WuliDancer
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- TL;DR
- 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)
(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)}##
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)
(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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