B Analysis of Simple Pendulum Motion

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The discussion focuses on the analysis of simple pendulum motion, detailing the construction and parameters involved, such as the pendulum length (L) and angles (α, β). It explains how to calculate the speed (Vα) at various angles using energy conservation principles, emphasizing that tension does not perform work during the pendulum's motion. The mathematical derivation involves geometric relationships and the sine theorem to establish height differences and speed calculations. A mentor notes the complexity of the explanation but confirms the final formula for speed is correct. The analysis highlights the interplay of geometry and physics in understanding pendulum dynamics.
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Simple pendulum motion is a type of object motion, and it has its unique (or unique beauty). As a "beauty of kinematics", the speed of the simple pendulum motion is naturally an important part. This article will tell you to use high school physics knowledge to find the speed at each point.
First, we construct a simple pendulum, the length of the rope remains unchanged and the object being swung is doing circular motion. If it is released from point D and reaches point F, we set:
The length of the rope (i.e the pendulum length) is L, ∠FCB=90°, ∠DCF=β, ∠ECF=α, the mass of the particle is m (Note: E is a moving point) (As shown in Figure 1)
Figure 1.png
Figure 2.png

(The left is Figure 1; the right is Figure 2.)
First, we must know the speed Vα corresponding to each α. We can find the height difference and then use the method to calculate the speed. (Note: Energy is a scalar, but speed is a vector, and the direction of speed is always perpendicular to the rope)
DrawDI⊥CF intersecting CF at point I, DI intersects CE at point H, draw EG⊥DI intersecting DI at G (As shown in Figure 2)
In △CDH,
∠CHD = 180°- ∠CHI = 180°- (180°- ∠CIH - ∠ICH) = 90°+ α
∠CDI = 180°- ∠CID - ∠ICD = 90°- β
According to the sine theorem,
CH = L·sin(π/2 - β) / sin(π/2 + α)
HE = L·(1-sin(0.5π - β) / sin(0.5π + α) = L·(1 - cosβ / cosα)
Through the eight-character shape, we get:
∠G = ∠ICH = α
∴GE=∆h=HE· cosα=L· (cosα - cosβ)
Using the conservation of mechanical energy ( tension of the rope is always perpendicular to the direction of motion, so the tension does no work), we get:
Vα=√2g∆h=√2gL (cosα - cosβ)

Mentor note: Last line rewritten in LaTeX
##V_{\alpha} = \sqrt{2gL(\cos(\alpha) - \cos(\beta)}##
 
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Assuming you mean ##v_\alpha=\sqrt{2gL(\cos\alpha-\cos\beta)}##, this appears to be correct, although an extremely long way of saying what's said. Do you have a question?
 
I just want to share my thoughts with everyone.
 
Hello everyone, Consider the problem in which a car is told to travel at 30 km/h for L kilometers and then at 60 km/h for another L kilometers. Next, you are asked to determine the average speed. My question is: although we know that the average speed in this case is the harmonic mean of the two speeds, is it also possible to state that the average speed over this 2L-kilometer stretch can be obtained as a weighted average of the two speeds? Best regards, DaTario
This has been discussed many times on PF, and will likely come up again, so the video might come handy. Previous threads: https://www.physicsforums.com/threads/is-a-treadmill-incline-just-a-marketing-gimmick.937725/ https://www.physicsforums.com/threads/work-done-running-on-an-inclined-treadmill.927825/ https://www.physicsforums.com/threads/how-do-we-calculate-the-energy-we-used-to-do-something.1052162/

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