A pendulum of length swings in a vertical plane

MATHEMATICS · Multi-part · CBSE · Mathematics · Applications of Derivatives · Applications of Derivatives

Problem

A pendulum of length 2.3 m2.3\,\text{m} swings in a vertical plane. The angle θ(t)\theta(t) (in radians) is measured from the vertical, and the height y(t)y(t) (in meters) of the bob above its lowest point satisfies y(t)=2.3 (1−cos⁡θ)y(t) = 2.3\, (1 - \cos \theta). At an instant when θ=0.40 rad\theta = 0.40\,\text{rad}, the angular velocity dθdt\frac{d\theta}{dt} is −0.35 rad s−1-0.35\,\text{rad}\,\text{s}^{-1} (the negative sign indicates that θ\theta is decreasing). (a) Derive an expression for dydt\frac{dy}{dt} in terms of θ\theta and dθdt\frac{d\theta}{dt}. (b) Calculate dydt\frac{dy}{dt} at the specified instant, including correct units and sign. (c) Determine whether the bob is rising or falling at this instant and explain how your calculation supports your conclusion. (d) The gravitational potential energy of the bob relative to its lowest point is U=mgyU = mgy, where mm is the bob’s mass and g=9.8 m s−2g = 9.8\,\text{m}\,\text{s}^{-2}. Express dUdt\frac{dU}{dt} in terms of mm, gg, θ\theta, and dθdt\frac{d\theta}{dt}, and interpret the physical meaning of your result when θ=0.40 rad\theta = 0.40\,\text{rad} and dθdt=−0.35 rad s−1\frac{d\theta}{dt} = -0.35\,\text{rad}\,\text{s}^{-1}.

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