Question

he "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 5.00 m long, and the upper end of the cable is fastened to the arm at a point 3.00 m 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 30.0° with the vertical. (b) Does the angle depend on the weight of the passenger for a given rate of revolution?

Answer 1

(a) 11.0 s

Let's start by writing the equations of the forces along the horizontal and vertical direction. We need to resolve the tension along the cable, T, along the two directions. Therefore we get:

- Vertical direction:

$T cos \theta = mg$ (1)

where

T is the tension in the cable

$\theta=30.0^{\circ}$ is the angle with the vertical

m is the mass of the passenger and the seat

$g=9.8 m/s^2$ is the acceleration of gravity

- Horizontal direction: here the component of the tension in the cable provides the centripetal force, so we can write

$Tsin \theta = m \omega^2 r$ (2)

where

$\omega$ is the angular velocity

$r=3.00 m + (5.00) sin 30^{\circ}=5.5 m$ is the distance from the central shaft (the radius of the circular path)

Dividing eq.(2) by eq.(1), we get

$tan \theta = \frac{\omega^2 r}{g}$

And re-arranging the equation, we can find the angular velocity:

$\omega = \sqrt{\frac{g tan \theta}{r}}=\sqrt{\frac{(5.5)(tan 30.0^{\circ})}{9.8}}=0.569 rad/s$

And the time for one revolution is equal to the period, so:

$T=\frac{2\pi}{\omega}=\frac{2\pi}{0.569}=11.0 s$

(b) No

As we have seen previously, the equation obtained for the angle is given by

$tan \theta = \frac{\omega^2 r}{g}$

where

$\omega$ is the angular velocity

r is the radius of the circular path

g is the acceleration of gravity

As we can see, in this equation m (the mass of the passenger) does not appear: this means that the angle does not depend on the mass (or the weight) of the passenger.

Answer 2

Answer:(a) 11.0 sLet's start by writing the equations of the forces along the horizontal and vertical direction. We need to resolve the tension along the cable, T, along the two directions. Therefore we get:- Vertical direction: (1)where T is the tension in the cable is the angle with the verticalm is the mass of the passenger and the seat is the acceleration of gravity- Horizontal direction: here the component of the tension in the cable provides the centripetal force, so we can write (2)where is the angular velocity is the distance from the central shaft (the radius of the circular path)Dividing eq.(2) by eq.(1), we getAnd re-arranging the equation, we can find the angular velocity:And the time for one revolution is equal to the period, so:(b) NoAs we have seen previously, the equation obtained for the angle is given bywhere is the angular velocityr is the radius of the circular pathg is the acceleration of gravityAs we can see, in this equation m (the mass of the passenger) does not appear: this means that the angle does not depend on the mass (or the weight) of the passeng

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