Question #11400124
What would be the linear displacement of the tire's axle as it moves forward from t=0sec to t=15 sec?
Starting from rest at t=0sec, a tire on a mountain bike has a constant angular acceleration. When t= 5 sec, the angular velocity of the wheel is 6.5 rad/sec. The angular acceleration continues until t= 15 sec, after which the angular velocity remains constant. The tire has a radius of 22 cm. Angular velocity (final) and acceleration were found through out this problem: ωf = 19.5 rad/s α = 1.3 rad/s^2 Along with total angular displacement: Θ = 146.25 rad And total displacement: Δs = 32.175 m
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linear displacement tire 039 s forward t 0sec Starting t 0sec mountain constant angular acceleration angular velocity rad sec angular acceleration continues angular velocity remains constant radius cm
Angular velocity final acceleration through problem
ωf rad s
α rad s 2
Along angular displacement
Θ 146 25 rad
And displacement
Δs 32 175