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 0.77 m. with the string hanging vertically, the object is given an initial velocity of 2.7 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 the 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 \(U_{f}=mgh\), where \(h = L(1 - \cos\theta)\) (\(L\) is the length of the string). By the conservation of mechanical energy \(K_{i}+U_{i}=K_{f}+U_{f}\), so \(\frac{1}{2}mv_{i}^{2}=mgL(1 - \cos\theta)\).

Step2: Solve for \(\cos\theta\)

Cancel out the mass \(m\) from both sides of the equation \(\frac{1}{2}v_{i}^{2}=gL(1 - \cos\theta)\). Then \(1-\cos\theta=\frac{v_{i}^{2}}{2gL}\). Given \(v_{i} = 2.7\space m/s\), \(L=0.77\space m\), and \(g = 9.8\space m/s^{2}\), we substitute the values: \(1-\cos\theta=\frac{2.7^{2}}{2\times9.8\times0.77}\).

$$
LATEXBLOCK0
$$

Then \(\cos\theta=1 - 0.483=0.517\).

Step3: Find \(\theta\)

Take the inverse - cosine of \(0.517\), \(\theta=\cos^{-1}(0.517)\). Using a calculator, \(\theta\approx59.5^{\circ}\)

Answer:

\(59.5^{\circ}\)