Question #197173

A wheel turning at 50rev/min has its speed increased to 65rev/min. If the increase occurred

in 20s, find the

(a) rate at which the angular speed changes.

(b) angular displacement during this time in

(i)

(ii)

(iii) radians

revolutions

degrees


1
Expert's answer
2021-05-23T16:34:07-0400

(a) We can find the angular acceleration from the kinematic equation:


ω=ω0+αt,\omega=\omega_0+\alpha t,α=ωω0t,\alpha=\dfrac{\omega-\omega_0}{t},α=(65 revmin50 revmin)1 min60 s2π rad1 rev20 s=0.078 rads2.\alpha=\dfrac{(65\ \dfrac{rev}{min}-50\ \dfrac{rev}{min})\cdot\dfrac{1\ min}{60\ s}\cdot\dfrac{2\pi\ rad}{1\ rev}}{20\ s}=0.078\ \dfrac{rad}{s^2}.

(b) We can find the angular displacement during this time from the kinematic equation:


θ=ω0t+12αt2.\theta=\omega_0t+\dfrac{1}{2}\alpha t^2.

(i)

θ=50 revmin1 min60 s2π rad1 rev20 s+120.078 rads2(20 s)2=120 rad.\theta=50\ \dfrac{rev}{min}\cdot\dfrac{1\ min}{60\ s}\cdot\dfrac{2\pi\ rad}{1\ rev}\cdot20\ s+\dfrac{1}{2}\cdot0.078\ \dfrac{rad}{s^2}\cdot(20\ s)^2=120\ rad.

(ii)

θ=120 rad1 rev2π rad=19 rev.\theta=120\ rad\cdot\dfrac{1\ rev}{2\pi\ rad}=19\ rev.


(iii)

θ=120 rad180π rad=6875.\theta=120\ rad\cdot\dfrac{180^{\circ}}{\pi\ rad}=6875^{\circ}.

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