Find the potential energy of a 2 kg ball 15 m in the air.
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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)
∝ =
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
∝ =
Now we have ,
∝ = , ∝ =
equating both values
We have,
where,
m = 2.1 × 10⁵kg, R = 86m, m = 3.5 × 10₄kg,
r = 31m, g = 9.8m/s² , t = 10800sec
Each thruster has to applied a force of 294.5N in tangential direction