Q50E - University Physics with Modern Physics

Question

The “Giant Swing” at a county fair consists of a vertical central shaft with a number of horizontal arms attached at its upper end. Each arm supports a seat suspended from a cable 5.00m long, and the upper end of the cable is fastened to the arm at a point 3.00m from the central shaft (Fig. E5.50). (a) Find the time of one revolution of the swing if the cable supporting a seat makes an angle of $\mathbf{30}\mathbf{.}\mathbf{0}\mathbf{°}$ with the vertical. (b) Does the angle depend on the weight of the passenger for a given rate of revolution?

Step-by-Step Solution

Answer

(a) 5.91

(b) Angle does not depend on the weight of the passenger for a given rate of revolution

Step 1: Identification of given data

Cable length is l = 5

The upper end of the cable is fastened to the arm at a point d = 3.00m from the central shaft 

Step 2: Significance of centripetal force and centrifugal force

The force applied to an item in curved motion that is pointed toward the axis of rotation or the centre of curvature is known as a centripetal force.

A fictitious force that moves in a circle and is directed away from the centre of the circle is called centrifugal force.

$F_c=\frac{m{v}^2}{r}or mr{\omega }^2$

Where, m is mass of body, v is velocity of body, r is radius of circular path and the $\omega$ is angular speed

Step 3: (a) Determining the time of one revolution of the swing

Equating the vertical forces

$T\sin 60°= mg$                                                                                                               …(i)

Equating the horizontal forces

$$\begin{gathered} T\cos 60°= F_c \\ =mr{\omega }^2 \end{gathered}$$                                                                                                          …(ii)

Dividing of equation (i) by the equation of (ii)

$$\begin{gathered} \tan 60°=\frac{g}{r{\omega }^2} \\ \omega =\sqrt{\frac{g}{r \tan 60°}} \end{gathered}$$                                                                                                    …(iii)

Where is radius of curvature and it is given as

$$\begin{gathered} r=d+l\sin 30° \\ =3 m +5m\times \sin 30° \\ =5.5 m \end{gathered}$$

Substitute all values in the equation (iii)

$$\begin{gathered} \omega =\sqrt{\frac{9.8m/s^2}{5m\times \tan 60°}} \\ =1.063rad/s \end{gathered}$$
 Time of one revolution is given by 

$t=\frac{2\mathrm{\pi }}{\omega }$ 

Substitute the $\omega$ value to get time

$$\begin{gathered} t=\frac{2\mathrm{\pi }}{1.063rad/s} \\ =5.91s \end{gathered}$$

Hence the time of one revolution of the swing is 5.91 s

Step 4: (b) Determining whether the angle depend on the weight of the passenger or not for a given rate of revolution

Equation (iii) is given as

 $\tan 60°=\frac{g}{r {\omega }^2}$

From the above equation here we can see that the term of mass is not present so that we can say that the angle does not depend on the weight of the passenger for a given rate of revolution