Angular Momentum and Acceleration
When someone or something travels in a straight line, it is quite easy to know how far someone has gone in a certain amount of time. But what if they begin moving to the side? What about when they have to turn around a? It can be a simple change, but one that demands a different kind of mathematics to figure out.
Let us imagine a road that curves away, turning 45 degrees in a turn whose radius has been found to be one half of a mile (0.5 mi). The speed limit on this road is 55 miles per hour. What is a vehicle’s angular momentum (how fast they are spinning) if they follow the speed limit around corner? That is simple,
ω =Δθ / Δt, where t is time, θ (lowercase theta) is the angle, Δ (uppercase delta) is change, and ω (lowercase omega) is angular momentum.
Sounds easy enough, right?
Well, we technically don’t have time. We have speed, but we don’t know the exact distance of the corner. We’re going to have to pull out another equation for this and modify it to give us the distance of this corner, so we can then find the time that passes in the corner from the speed. That will then allow us to solve our initial problem. The equation should be fairly recognizable,
C = 2πr, where C is the circumference, π is 3.141594265… and r is the radius of the circle.
Plugging in, C = 2π(0.5mi), and the circumference of our circle is 3.141594265… let’s just say it is π. this is a step towards our answer, but our corner is only 45 degrees. 3.141594265 is the entire circumference of our 360° circle, but we only need 45°. Finding that is simple!
(45 / 360) can be simplified down to (1 / 8) and (1 / 8) of π is (3.141594265 / 8)= 0.3927 miles!
How long does it take for us to go 0.3927 miles at 55mph? Again, simple!
(0.3927 miles x 1 hour) / 55 miles = 0.00714 hours, or about 0.4284 minutes, or about 25.7 seconds
Now that we finally have time, we can plug in that time into our initial equation.
Before continuing, though, we must convert degrees to radians. This is simple: deg. x π / 180= radians,therefore 45° is 0.785398 radians
ω = Δθ / Δt is now ω =0.03056 / 25.7 ≈ 0.03056 rad/s. This is a very slow angular speed, which is good in this case, because you don’t want to make drivers nervous about going too fast.
Let us up the ante a bit. What if we wanted to find the angular acceleration of a baseball being thrown? Professional baseball players throw a ball in an arc, or part of a circle. Most professional baseball games use a radar to measure the speed of a ball’s pitch. Let us say that a ball was pitched in a 170° arc from a pitchers 26 inch long arm, and was measured leaving the pitcher’s hand at 100 miles per hour, and pitching the ball from rest (no speed) only took 0.1s (one tenth of a second).
The formula α =(ωf-ωi) / Δt gives us angular acceleration, where α (lowercase alpha) is angular acceleration, ω (lowercase omega) is angular momentum in radians per second, and t is time.
Well, ω is Δθ / Δt, and ωi means initial, and ωf means final. Our initial angular momentum is 0, because it is not moving. ωf is where we have to convert the ball’s speed as it leaves the pitcher’s hand (100 mph) to angular velocity (rad/s). Let us assume the ball can continue travelling in a 100mph circle so we can apply the ball’s speed as radians per second.
C = 2πr, so C = 2π (26 inch), so the circumference is approximately 163.3628 inches. We can assume that the ball can travel around this circumference at 100 miles per hour, or…
(100 miles / 1 hour) x (5280 feet / 1 miles) x (12 inches / 1 foot) x (1 hour / 360 seconds) = 17,600 inches per second, or
(17600 inches / 1 seconds) x (1 revolution / 163.3628 inches) x (6.2882.. radiations / 1 revolution) ≈ 676.9236 radians per second.
That was quite a bit of conversion, but we now have all we need to solve the problem.
α =(ωf - ωi) / Δt =(676.9236) - (0) / 0.1 =676.9236 / 0.1 = 6769.236 rd. / s2 (radians per second squared). This is an immense amount of speed and acceleration, just as one would expect from a professional!
Robert Wich is a FPLA certified tutor, who is a passionate engineer and an ace auto enthusiast. When he is not teaching at FPLA, you may find him tinkering with his old model truck. Here, he shares his passion for speed and thrill.
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