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harkovskaia [24]
3 years ago
10

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 t

hen 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
Physics
2 answers:
DedPeter [7]3 years ago
5 0

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}

neonofarm [45]3 years ago
3 0

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.

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