# Rotational Energy problem

1. Mar 11, 2005

### futron

Can anyone help me on the following problem? I'm not sure how to relate the 90-degree rotation with the rest of what's being asked.

A thin uniform rod is initially positioned in the vertical direction, with its lower end attached to a frictionless axis that is mounted on the floor. The rod has a length of 1.20 m and is allowed to fall, starting from rest. Find the tangential speed of the free end of the rod, just before the rod hits the floor after rotating through 90°.

2. Mar 11, 2005

### Gamma

Consider the rod after it has fallen through an angle theta. Use conservation of energy. You have the gravitational potantial energy, rotational kinetic energy and the translational kinetic energy to consider when writing conservation of energy equation. Using this you will be able to find the angular velocity at angle theta. Substitute theta = 90 to find the angular speed at 90 degrees. Use Tangantial speed = length * angular speed to find your required answer.

3. Mar 12, 2005

### futron

I've E=(1/2)mv^2+(1/2)Iw^2+mgh, but I'm still unclear as to how theta fits in to this equation.

4. Mar 12, 2005

### Gamma

Let's assume that the gravitational potantial energy is zero at the bottom level of the rod.

Initial Energy = mg L/2 ( L, the length of the rod)

Final energy = $$mg L/2 * cos \theta + 1/2 m (L/2*w)^2 + 1/2 I w^2$$

The second term in the right is same as your 1/2 m v^2

But,

$$1/2 m v^2 = 1/2 m ( v_{radial} ^2 + v_{tangantial}^2)$$

But if you look at the motion of the rod, radial velocity = 0.

therefore, $$1/2 m v^2 = 1/2 m v_{tangantial}^2)$$

5. Mar 12, 2005

### Andrew Mason

Since the bottom is frictionless, there is no horizontal component of force on the rod. There is only gravity and a vertical normal force. That normal force produces the torque on the rod that causes it to rotate. It also reduces the downward acceleration. The centre of mass of the rod just falls straight down.

The rotational motion is given by the torque of the centre of mass about the end on the table:

$$\tau = mgcos\theta L/2 = I\alpha$$

where $\theta$ is the angle made by the rod to the horizontal surface.

The change in potential energy of the rod during the fall is mgL/2. This is equal to the work done by the torque, which is converted entirely to rotational energy:

$$mgL/2 = \frac{1}{2}I\omega^2$$

Just work out the angular speed $\omega$ and the tangential speed from that. (You have to use the moment of inertia of a thin rod about its end).

AM

6. Mar 12, 2005

### Staff: Mentor

I believe the pivot is mounted on the floor and does not move. The rod merely pivots about the fixed axis; the center of mass does not fall straight down.

Nonetheless, your solution is correct. (If the pivot were actually freely moving, then the rod could not be modeled as being in pure rotation about the pivot. )

7. Mar 12, 2005

### Andrew Mason

If the centre of mass does not fall straight down, what force moves the centre of mass horizontally?

The pivot is freely moving horizontally. I don't see why the rod cannot be modeled as pure rotation (about the end that is on the surface) in the inertial frame of the floor.

It can also be modeled as a combination of rotation about the centre of mass (torque provided by the normal force) + the acceleration of the centre of mass. But the latter is more complicated to do:

$$mg - mgsin\theta = ma$$ and:

$$mgsin\theta cos\theta L/2 = I\alpha = \frac{1}{12}mL^2\alpha$$

You then work out the rotational speed of the rod about its centre of mass to find the tangential speed of the end relative to the centre of mass. You then add the speed of the centre of mass to get the speed of the end relative to the floor - not a trivial problem to solve.

AM

EDIT: Sorry, I see that I misread the problem. I interpreted it as a frictionless surface. In re-reading, I see that end on the floor is fixed to the floor, so there is horizontal force. You can use the above for the case where the end is not horizontally fixed. The solution is the same, however, because the horizontal force does no work.

AM

Last edited: Mar 12, 2005