Rectilinear Motion with Constant Acceleration

[Foghorn]

Homework Statement

(1) If you measure the acceleration of the falling mass to be 7.2 m/s^2, should you be concerned that you somehow made a mistake? What if you found the acceleration to be 11.5 m/s^2?

For this question, a falling mass is attached to a piece of paper, which is subsequently pulled through a tape timer. The timer prints dots on the paper at even time intervals, and these dots represent the position of the mass as a function of time.

(2) What two measurements and one equation will be needed to find the acceleration of the glider in part 2? Remember that the photogates will only give the time for the glider to pass through them, and the time interval between photogates cannot be measured.

Here, an air track glider is pulled through photogates by a hanging mass attached to the glider through a pulley system. The photogates will measure the time the photogate is blocked; the first photogate will display its time (t1) immediately, while the second photogate will measure its time (t2) but not display it. Finally, the display will show the sum of t1 and t2.

Homework Equations

v = delta x/delta t ;; delta x is distance traveled, delta t is time interval
a = delta v/t ;; delta v is avg. velocity
v = v_o + at

The Attempt at a Solution

For the first question,

Indeed, such a measurement would trouble me. I would wonder why the acceleration was not measured to be somewhere near 9.8 m/s^2, as once the object is in free-fall it should be accelerating at such a magnitude. Obviously, I would check my calculations and insure that they were performed correctly and that they were using the actual data obtained from the experiment. Perhaps one could chalk such deviations up to air resistance and friction from the tape machine, but I doubt they would introduce that much error. Both 7.2 m/s^2 and 11.5 m/s^2 seem like unusually unlikely results, even considering a minor lack of precision or accuracy within the measuring device and in the calculations.

I'm not sure if such an answer is within reason, but oh well.

For the second question, I'm not entirely sure where to start. I thought I might be able to measure the distance between photogates and from there determine avg. velocity, and subsequently acceleration, but it calls for only two measurements and one equation.

I'm not sure if the question is worded strangely, but perhaps initial and final velocities in addition to the time, combined with the kinematic equation v = v_o + at would allow one to derive acceleration.

 

[D H]

For the first question,

Indeed, such a measurement would trouble me.

I, too, would find an observed acceleration of 7.2 m/s² troublesome for a feather. Think about it.

 

[Foghorn]

It seems to me that a feather, under realistic conditions, would display reduced acceleration. Wouldn't the force of air resistance counteract gravity?

 

[D H]

Exactly. Suppose you hold a feather still and then let it drop. The very instant you let the feather free it will accelerate downward at 9.81 m/s². A brief instant later it will have a non-zero downward velocity with respect to the air -- and hence an upward drag resistance that somewhat counteracts gravity. In short order, the feather will be falling at the upward drag will equal the force of gravity. The feather will stop accelerating.

Suppose you could measure acceleration directly, and do so instantaneously. In that case it would be surprising to see anything falling at other than 9.81 m/s² -- at the very moment it is released, that is. However, while we can directly measure position and velocity, we do not know how to measure acceleration. Multiple measurements must be taken over a non-zero span of time, and that means the inferred acceleration for a feather will always be less than g.

Unless of course you drop the feather in a vacuum chamber from which the air has been evacuated.