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1.7 Falling objects  (Page 5/10)

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Find g From data on a falling object

The acceleration due to gravity on Earth differs slightly from place to place, depending on topography (e.g., whether you are on a hill or in a valley) and subsurface geology (whether there is dense rock like iron ore as opposed to light rock like salt beneath you.) The precise acceleration due to gravity can be calculated from data taken in an introductory physics laboratory course. An object, usually a metal ball for which air resistance is negligible, is dropped and the time it takes to fall a known distance is measured. See, for example, [link] . Very precise results can be produced with this method if sufficient care is taken in measuring the distance fallen and the elapsed time.

Figure has four panels. The first panel (on the top) is an illustration of a ball falling toward the ground at intervals of one tenth of a second. The space between the vertical position of the ball at one time step and the next increases with each time step. At time equals 0, position and velocity are also 0. At time equals 0 point 1 seconds, y position equals negative 0 point 049 meters and velocity is negative 0 point 98 meters per second. At 0 point 5 seconds, y position is negative 1 point 225 meters and velocity is negative 4 point 90 meters per second. The second panel (in the middle) is a line graph of position in meters versus time in seconds. Line begins at the origin and slopes down with increasingly negative slope. The third panel (bottom left) is a line graph of velocity in meters per second versus time in seconds. Line is straight, beginning at the origin and with a constant negative slope. The fourth panel (bottom right) is a line graph of acceleration in meters per second squared versus time in seconds. Line is flat, at a constant y value of negative 9 point 80 meters per second squared.
Positions and velocities of a metal ball released from rest when air resistance is negligible. Velocity is seen to increase linearly with time while displacement increases with time squared. Acceleration is a constant and is equal to gravitational acceleration.

Suppose the ball falls 1.0000 m in 0.45173 s. Assuming the ball is not affected by air resistance, what is the precise acceleration due to gravity at this location?

Strategy

Draw a sketch.

The figure shows a green dot labeled v sub zero equals zero meters per second, a purple downward pointing arrow labeled a equals question mark, and an x y coordinate system with the y axis pointing vertically up and the x axis pointing horizontally to the right.

We need to solve for acceleration a size 12{a} {} . Note that in this case, displacement is downward and therefore negative, as is acceleration.

Solution

1. Identify the knowns. y 0 = 0 ; y = –1 .0000 m ; t = 0 .45173 ; v 0 = 0 size 12{v rSub { size 8{0} } =0} {} .

2. Choose the equation that allows you to solve for a size 12{a} {} using the known values.

y = y 0 + v 0 t + 1 2 at 2 size 12{y=y rSub { size 8{0} } +v rSub { size 8{0} } t+ { {1} over {2} } ital "at" rSup { size 8{2} } } {}

3. Substitute 0 for v 0 size 12{v rSub { size 8{0} } } {} and rearrange the equation to solve for a size 12{a} {} . Substituting 0 for v 0 size 12{v rSub { size 8{0} } } {} yields

y = y 0 + 1 2 at 2 . size 12{y=y rSub { size 8{0} } + { {1} over {2} } ital "at" rSup { size 8{2} } "." } {}

Solving for a size 12{a} {} gives

a = 2 y − y 0 t 2 . size 12{a= { {2 left (y - y rSub { size 8{0} } right )} over {t rSup { size 8{2} } } } "." } {}

4. Substitute known values yields

a = 2 ( − 1 . 0000 m – 0 ) ( 0 . 45173 s ) 2 = − 9 . 8010 m/s 2 , size 12{a= { {2 \( - 1 "." "0000 m–0" \) } over { \( 0 "." "45173 s" \) rSup { size 8{2} } } } = - 9 "." "8010 m/s" rSup { size 8{2} } ,} {}

so, because a = − g size 12{a= - g} {} with the directions we have chosen,

g = 9 . 8010 m/s 2 . size 12{g=9 "." "8010 m/s" rSup { size 8{2} } } {}

Discussion

The negative value for a size 12{a} {} indicates that the gravitational acceleration is downward, as expected. We expect the value to be somewhere around the average value of 9 . 80 m/s 2 size 12{9 "." "80 m/s" rSup { size 8{2} } } {} , so 9 . 8010 m/s 2 size 12{9 "." "8010 m/s" rSup { size 8{2} } } {} makes sense. Since the data going into the calculation are relatively precise, this value for g size 12{g} {} is more precise than the average value of 9 . 80 m/s 2 size 12{9 "." "80 m/s" rSup { size 8{2} } } {} ; it represents the local value for the acceleration due to gravity.

Applying the science practices: finding acceleration due to gravity

While it is well established that the acceleration due to gravity is quite nearly 9.8 m/s 2 at all locations on Earth, you can verify this for yourself with some basic materials.

Your task is to find the acceleration due to gravity at your location. Achieving an acceleration of precisely 9.8 m/s 2 will be difficult. However, with good preparation and attention to detail, you should be able to get close. Before you begin working, consider the following questions.

What measurements will you need to take in order to find the acceleration due to gravity?

What relationships and equations found in this chapter may be useful in calculating the acceleration?

What variables will you need to hold constant?

What materials will you use to record your measurements?

Upon completing these four questions, record your procedure. Once recorded, you may carry out the experiment. If you find that your experiment cannot be carried out, you may revise your procedure.

Once you have found your experimental acceleration, compare it to the assumed value of 9.8 m/s 2 . If error exists, what were the likely sources of this error? How could you change your procedure in order to improve the accuracy of your findings?

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Source:  OpenStax, Sample chapters: openstax college physics for ap® courses. OpenStax CNX. Oct 23, 2015 Download for free at http://legacy.cnx.org/content/col11896/1.9
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