Question

How much constant force should be applied tangential to equator of the earth to stop its rotation in one day?
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Answer 1

Answer:

Explanation: τ = Iα

τ = F ∙ R

hence FR = Iα

F = [(Iα) / R]                        

(1) for earth,

I = (2/5)MR2  ω1 = 2π rad/day

ω2 = 0.....earth comes to rest

T = 1 day = 86400 sec

As   α = (dω / dt)

= [(2π – 0) / (86400)2]

= [(2π) / (86400)2] ------------ [α = [{(ω2 – ω1) / (t2 – t1)} = (2π / T2)]

from (1)

F = (α/R) ∙ (2/5)MR2­­ = (2/5)MR ∙ α

F = (2/5) × [(2π) / (86400)2] × 6 × 1024 × (6400 × 103)­----Radius = 6400 km

mass = 6 × 1024 kg

F = 1.3 × 1022 N

hope it helped you.