Measuring Centripetal Acceleration Geometrically

[scott_alexsk]

Is there a way to measure centripetal acceleration independent of a force measurement? If I had just a ruler, and a ball was rolling along a circular path, could I 'measure' the centripetal acceleration itself, without resorting to finding the radius, period, etc. Someone once told me that this is a special case of acceleration, with rotating frames, but then what does the acceleration mean, in the rate of change of the angle of inclination of the frame?

Thanks,
-Scott

 

[ZapperZ]

This is a bit confusing.

Maybe you can illustrate what you mean by doing this with linear motion and demonstrate what you wanted to do with the regular linear acceleration, presuming that what you want to do is clear with this case.

Zz.

 

[PatPwnt]

v^2 / r How can you measure something without units of measurement? Acceleration is a change in the velocity. Acceleration is there because there is a change in direction in circular motion.

 

[pervect]

Is there a way to measure centripetal acceleration independent of a force measurement? If I had just a ruler, and a ball was rolling along a circular path, could I 'measure' the centripetal acceleration itself, without resorting to finding the radius, period, etc. Someone once told me that this is a special case of acceleration, with rotating frames, but then what does the acceleration mean, in the rate of change of the angle of inclination of the frame?

Thanks,
-Scott

I assume you're interested in the Newtonian case.

For convenience, set up a Cartesian coordinate system (x,y,z).

Measure x(t), y(t), and z(t), the position of the object as a function of time.

The object's acceleration vector will then be d^2 x / dt^2, d^2 y/ dt^2, d^2 z / dt^2. (This works for any path, including a circular one).

The magnitude of this vector will be the total acceleration.

Note that you'll need at least two rulers if the ball is on a plane, i.e. you'll need to measure x and y.

 

[scott_alexsk]

Thank you prevect. That is a very useful (and simple) definition.

-Scott