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Math  /  Geometry

QuestionProblem 12.76 The given angular acceleration remains valid even if the pendulum cord is replaced by a massless rigid bar. For this case, let L=5.3ftL=5.3 \mathrm{ft} and assume that the pendulum is placed in motion at θ=0∘\theta=0^{\circ}. What is the minimum angular velocity at this position for the pendulum to swing through a full circle?
Answer θ˙min⁡=4.930rad/s\dot{\theta}_{\min }=4.930 \mathrm{rad} / \mathrm{s}

Studdy Solution
Substitute the values g=32.2 ft/s2 g = 32.2 \, \text{ft/s}^2 and L=5.3 ft L = 5.3 \, \text{ft} :
θ˙min⁡=4×32.25.3 \dot{\theta}_{\min} = \sqrt{\frac{4 \times 32.2}{5.3}} θ˙min⁡=128.85.3 \dot{\theta}_{\min} = \sqrt{\frac{128.8}{5.3}} θ˙min⁡=24.3019 \dot{\theta}_{\min} = \sqrt{24.3019} θ˙min⁡≈4.930 rad/s \dot{\theta}_{\min} \approx 4.930 \, \text{rad/s}

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