Question

u are holding the axle of a bicycle wheel with radius30 cm and mass 1.05 kg. You get the wheel spinning at arate of 77 rpm and then stop it by pressing the tire againstthe pavement. You notice that it takes 1.37 s for the wheelto come to a complete stop. What is the angular accelerationof the wheel

Answer 1

Answer:

-5.9 rad/s^{2}

Explanation:

radius (r) = 30 cm = 0.3 m

mass (m) = 1.05 kg

initial speed (u) = 77 rpm

final speed (v) = 0 rpm

time (t) = 1.37 s

angular acceleration =$\frac{(final speed-initial speed)rad/s}{time}$

therefore

initial speed (U) = 77 rpm = 77 x (2π/60) = 8.06 rad/s

final speed (v) = 0 rpm = 0 rad/s

angular acceleration = $\frac{0-8.06}{1.37}$ = -5.9 rad/s^{2}

Answer 2

Answer:

-5.886 rad/s^2.

Explanation:

radius, r = 30 cm

= 0.3 m

mass, m

= 1.05 kg

initial speed, wo = 77 rpm

Converting from rpm to rad/s,

= 77 rpm * 2pi rad * 1 min/60 s

= 8.063 rad/s

final speed, wi = 0 rad/s

time, t = 1.37 s

angular acceleration = Δw/Δt

= (wi - wo)/t

= 8.063/1.37

= -5.886 rad/s^2.