Question

a pendulum consists of a small object hanging from the ceiling at the end of a string of negligible mass. the string has a length of 0.74 m. with the string hanging vertically, the object is given an initial velocity of 2.5 m/s parallel to the ground and swings upward in a circular arc. eventually, the object comes to a momentary halt at a point where the string makes an angle θ with its initial vertical orientation and then swings back downward. find the angle θ.

Explanation:

Step1: Apply conservation of mechanical energy

The initial kinetic energy \(K_{i}=\frac{1}{2}mv_{i}^{2}\), and the initial potential energy \(U_{i} = 0\) (taking the initial vertical position as the zero - potential level). At the highest point, the kinetic energy \(K_{f}=0\) (momentary halt), and the potential energy \(U_{f}=mgh\). By the conservation of mechanical energy \(K_{i}+U_{i}=K_{f}+U_{f}\), so \(\frac{1}{2}mv_{i}^{2}=mgh\).

Step2: Relate the height \(h\) to the length \(L\) and the angle \(\theta\)

We know that \(h = L(1-\cos\theta)\) (from the geometry of the pendulum, where \(L\) is the length of the string). Substituting \(h = L(1 - \cos\theta)\) into \(\frac{1}{2}v_{i}^{2}=gh\) (the \(m\) cancels out from both sides of \(\frac{1}{2}mv_{i}^{2}=mgh\)), we get \(\frac{1}{2}v_{i}^{2}=gL(1-\cos\theta)\).

Step3: Solve for \(\cos\theta\)

Rearrange the equation \(\frac{1}{2}v_{i}^{2}=gL(1-\cos\theta)\) for \(\cos\theta\):

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Given \(v_{i} = 2.5\space m/s\), \(L=0.74\space m\), and \(g = 9.8\space m/s^{2}\)

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Step4: Solve for \(\theta\)

Take the inverse - cosine of \(0.569\): \(\theta=\cos^{-1}(0.569)\approx55^{\circ}\)

Answer:

\(55^{\circ}\)