Question
A bicycle has wheels with a diameter of 0.630 m. It accelerates uniformly and the rate of rotation of its wheels increases from 197 rpm to 300 rpm in a time of 22.5 s. Find the linear acceleration of the bicycle
Answers
R=v₀•t
t=R/v₀
197 rpm = 197/60 =3.28 rev/s
300 = 5 rev/s
ω=ω₀+εt
ε= (ω-ω₀)/t =(5-3.28)/22.5=0.0076 rad/s
a = ε•R=0.0076•0.63=0.0048 m/s²
t=R/v₀
197 rpm = 197/60 =3.28 rev/s
300 = 5 rev/s
ω=ω₀+εt
ε= (ω-ω₀)/t =(5-3.28)/22.5=0.0076 rad/s
a = ε•R=0.0076•0.63=0.0048 m/s²
you forgot to convert form from frequency to angular velocity where
197rpm=3.28rev/s*2pi
apart from that the steps are accurate
197rpm=3.28rev/s*2pi
apart from that the steps are accurate
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