A pulley is turning at 30 revolutions per second slows down uniformly to 20 revolutions per second in 2 seconds .
Calculate the number of revolutions it make to a complete stop.
ω=ω0−ϵt→ϵ=(ω0−ω)/t=(2πν0−2πν)/t=\omega=\omega_0-\epsilon t \to \epsilon=(\omega_0-\omega)/t=(2\pi\nu_0-2\pi\nu)/t=ω=ω0−ϵt→ϵ=(ω0−ω)/t=(2πν0−2πν)/t=
=(2⋅3.14⋅30−2⋅3.14⋅20)/2=31.4 (rad/s2)=(2\cdot3.14\cdot30-2\cdot3.14\cdot20)/2=31.4\ (rad/s^2)=(2⋅3.14⋅30−2⋅3.14⋅20)/2=31.4 (rad/s2)
ϕ=ω02/(2ϵ)=(2⋅3.14⋅30)2/(2⋅31.4)=565.2 (rad)\phi=\omega_0^2/(2\epsilon)=(2\cdot3.14\cdot30)^2/(2\cdot31.4)=565.2\ (rad)ϕ=ω02/(2ϵ)=(2⋅3.14⋅30)2/(2⋅31.4)=565.2 (rad)
N=ϕ/(2π)=565.2/(2⋅3.14)=90 (rev)N=\phi/(2\pi)=565.2/(2\cdot3.14)=90\ (rev)N=ϕ/(2π)=565.2/(2⋅3.14)=90 (rev). Answer
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