A bicycle has wheels with a diameter of 0.630 m. It accelerates uniformly and the rate of rotation of its wheels increases from 225 rpm to 260 rpm in a time of 28.7 s. Find the linear acceleration of the bicycle.

the linear acceleration of the bicycle is 0.0403m/s²

Explanation:

First convert the angular given speeds to radians per second :  

ω₀ = 225rev/min×(2πrad/rev)×(1.00min/60.0sec)

= 23.56 rad/s

ω =  269rev/min×(2πrad/rev)×(1.00min/60.0sec)

= 27.23 rad/s

The angular acceleration is needed to find the tangential (linear) acceleration, it is :  

ω = ω₀ + αt

α = (ω - ω₀) / t

= (27.23rad/s - 23.56rad/s) / 28.7s

= 0.128rad/s²

The linear acceleration is :

linear acceleration = angular acceleration * r

a = rα

r = d/2

r = 0.630/2 = 0.315

a = rα

= (0.315m)(0.128rad/s²)

= 0.0403m/s²

Hence, the linear acceleration of the bicycle is 0.0403m/s²

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