# Rod rotating round it's end

1. Jul 10, 2015

### Karol

1. The problem statement, all variables and given/known data
A rod of length L and mass m which is free to rotate round it's edge is held, initially, horizontally.
What are: the angular velocity and acceleration at angle θ
The force at the nail (the axis)

2. Relevant equations
Torque and moment of inertia: $M=I\alpha$

3. The attempt at a solution
Conservation of energy:
$$mg\frac{L}{2}\cos\theta=\frac{1}{3}mL^2\omega\;\rightarrow \omega=\frac{3}{2}gl\cos\theta$$
$$M=I\alpha:\; mg\frac{L}{2}\sin\theta=\frac{1}{3}mL^2\alpha\;\rightarrow \alpha=\frac{3g}{2L}\sin\theta$$
The linear, tangential acceleration $a_t=\alpha\frac{L}{2}$
I take forces in the tangential direction, F is the reaction:
$$mg\frac{L}{2}\sin\theta-F=m\cdot \frac{3g}{2L}\sin\theta \frac{L}{2}\;\rightarrow\; F=\frac{1}{4}mg\sin\theta$$
But i am not sure.
The reverse of the axial component of gravity $mg\cos\theta$ is the second component of the reaction in the nail.

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2. Jul 10, 2015

### haruspex

You seem to be mixing energy and angular momentum.

3. Jul 10, 2015

### Karol

$$mg\frac{L}{2}\cos\theta=\frac{1}{2} \frac{1}{3}mL^2\omega^2\;\rightarrow \omega^2=\frac{3g\cos\theta}{L}$$

4. Jul 10, 2015

### haruspex

Right.

5. Jul 10, 2015

### Karol

But what about the main question, the force on the nail?

6. Jul 10, 2015

### haruspex

The force will depend on the rate of rotation (centripetal force), which you seem to have omitted.

7. Jul 10, 2015

### Karol

The centripetal force on the COM at the distance L/2: $m\omega^2\frac{L}{2}$ is added:
$$mg\frac{L}{2}\sin\theta-F=m\cdot \frac{3g}{2L}\sin\theta \frac{L}{2}\;\rightarrow\; F=\frac{1}{4}mg\sin\theta$$
$$F_{tot axial}=\frac{1}{4}mg\sin\theta+m\omega^2\frac{L}{2}$$

8. Jul 11, 2015

### haruspex

F is the component of the axial force parallel to the motion of the mass, i.e. tangential, no?
You need to find the total radial force (including the centripetal force) and add that vectorially.

9. Jul 11, 2015

### Karol

Yes, i made a mistake.
$$\left\{ \begin{array}{l} F_{tan}=\frac{1}{4}mg\sin\theta \\ F_{rad}=mg\cos\theta+m\omega^2\frac{L}{2} \end{array} \right.$$

10. Jul 11, 2015

### haruspex

That looks right.

11. Jul 11, 2015

### Karol

Thanks

12. Jul 11, 2015

### SammyS

Staff Emeritus
Karol,

Should this title actually be "Rod rotating round it's end" ? -- not edge

13. Jul 11, 2015

### Karol

Yes, correct, i don't know physics AND english well...

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