Answer to Question #226629 in Mechanical Engineering for Papi Chulo

Question

1.1 The speed of a shaft increases from 93 r/min to 305 r/min with an angular acceleration of 2,4 rad/s2. How many revolutions does the shaft make during the first 3 seconds of acceleration? How many revolutions does the shaft make in the last 4.25seconds. Show all calculations[10]

1.2 A motor that runs at 1240 rev/minute is required to drive a belt at 20 m/s. What diameter belt pulley is required?[3]

Also clearly show the following values:

ωi

ωf

θ

r

d

ωf

Answer 1

1.1 Let us find out how many revolutions does the shaft make during the first 3 seconds of acceleration:

$$n(t)=n_it+\frac{\alpha t^2}{2},\\\space\\ n_f=n(3)=(93/60)3+\frac{(2.4/(2\pi)) ·3^2}{2}=6.$$

Now, find the total time required to increase the speed from 93 to 305 r/min:

$$n(T)=n_0T+\frac{\alpha T^2}{2},\\\space\\ n(T)=(305/60)\text{ r/s},\\\space\\ (305/60)=(93/60)T+\frac{(2.4/(2\pi))·T^2}{2},\\\space\\ 0.191T^2+1.55T-5.083=0,\\\space\\ T=\frac{-1.55+\sqrt{1.55^2-4·0.191·(-5.083)}}{2·0.191}=5.013\text{ s}.$$

Therefore, the time before the last 4.25 s is

$$\tau=5.013-4.25=0.76\text{ s}.$$

How many revolutions during the 0.76 s:

$$n(0.76)=(93/60)0.76+\frac{(2.4/(2\pi)) ·0.76^2}{2}=1.$$

How many revolutions during the 5.013 s:

$$n(5.013)=(93/60)5.013+\frac{(2.4/(2\pi)) ·5.013^2}{2}=12.\\\space\\ n_\text{last 4.25 s}=n(5.013)-n(0.76)=11.$$

1.2 Find the angular velocity:

$$\omega=2\pi n.$$

The velocity:

$$v=\omega R=2\pi n R,\\\space\\ R=\frac{v}{2\pi n}=\frac{20}{2\pi·(1240/60)}=0.15\text{ m}.$$