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 5.00 m long, and the upper end of the cable is fastened to the arm at a point 3.00 m from the central shaft. (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?

Respuesta :

Answer :

(A) Time period [tex]T = 6.189[/tex] sec

(B) Angle is independent of weight.

Explanation:

Given :

Length of fastened arm [tex]= 3[/tex] m

Length of cable [tex]L = 5[/tex] m

(A)

Time of one revolution is [tex]T = \frac{2\pi R }{v}[/tex]

Total radius of circle is [tex]R = 3 + 5\sin 30 = 5.5[/tex] m

Total force in horizontal direction X- direction

  [tex]\Sigma F_{x} = m a _{rad} = F_{T} \sin 30 = m\frac{v^{2} }{R}[/tex]

Since [tex]v = \sqrt{\frac{\sin 30 F_{T} R}{m} }[/tex]

Total force in Y - direction,

  [tex]F_{T} \cos 30 = mg[/tex]

Comparing both equation,

   [tex]v = \sqrt{\frac{\tan 30 m g R}{m} }[/tex]

   [tex]v = \sqrt{\tan 30 g R }[/tex]

Where [tex]g = 9.8 \frac{m}{s^{2} }[/tex]

  [tex]v = \sqrt{31.12} = 5.58\frac{m}{s}[/tex]

So time period,

 [tex]T = \frac{2\times \pi \times 5.5}{5.58}[/tex]

 [tex]T = 6.189[/tex] s

(B)

Mass is cancelled out, so the angle is independent of weight.

The time for one revolution of the swing is 6.2 s.

The angle does not depend on the weight of the passenger.

The given parameters;

  • length of the fastened arm, r₁ = 3 m
  • length of the cable, L = 5 m
  • angle of inclination, θ = 30⁰

The total radius of the circular path is calculated as follows;

R = r₁  + Lsin(θ)

R = 3   +  5sin(30)

R = 5.5 m

The normal force on the swing is calculated as follows;

[tex]Tcos(\theta) = mg\ \ --(1)[/tex]

The parallel force on the swing is calculated as follows;

[tex]Tsin(\theta) = \frac{mv^2}{R} \ --(2)[/tex]

divide (2) by (1);

[tex]\frac{Tsin(\theta)}{Tcos(\theta)} = \frac{mv^2}{Rmg} \\\\tan(\theta) = \frac{v^2}{Rg} \\\\v^2 = Rg tan(\theta)\\\\v = \sqrt{Rgtan(\theta)} \\\\v = \sqrt{5.5 \times 9.8 \times tan(30)} \\\\v = 5.58 \ m/s[/tex]

The time for one revolution of the swing is calculated as follows;

[tex]T = \frac{2\pi R}{v} \\\\T = \frac{2\pi \times 5.5}{5.58} \\\\T = 6.2 \ s[/tex]

(b) The angle of the rotation is given as;

[tex]tan(\theta) = \frac{v^2}{Rg}[/tex]

where;

  • v is the speed of the swing
  • R is the radius of the circular path

Thus, the angle does not depend on the weight of the passenger.

Learn more here:https://brainly.com/question/13493114