009

### D H

Staff Emeritus
This is not Carroll's paradox. Carroll's paradox describes an unreal situation. The OP describes something that is real.

7. Aug 23, 2009

### Mapes

Ah, right you are; I spoke (wrote) too soon.

8. Aug 23, 2009

### rcgldr

Note that the tangental force at the end of the rod is partially opposed by a force at the hinged end of the rod, and I don't see where this opposing force at the hinged end of the rod is taken into account in the original post.

9. Aug 24, 2009

### redoxes

Yep, I have realized that I ignored the constraint force, but I still do not understand how to calculate the constraint force f',it is f'=3f/2-f=f/2, isn't it? Is there another way to calculate it ? I only learn a few thing about constraint force before. Constraint force can adjust to the condition automatically, is that right?
Thanks!

10. Aug 24, 2009

### arildno

You have:
$$\frac{dF}{dr}=\lambda\alpha{r}=\frac{3f\lambda}{Rm}r$$

This yields:
$$F(r)=\frac{3f\lambda{r}^{2}}{2mR}+C$$
where C is an undetermined constant.
Using the relation F(R)=f, we get:
$$C=f-\frac{3}{2}fR\frac{\lambda}{m}=-\frac{1}{2}f$$
and that is the constraint force at the hinge.

Your fallacy lies in assuming that constraint force to be 0.

11. Aug 24, 2009

### Mapes

You could sum the moments around the center of mass and apply

$$M_C=\dot{H}_C=I_C\alpha=ml^2\alpha/12$$.

12. Aug 27, 2009

### redoxes

I am afraid I could not agree with you, because under your assumption,there will be F(0)=C Which is not equal to zero at the fixed end, and this will make the rod fly off, so it sounds impossible. I would like to think:

$$f+C=F=\frac{3f\lambda R^{2}}{2mR}=\frac{3}{2}f$$

and therefore: $$C=\frac{1}{2}f$$

Is that right ?