Angular Momentum

[Lloyd Burton]

I am trying to calculate the effect of gravity (in terms of angular momentum) acting upon a body of say, 1 kg when it is rotated (against gravity) through 180 degrees from the 6 o'clock position to the 12 o'clock position. Let the radius be 1 meter and the time of travel be 1 second. Help, anyone?

[ME]

What do you already have figured out?

[Lloyd Burton]

Hi Marcello,
Would you go along with this reasoning?
Calculations for L
Angular Momentum = L; P = linear momentum; m = mass = 1kg; r = radius = 1000mm; V= velocity; s= seconds; g = 9.8 kg.m/s/s; t = 1 second;
L = P x r = m x v x r

m = mass = 1kg;

Next find v.
V = g x t = 9.8m/s/s x 1 s = 9.8m/s
Thus v = 9.8ms.

Now find r.
First I divided the arc of fall into 16 parts, each marking the position of the rising mass at 1/16 second intervals. Then I measured the distance of each point from the vertical line through the centre of the circle, added them together and divided by 16 to get the average length. A small error in measurement was expected.

Sum of distances: 10380mm
Average distance: 648.75 mm = .64875 metres.
Thus r = .64875 metres.

To find L, substitute these quantities into formula:
L = P x r = m x v x r
Thus L = 1kg x 9.8m/s x .64875m = 6.36 kg.m/s.
L = 6,36 kg.m/s.
So the Angular Momentum is 6.36 kg.m/s (approx.)
I expect there must be an easier way of doing it.
Cheers,
Lloyd.

[Tarsier79]

Hi Lloyd

Sum of distances: 10380mm
Average distance: 648.75 mm = .64875 metres.
Thus r = .64875 metres.

The radius of your arc is 1m. The length of the arc (180 degrees) you are measuring is 3.14m

Next find v.
V = g x t = 9.8m/s/s x 1 s = 9.8m/s
Thus v = 9.8ms.

Why? Velocity is a function of more than just gravity and time. Gravity pushes directly down, but a pendulum bob spends virtually no time travelling directly with, or directly against gravity.

[Tarsier79]

I am trying to calculate the effect of gravity (in terms of angular momentum) acting upon a body of say, 1 kg when it is rotated (against gravity) through 180 degrees from the 6 o'clock position to the 12 o'clock position. Let the radius be 1 meter and the time of travel be 1 second.

Oops. Didn't read your first post properly. So you have a constant speed? (in which case I don't know what you are asking) or do you want the time to be 1 second including the decelleration due to gravity?

[ME]

I think I'm as confused as Tarsier here :-)

Your ingredients don't seem to match, so I don't think I follow your reasoning.
Especially difficult to figure out is what happened to those 16 parts which add-up to 10.38 meter... ((10.38/16=0.64875) ≈ (2/pi=0.6366)).
To add to Tarsier's guess: the radius of 0.6366m indicates you tried to match a circumference of 2m with a straight drop of (2*1000mm= 2m) ?
A 1 second straight down drop under gravity (9.8 m/s2) will fall a distance of 4.9 m, and obtains a velocity of 9.8 m/s. (I don't see where this calculation should fit).

Anyway, what kind of radius would you like to find?

I am trying to calculate the effect of gravity (in terms of angular momentum) acting upon a body of say, 1 kg when it is rotated (against gravity) through 180 degrees from the 6 o'clock position to the 12 o'clock position.

The influence of gravity is g*Sine(θ) at any orientation θ when it would rotate on a weightless beam. The impact could be calculated by integration, or (my advice) just simulation.
If you already have a setup you could also measure the 1-second result (with a camera for instance) under variable velocity conditions at the 6 o'clock position.
The following formula should provide the minimum velocity to make it go over the top: v = √(2·g·∆h) = √(4·g·r)
See: http://www.besslerwheel.com/forum/viewtopic.php?t=6960 ?