1. May 28, 2005

### ac2707

I need help with this :grumpy:
say there's a truck travellin at a constant speed along a winding flat road. the car body is connected to the axles with springs. Thus, the car body is assumed to roll at a 10degree angle from the vertical of a horizontal 'roll axis' which is fixed on the axles. An accelerometer is fixed in the car body at a distance 30cm from the center of gravity. Find the realtionship between the Acceleration measured by the accelerometer and the normal component of the required acceleration.
I know i'll have to use inertial acceleration A = A + a + wdot x v + 2w x v+ w x (w x v)...but I still don't get it

2. May 28, 2005

### OlderDan

I'm guessing the accelerometer in question is a linear accelerometer oriented to measure horizontal acceleration, perpendicular to the horizontal "roll axis" The 30cm information does not seem to be of any use, but hints that you are suppose to make some adjustment for the radius of curvature of the accelerometers path compared to the radius of the path of the CM. It should be simple enough to express the actual acceleration of the device in terms of its speed and path radius. Then you need to determine the effect the 10 degree roll is going to have on the acceleration the device will read because it tilts with the roll.

Your expression for acceleration seems to be a bit more complex than what you need for this problem. I would think the centripetal acceleration related to the speed and path radius is all you need. I assume the devce would be calibrated to read zero while the truck is moving straight and level at constant speed.

3. May 29, 2005

### ac2707

well u got it right..
so the actual acceleration of the device would be V^2/r because it is on the normal axis ...but how do I determine the effect of the roll??
I think I presented the problem inaccurately at the first time..I have an image here you might want to refer to

4. May 29, 2005

### OlderDan

Ah.. the missing diagram tells a lot. The location of the device combined with the tilt of the truck has to be used to determing the radius of curvature of the path of the device, and its speed. There will be a common angular speed for all locations in the truck, which you can calculate from the speed and radius of curvature of point C, so you will want to express the device acceleration in terms of its path radius and angular speed. Once you have that accomplished, the tilt of the truck is going to affect the reading because acceleration is a horizontal vector, and the device will be reading one component that is inclined at the 10 degree declination of the device. Break the acceleration into perpendicular components, one of which is parallel to the active axis of the device.

Then the question is will gravity affect the reading? If the active axis of the device were simply rotated to a vertical position, would it read the acceleration of gravity? If so, the 10 degree tilt is going to introduce a component of gravity to the reading.

Putting it all together, I think a reaonable model of the device is that it will provide whatever force is needed to some internal mass to offset any forces applied in the direction of its active axis. It is calibrated to divide the force by the mass and report the acceleration. If you think about a little mass internal to the device, and find the force the device must supply in the direction of its active axis to support a component of weight and produce a centripetal acceleration, I think you will have what you need. There will be force components perpendicular to the active axis provided by the device, but they will not affect the reading.

5. May 29, 2005

### ac2707

so the angular speed should be V/r ? if so, how do I espress the device acceleration ? is it a = theta(dot)^2 x r (normal axis) + theta(double dot) x r (tangential axis)?. theta as in 10 degrees..
and yes..if the accelerometer is tilted 90 degress (or vertical) it will read the acc of gravity.

6. May 29, 2005

### learningphysics

Imagine that the car is going around a circular track, and there is a vertical axis at the center of the circle.

The acceleration of point C is $$a_c = w^2 R_c$$ where $$R_c$$ is the distance from C to this vertical axis. Now calculated $$R_{acc}$$ where $$R_{acc}$$ is the distance from the accelerator to this same axis. Get $$R_{acc}$$ in terms of $$R_c$$So the acceleration of the accelerometer is $$w^2 R_{acc}$$.

You need the component parallel to the y_axis on your diagram ie $$w^2 R_{acc} cos\theta$$. Finally you have to add to this the component of g along the y_axis ie $$g sin\theta$$

So now you'll have the accelerometer reading in terms of $$R_c$$. Finally substitute $$R_c = a_c/w^2$$ into this equation, in order to relate the reading to the acceleration of C (which is what I think the question is asking).

7. May 29, 2005

### OlderDan

Yes, the angular speed is V/r, where r is the radius of the path of point C on the tilt axis, or what learingphysics has labeled as $$R_c$$. Let's use the big R and write

$$\omega = V/R_c$$

The tricky part is figuring out the radius of the path of the accelerometer, $$R_{acc}$$ . I am assuming you know the values of a, b, and c labelled in the figure. What you need to do is figure out the distance from the vertical axis above C out to a point above the y axis in terms of b, c, and the tilt angle. Once you do that you will know the distance from the center of rotation to that point, and a simple Pythagorean calculation using that distance and the distance a wll give you $$R_{acc}$$. Can you do this geometry?

Once you know $$R_{acc}$$ and $$\omega$$ the acceleration is the simple expression learningphysics posted

$$a_{acc} = \omega^2 R_{acc}$$

Find the component of that acceleration and of $$\overrightarrow g$$ in the direction of the active axis and you have the answer.