Question

1. a toy airplane is suspended from the ceiling by a 0.75 m string. the plane revolves in a horizontal circle at a constant speed as shown,

a. what is the speed of the plane? (3 marks)

b. what is the period of revolution and frequency of the airplane? (2 marks)

Explanation:

Step1: Analyze the forces

Let the tension in the string be \(T\). In the vertical direction, \(T\cos\theta = mg\). In the horizontal direction, \(T\sin\theta=\frac{mv^{2}}{r}\), where \(r = L\sin\theta\) (\(L = 0.75m\) is the length of the string).
From \(T\cos\theta = mg\), we get \(T=\frac{mg}{\cos\theta}\). Substitute \(T\) into \(T\sin\theta=\frac{mv^{2}}{r}\): \(\frac{mg}{\cos\theta}\sin\theta=\frac{mv^{2}}{L\sin\theta}\).
Simplify the equation: \(v=\sqrt{gL\sin^{2}\theta\tan\theta}\).
Given \(g = 9.8m/s^{2}\), \(L = 0.75m\), \(\theta=27^{\circ}\), \(\sin27^{\circ}\approx0.454\), \(\tan27^{\circ}\approx0.5095\).
\(v=\sqrt{9.8\times0.75\times(0.454)^{2}\times0.5095}\approx1.2m/s\).

Step2: Calculate the period \(T\)

The formula for the period \(T=\frac{2\pi r}{v}\), \(r = L\sin\theta=0.75\times0.454 = 0.3405m\), \(v\approx1.2m/s\).
\(T=\frac{2\pi\times0.3405}{1.2}\approx1.8s\).

Step3: Calculate the frequency \(f\)

The formula for frequency \(f=\frac{1}{T}\), \(T\approx1.8s\), so \(f=\frac{1}{1.8}\approx0.56Hz\).

Answer:

a. The speed of the plane is approximately \(1.2m/s\).
b. The period of revolution is approximately \(1.8s\) and the frequency is approximately \(0.56Hz\).