The wheel on an upside-down bicycle moves through 17.3 rad in 5.41 s. What is the wheel’s angular acceleration if its initial angular speed is 2.1 rad/s? Answer in units of rad/s 2 .

0.406 rad/s^2.

Explanation:

Equating equations of linear motion to angular motion;

ω = θ/t

ωf^2 = ωi^2 + 2a*θ

θ = ωi*t + (1/2)*a*t^2

where,

ωi = initial angular velocity (rad/s),

ωf = final angular velocity (rad/s),

θ = angular distance (rad),

a = acceleration due to gravity (rad/s^2),

and t = time taken for motion.

θ = ωi*t + (1/2)*a*t^2,

17.3 = (2.1*5.41) +(1/2)*a*(5.41)^2

17.3 = 11.361 + 14.634*a

17.3 - 11.361 = 14.634*a

14.634*a = 5.939

a = 5.939/14.634

= 0.406 rad/s^2.

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