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Rolling ball down ramp to calculate acceleration of gravity

Thu 3 Dec, 2015 06:54 pm
I rolled a golf ball down a ramp using a yard stick, and I measured the time (.9 seconds), the angle (27 degrees),
yet my acceleration of gravity calculation was off by a factor of 2.

Since it covered 3 feet in .9 seconds, I calculate the avg velocity to be 3.33 and assuming that the average is the intial velocity
plus the final velocity divided by 2, I get 6.66 for the final velocity

So the final velocity is the change of velocity, so 6.66 feet per second / .9 seconds gives 7.407

I then divided that by the sin of 27.05 degrees (.4548), getting 16.285 feet / seconds squared.
It should be 32, so can anyone explain where my error is?

Thank you.
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Type: Question • Score: 2 • Views: 5,527 • Replies: 86

fresco

1
Fri 4 Dec, 2015 02:37 am
@sjlcomp,
I don't see how you can use 'average velocity' which is TOTAL DISTANCE/TOTAL TIME never (u+v)/2

If you dont know v, then use d=0.5 g.sin(s).t sqd
assuming u=0 s= angle of slope.

Using your figures, 3=0.5 g sin(27)x 0.81

so g =3/(0.5 x 0.81 x sin 27) which still gives answer of around 16 !

I therefore suggest that your measurements are inaccurate. In particular, check that you are measuring t from u= zero. Also consider the surface of the ball which if not smooth may give a retarding effect equivalent to a decceleration.

sjlcomp

1
Fri 4 Dec, 2015 03:41 pm
@fresco,
Thanks for you reply.

I figure that my time measurement is significantly off, even though I did a number of trials timing another event (another digital clock) and I was off less than .1 seconds, which had me add this error amount to my time measurement.
Still, if my time measurement is off by around .2 seconds it would explain it.

I have thought about using a Basic Stamp with light sensors to count time from when the light sensor detects an interruption...you get the idea.

I used total dist / total time to get the avg velocity so I could determine the final velocity...

If I ever build or get a device to measure the time automatically I will post the results here.

fresco

1
Sat 5 Dec, 2015 01:09 am
@sjlcomp,
How does average velocity give final velocity ?
And you don't need final velocity using d=u.t+0.5 a.t*2
engineer

1
Sat 5 Dec, 2015 08:56 am
@fresco,
No, since the velocity increases linearly, you can use average velocity and get the correct answer. v = at, so if v starts at zero, average velocity = 1/2 at and distance traveled is average velocity times time = 1/2 a t^2.

The reason your experiment doesn't work is likely due to friction from the board slowing the ball down.
maxdancona

1
Sat 5 Dec, 2015 09:16 am
@engineer,
The function you are using is for an object that is sliding down a friction-less ramp, not rolling.

The ball in your experiment is rolling. This means that it is spinning faster and faster (or gaining angular momentum). The fact that the ball is also gaining angular momentum means that there is an additional term for torque.

Here is a good explanation at a University level... https://iweb.tntech.edu/murdock/books/v2chap2.pdf

I will try to see if I can find a simpler explanation.

maxdancona

1
Sat 5 Dec, 2015 09:41 am
@maxdancona,
This explanation might be a bit simpler to understand...

http://www.wired.com/2014/07/a-rolling-object-accelerating-down-an-incline/ 0 Replies

fresco

1
Sat 5 Dec, 2015 10:02 am
@engineer,
I have never come across the need to use (or teach) a concept of 'average velocity' in this case. ...only the derivation of that formula from the area under the velocity time graph.(UK school physics).
0 Replies

Kolyo

1
Sat 5 Dec, 2015 01:14 pm
@maxdancona,
maxdancona wrote:

I will try to see if I can find a simpler explanation.

Energy concepts make for a much easier explanation than vectors for problems like these. You just need to analyze the situation in terms of gravitational potential energy, rotational kinetic energy, and translational kinetic energy.

maxdancona

1
Sat 5 Dec, 2015 01:24 pm
@Kolyo,
Yes, you are right Kolyo.

The potential energy of the golf ball at the top of the ramp should equal sum of the rotational kinetic energy and the translational kinetic energy at the bottom.
0 Replies

sjlcomp

1
Sun 6 Dec, 2015 10:30 pm
@maxdancona,
Thanks, all, for the replies and for the links.

I would like to attack the problem in reverse, to ask the question from a different perspective.

This revised problem already knows the acceleration of gravity to be 32 ft / sec squared, and is trying to calculate the time. If we use a 30 degree angle to get an easy sin of .5, and a 16 foot long ramp, can we say that 1/2 A * T-squared * sin theta is the time squared? In this case 16 t squared = 16, meaning the time squared is 1, as is time. If everyone agrees with this (if wrong, someone please correct it), my question is MUST an additional FACTOR be used because the ball is rolling, with no slippage, but we need to consider torque or moment of inertia, or some other calculation for friction?

My guess is that there is a slight, probably insignificant difference if we are rolling a marble or golf ball, but if we used a 100 pound bowling ball of 2 feet in diameter... would anyone like to work out 2 examples illustrating this?

1 - calculate the time for a small ball to roll down a 16 foot long ramp, at a 30 degree angle. Then do it again using the bowling ball on the same ramp...is the time significantly different?

Thank you.

Kolyo

1
Sun 6 Dec, 2015 11:51 pm
@sjlcomp,
Some balls may descend faster than others, but it has nothing to do with total mass.

Are you familiar with the energy concepts I wrote about?

The gravitational potential energy of the ball at the top of the ramp ends up in two places once the ball is at the bottom of the ramp: (1) the translational motion of the ball's center of mass through space; (2) the rotational motion of the ball around its axis of rotation.

If less of the energy goes into rotation, more will go into translation through space, and the ball will move faster. The distribution of mass in the ball -- not the total mass -- will determine how much of the energy goes into rotation. Consider the case of a golf ball versus a marble. The golf ball has all its mass on the outside, so more energy has to go into rotating it, because it's harder to rotate mass that is further from the axis of rotation. The marble is more compact, with some of its mass near the center, so less of the energy will go into rotating it, which means more of the energy will go into propelling it through space, and it will descend faster.

(Of course, all this assumes no air resistance. The heavier objects will of course be better at overcoming air resistance, and they will move faster. But as you said, their advantage is pretty negligible.)
maxdancona

1
Tue 8 Dec, 2015 08:29 am
@Kolyo,
Kolyo is correct (other than that I have nothing more to add).
0 Replies

puzzledperson

-1
Tue 22 Dec, 2015 03:43 am
@sjlcomp,
Let's try a thought experiment. We have two ramps. Ramp A has a 45 degree incline. Ramp B has a very, very slight incline. We measure how long it takes a ball to move the same distance (say, along 20 feet worth of ramp). The ball will take much longer to move 20 feet down ramp B than it will to move 20 feet down ramp A.

Gravity is pulling straight down on the ball in both cases, and by straight down I don't mean down the ramp, I mean toward the Earth. Neither ramp moves the ball along the gravity vector, but ramp A is much closer to that vector than ramp B.

Any vector can be considered to be the vector sum of two component vectors. For example, a particular 45 degree vector could be the vector sum of a horizontal vector and a vertical vector.

The only part of the gravity force vector that causes forward motion down the ramp is the component of the vector pointing in the forward direction. If the board were perfectly flat that component would be zero and all of the gravity force vector would be a single component pointing straight down towards the Earth and contributing only to static deformation of the ball and the ramp.

So fundamentally, you have to account for the difference in the component of the gravity vector applied to motion. In the case of a ball dropped out a window of a tower, that component is wholly coincident with the gravity force vector itself. In every other case it is less.

Beyond that, you have to consider frictional forces which convert gravitational energy into conductive, convective and radiative heat energy that increases the temperature of the ball, ramp, and air (or radiated into space). That will depend on the angle of the ramp also, as well as the materials which the ball and ramp are made of, and the weight (not mass) of the ball, since that can cause dynamic deformation, or bring more of the ball surface into contact with the ramp, or bring the ball microscopically closer to the ramp, any of which may aggravate the molecular attraction which constitutes friction. Modern golf balls are dimpled and that irregular rough surface in contact with a rough wooden surface would tend to increase friction.
layman

-2
Tue 22 Dec, 2015 02:51 pm
@puzzledperson,
Quote:
In the case of a ball dropped out a window of a tower, that component is wholly coincident with the gravity force vector itself. In every other case it is less.

Yeah, and forward motion, alone, doesn't change anything, eh?

If you take a bulled, hold it at shoulder height and drop it, it will quickly hit the floor.

If you put that bullet in a rifle, and fire it from shoulder height along a plane that is "level" (90 degrees from "down") it will hit the ground just as fast. It may go a half a mile before it does, but that's only because it's travelling at a high speed.

On the other hand, if you were standing on, say, a diving board, and shot the rifle straight down when the end of it's nozzle was shoulder height, it would hit the ground much faster than one falling under the force of gravity alone.

puzzledperson

1
Thu 24 Dec, 2015 07:09 pm
@puzzledperson,
Just to simplify my previous comment:

The acceleration of a ball down a ramp will always be less than the acceleration of an object in freefall under the influence of gravity. This has nothing to do with the conversion of translational energy into rotational energy, as suggested by several writers above. The same thing would be true if the ball didn't rotate at all. You can make the acceleration as close to zero as you like by making the incline of the ramp as close to flat as you can. Conversely, the closer the incline is to 90 degrees, the closer acceleration will be to the gravitational acceleration constant.

cicerone imposter

1
Thu 24 Dec, 2015 07:37 pm
@puzzledperson,
Think friction; more surface means slower.
puzzledperson

1
Thu 24 Dec, 2015 07:43 pm
@layman,
layman wrote: "If you take a bulled, hold it at shoulder height and drop it, it will quickly hit the floor. If you put that bullet in a rifle, and fire it from shoulder height along a plane that is "level" (90 degrees from "down") it will hit the ground just as fast."

Not true. The amount of time a bullet fired in this fashion to hit the ground will vary with, among other things, the amount of propellant. Rounds of the same identical caliber will travel further and take different amounts of time to hit the ground if they are loaded with different amounts of powder. Yet they all take the same amount of time to drop from shoulder height in freefall.

A bullet dropped from six feet will take 3/8 of a second to hit the ground.

A bullet fired from a rifle will be enormously effected by drag (air resistance). You would need to specify a bullet weight, shape, and muzzle velocity, as well as wind speed and direction, to be able to calculate how far the bullet will take to hit the ground when fired horizontally from shoulder height. In general it won't match the time it takes the same bullet to drop to the ground if held in the hand six feet above ground and let go to descend in freefall.
roger

0
Thu 24 Dec, 2015 07:46 pm
@puzzledperson,
I think layman had it right.
puzzledperson

1
Thu 24 Dec, 2015 07:46 pm
@cicerone imposter,
cicerone imposter wrote: "Think friction; more surface means slower."

You can measure how long it takes to move along 20 feet of ramp regardless of the inclination. That's the same amount of surface.
0 Replies

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