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A space station has a large ring-like component that rotates to simulategravity for the crew. This ring has a massM= 2.1×105kg and a radiusofR= 86.0 m and can be modeled as a thin hoop. Before spinning upthe ring section, crew members Dave and Frank dock their ships, eachwith massm= 3.5×104kg on two docking ports located on oppositesides of the center of the ring. The docking ports are locatedr= 31.0 mfrom the center of the ring. When the station’s computer begins to spinup the ring it has to spin both the ring and the ships. The ships canbe treated as point particles. Two identical thrusters on the edge of thering are used to spin up the ring with a constant angular acceleration.The ring-ship system takes 3 hours to reach the angular speed at which it simulates Earth’s gravity forthe crew on the edge of the ring. What constant force must each of the two thrusters apply to reach thisrotational speed?

Answer :

Olajidey

Answer:

Each thruster has to applied a force of 294.5N in tangential direction

Explanation:

Mass of the ring ,M =2.1×105kg

Mass of the ship ,m = 3.5×104kg

Radius of the ring R = 86.0 m

distance of ship from center of the ring, r =31.0 m

Let force applied by each thruster be F

Time taken to reach  gravity ,t = 3hrs = 3600× 3 =10800sec

The movement of ring make the object kept at the edge feel a force of centrifuge in outward direction.

Centrifugal force = weight of the object on earth

Assume the ring is moving with angular speed ω

Centripetalforce of the object kept at ring

m₁R ω²=m₁g  (m₁=mas of object)

⇒Rω² = g

⇒ω = √g/R

The ring start from 0 angular speed with constant angular acceleration

Let the constant angular acceleration be ∝

∝ = ω  / t

(ω = √g/R)

∝ = [tex]\frac{1}{t } \sqrt{\frac{g}{R} } }[/tex]

Consider Torque on the ring and ship system

T = FR + FR = 2FR

Moment of inertia of ring ship system

I = I(ring)+I(ship)+I(ship)

= MR² + mr² + mr² = MR² + 2mr²

angular acceleration of the ring ship system

∝ = [tex]\frac{2FR}{MR^2 + 2mr^2}[/tex]

Now we have ,

∝ = [tex]\frac{1}{t} \sqrt{\frac{g}{R} }[/tex]  , ∝ = [tex]\frac{2FR}{MR^2+2mr^2}[/tex]

equating both values

We have,

[tex]F = \frac{MR^2+2mr^2}{2Rt} \sqrt{\frac{g}{R} }[/tex]

where,

m = 2.1 × 10⁵kg, R = 86m, m = 3.5 × 10₄kg,

r = 31m, g = 9.8m/s² , t = 10800sec

[tex]F = \frac{(2.1 \times 10^5 \times86^2)+(2\times3.5v10^4\times31^2) }{2\times86\times10800} \times\sqrt{\frac{9.8}{86} } \\\\F = 294.47N\\\approx294.5N[/tex]

Each thruster has to applied a force of 294.5N in tangential direction

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