### How can we experminetly determine the acceleration due to gravity?

Posted:

**Sun Mar 05, 2017 9:19 am UTC**can any one knows how to determine value of g at home?

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Posted: **Sun Mar 05, 2017 9:19 am UTC**

can any one knows how to determine value of g at home?

Posted: **Sun Mar 05, 2017 10:20 am UTC**

Rolling balls down inclined planes of varying angles is a pretty good way to go. If we neglect rolling and air resistance (which may be done if the balls are hard, round, and smooth, the ramp is somewhat smooth, and the speed is never very high), the acceleration is a = (g·ɵ̂)ɵ̂, where g is the local acceleration due to gravity (which you are trying to calculate) and ɵ̂ is the unit vector in the direction of the inclined plane. The magnitude of acceleration is a = g sin θ, where g is the magnitude of the local acceleration of gravity and θ is the measure of the angle the inclined plane makes with the horizontal (the "horizontal plane" is normal to gravity, i.e. it is level ground). Rearranging gives g = ^{a} ⁄ _{sin θ}. So if you test at multiple angles and plot the results with sin θ on the horizontal axis and a on the vertical axis, you should get (approximately) a straight line, and its slope is g.

Calculating these values only requires a ruler, a stopwatch, a pen or some tape (to make marks on the inclined plane), and some multiplication and division. To calculate sin θ, take the ratio of the inclined plane's height h (vertical distance from bottom to top) to its length l (distance along ramp from bottom to top): sin θ =^{h} ⁄ _{l}. To calculate the magnitude of acceleration, first mark two points on the plane, then measure the distance between them with the ruler and call it Δx. For each test, place the ball at the top mark and start a timer simultaneously with releasing it. Stop the timer when the ball passes the second mark, and call the difference in time Δt. The acceleration is then given by the kinematic equation a = ^{2 Δx} ⁄ _{Δt}².

There is obviously a lot of stuff you could do to improve accuracy, but this isn't bad. You will need to be careful that the balls roll straight down, so it may help to use narrow ramps, and the surface must be even. You will also need to be careful with your measurements using the stopwatch (photogates are clearly better here if you have them). And you will need to make sure the balls roll without slipping, as slipping will be the main source of drag for hard balls at low speeds.

Calculating these values only requires a ruler, a stopwatch, a pen or some tape (to make marks on the inclined plane), and some multiplication and division. To calculate sin θ, take the ratio of the inclined plane's height h (vertical distance from bottom to top) to its length l (distance along ramp from bottom to top): sin θ =

There is obviously a lot of stuff you could do to improve accuracy, but this isn't bad. You will need to be careful that the balls roll straight down, so it may help to use narrow ramps, and the surface must be even. You will also need to be careful with your measurements using the stopwatch (photogates are clearly better here if you have them). And you will need to make sure the balls roll without slipping, as slipping will be the main source of drag for hard balls at low speeds.

Posted: **Sun Mar 05, 2017 2:59 pm UTC**

But for an object rolling without slipping, some of the gravitational potential energy is transformed into rotational energy. If I do the calculations right, the proportion depends on the ratio of the moment of inertia of the object and the product of its mass and the square of its rolling radius, with g = (1 + M/(mr^{2})) a / sin(theta), where M is the moment of inertia of the object, r is its radius, and m is its mass. (I write moment of inertia as M rather than I, because of font issues.)

From wikipedia, I see M/(mr^{2}) is 2/5 for a solid ball of uniform density, 2/3 for a spherical shell, 1/2 for a solid cylinder, and 1 for a cylindrical shell.

From wikipedia, I see M/(mr

Posted: **Sun Mar 05, 2017 3:07 pm UTC**

muheedmir007 wrote:can any one knows how to determine value of g at home?

Why don't you try doing your homework at home?

Posted: **Mon Mar 06, 2017 11:14 am UTC**

DavidSh wrote:But for an object rolling without slipping, some of the gravitational potential energy is transformed into rotational energy. If I do the calculations right, the proportion depends on the ratio of the moment of inertia of the object and the product of its mass and the square of its rolling radius, with g = (1 + M/(mr^{2})) a / sin(theta), where M is the moment of inertia of the object, r is its radius, and m is its mass. (I write moment of inertia as M rather than I, because of font issues.)

From wikipedia, I see M/(mr^{2}) is 2/5 for a solid ball of uniform density, 2/3 for a spherical shell, 1/2 for a solid cylinder, and 1 for a cylindrical shell.

The rotation is provided by friction. The acceleration of the center of mass is independent of this.

Posted: **Mon Mar 06, 2017 12:42 pm UTC**

Eebster the Great wrote:DavidSh wrote:But for an object rolling without slipping, some of the gravitational potential energy is transformed into rotational energy. If I do the calculations right, the proportion depends on the ratio of the moment of inertia of the object and the product of its mass and the square of its rolling radius, with g = (1 + M/(mr^{2})) a / sin(theta), where M is the moment of inertia of the object, r is its radius, and m is its mass. (I write moment of inertia as M rather than I, because of font issues.)

From wikipedia, I see M/(mr^{2}) is 2/5 for a solid ball of uniform density, 2/3 for a spherical shell, 1/2 for a solid cylinder, and 1 for a cylindrical shell.

The rotation is provided by friction. The acceleration of the center of mass is independent of this.

If that were true, then cylinders of the same mass but a different moment of inertia would roll down with the same acceleration and same final speed. But they don't.

Posted: **Mon Mar 06, 2017 2:23 pm UTC**

It would also violate conservation of energy.

Like, you can calculate the acceleration of gravity from rolling cylinders but you need to take into account the rotational energy (as well as the linear kinetic enrgy).

Like, you can calculate the acceleration of gravity from rolling cylinders but you need to take into account the rotational energy (as well as the linear kinetic enrgy).

Posted: **Mon Mar 06, 2017 2:58 pm UTC**

I believe that one of the most accurate ways to measure g is also one of the simplest; drop something and measure its acceleration. I forget the exact name of the machine, but I remember how it works. A mirror is dropped in a vacuum chamber. Then a laser bounces pulses of light off the mirror and into a light sensor. The delay between the laser sending the light and the sensor detecting the light tells you the mirror's location. The entire machine is light and small enough to be transported by someone using a dolly. It is used to measure the differences in g in different location, although the paper I cited used it to measure G instead.

Posted: **Mon Mar 06, 2017 3:00 pm UTC**

Well, a vacuum chamber and laser measurement equipment is kind of difficult to get at home.

Posted: **Mon Mar 06, 2017 3:09 pm UTC**

[bangs head on table] I made a fool of myself because I skipped the last 2 words of the OP! 2 WORDS!

Moving on, if you have a good camera you could film yourself dropping a ball in front of a yard stick. It works the same in principle and the math would not be particularly difficult.

Moving on, if you have a good camera you could film yourself dropping a ball in front of a yard stick. It works the same in principle and the math would not be particularly difficult.

Posted: **Mon Mar 06, 2017 3:46 pm UTC**

Naw, you were fine...

If you're up to going over to Home Depot for lengths of slope/short round meyal bars, whatever's-your-equivalent-of-Maplin for electronic components, sports stores for stopwatches and/or any electronic retailer for a webcam, just forget about all that expense and pop over to the local campus and pester their Physics Department to come on over.

Tell 'em that g is unexpectedly high/low in your house, and they'll surely be tripping over themselves (physics geeks being notoriously absentminded and clumsy) to investigate! Then blame the lack of interesting results on the aliens who live in your attic that nobody else can see.

...e.g. the OP's home.jewish_scientist wrote:The entire machine is light and small enough to be transported by someone using a dolly. It is used to measure the differences in g in different location

If you're up to going over to Home Depot for lengths of slope/short round meyal bars, whatever's-your-equivalent-of-Maplin for electronic components, sports stores for stopwatches and/or any electronic retailer for a webcam, just forget about all that expense and pop over to the local campus and pester their Physics Department to come on over.

Tell 'em that g is unexpectedly high/low in your house, and they'll surely be tripping over themselves (physics geeks being notoriously absentminded and clumsy) to investigate! Then blame the lack of interesting results on the aliens who live in your attic that nobody else can see.

Posted: **Mon Mar 06, 2017 3:53 pm UTC**

jewish_scientist wrote:[bangs head on table] I made a fool of myself because I skipped the last 2 words of the OP! 2 WORDS!

No no no, you only made a fool of yourself by not taking gmalivuk's message into account. It's a great answer considering the circumstances.

You can also put a thermometer and a small wire spool in a vacuum flask, attach a heavy object to the wire and let it come down. Make sure the spool provides enough friction to let the object travel down slowly, so the potential energy is efficiently converted to heat. You need to measure: the mass of the object and both the temperature in the vacuum flask and the height of the object before and after.

Yet another way is to buy three 1-litre bottles of <your favourite beverage>. Drink one of them and attach it to a force meter. Next attach a full bottle to the force meter. Now you need to measure the density of the beverage by pouring exactly 1kg (depending on the beverage this may require the 3rd bottle) into a measuring jug on a scale (don't forget to account for the weight of the jug) and read off the exact volume. Given the force difference and the mass of the beverage you can easily calculate g.

Soupspoon wrote:Then blame the lack of interesting results on the aliens who live in your attic that nobody else can see.

Physicists scare the aliens and ghosts away.

Posted: **Mon Mar 06, 2017 4:02 pm UTC**

That reminds me a lot of the barometer question.

Posted: **Mon Mar 06, 2017 4:37 pm UTC**

There's also our own thread about the mass of a feather.

Posted: **Mon Mar 06, 2017 5:37 pm UTC**

Zohar wrote:That reminds me a lot of the barometer question.

Same here. I immediately thought of using a simple pendulum of known length.

Posted: **Mon Mar 06, 2017 6:54 pm UTC**

I'd get a good video camera (big front lens and high frame rate, specifically) and use that to film a mass being towed along a table using a spring balance, a length of cord, a pulley and a weight. The larger the mass relative to the pulley and the rollers, the better your results should be ... I think ... so maybe use tiny LEGO skates under a housebrick. Include a watch in the video, in focus, with the seconds showing (or tenths or even hundredths if you can and want to). Behind the track of the moving mass, have a metre ruler visible. You'll need depth of field and minimal parallax error here, so position your camera well back and zoom in. This'll reduce the light hitting your camera (inverse square law) so be sure to do this somewhere really brightly lit, like the middle of a floodlit stadium or Nevada. Repeat the brick-draw with various weights on the end of the cord. By going through your videos frame-by-frame looking for the changes in the stopwatch reading and then at the brick and ruler, you can determine the distance moved in each second or smaller part thereof (accuracy limited by low frame-rates, parallax error and lousy watches) and the force pulling that mass along in each case. From your distance and time figures you can determine acceleration and compare that to the forces recorded to deduce a relationship between force (in whatever units your spring balance shows) and acceleration (in markings on your ruler per beat of your stopwatch). Having done so, take hold of the cord and lift the brick and its skates off the table. Note the tension reading. Check your graph. There ya go.

The greater the spread of weights used, the better, but if you use more than two bricks to pull that one along you'd better have a long table, and a high one too because once your weight hits the deck things change. I'd suggest making sure you include weights both sides of "a brick" and near to "a brick." You don't have to use a brick, of course. You could use three 2 l drink bottles on a skateboard on a raised girder as your towed mass and a basket with 1 x 500 ml, 1 x 1 l, 1 x 1 l + 1 x 500 ml, 1 x 2 l, 1 x 2 l + 1 x 500 ml, 1 x 2 l + 1 x 1 l, 1 x 2 l + 1 x 1 l + 1 x 500 ml and 2 x 2 l drink bottles as your weights. Don't go assuming the weight is proportional to the total volume of drink. There's the basket and the material of the bottles to consider, and the weight of the vertical part of the cord (which changes as the cord runs over the pulley) to consider, so use that spring balance.

So, what does "experminetly" mean?

The greater the spread of weights used, the better, but if you use more than two bricks to pull that one along you'd better have a long table, and a high one too because once your weight hits the deck things change. I'd suggest making sure you include weights both sides of "a brick" and near to "a brick." You don't have to use a brick, of course. You could use three 2 l drink bottles on a skateboard on a raised girder as your towed mass and a basket with 1 x 500 ml, 1 x 1 l, 1 x 1 l + 1 x 500 ml, 1 x 2 l, 1 x 2 l + 1 x 500 ml, 1 x 2 l + 1 x 1 l, 1 x 2 l + 1 x 1 l + 1 x 500 ml and 2 x 2 l drink bottles as your weights. Don't go assuming the weight is proportional to the total volume of drink. There's the basket and the material of the bottles to consider, and the weight of the vertical part of the cord (which changes as the cord runs over the pulley) to consider, so use that spring balance.

So, what does "experminetly" mean?

Posted: **Mon Mar 06, 2017 11:01 pm UTC**

For the record, one of my favourite practicals from the first year of my (physics) degree was measuring g to a precision of better than 1 part in 1000 using a pendulum (and a torsion pendulum as well for reasons I can't remember). We used the method of exact fractions to get the precision so high; I think my group got to about 1 part in 5000/7000 or so which wasn't particularly unusual.

Turns out the gravity at my university is about 0.5% higher than the standard value. The lab book mentioned this because, obviously, they were worried about students thinking they'd hecked up.

Turns out the gravity at my university is about 0.5% higher than the standard value. The lab book mentioned this because, obviously, they were worried about students thinking they'd hecked up.

Posted: **Mon Mar 06, 2017 11:04 pm UTC**

Sableagle wrote:So, what does "experminetly" mean?

Adjectival form of "previously having had a quality proportional to that of kitten".

Posted: **Mon Mar 06, 2017 11:08 pm UTC**

eSOANEM wrote:Turns out the gravity at my university is about 0.5% higher than the standard value.

It's that "freshman 15," y'know.

Posted: **Tue Mar 07, 2017 6:05 pm UTC**

I forgot to mention that you have to repeat the experiment every two hours for thirty days to make sure you're accounting for fluctuations caused by the relative positions of the Sun and Moon and by local tides if you're anywhere near a large body of water or other liquid.

Posted: **Tue Mar 07, 2017 6:59 pm UTC**

Sableagle wrote:if you're anywhere near a large body of water or other liquid.

So stay away from me after I've been to the pub.