Answer to Question #226632 in Mechanical Engineering for Papi Chulo

Question

An endless belt 4 m long passes over a pulley of 300 mm diameter making 80 rev/minute. The pulley accelerates to 240 rev/minute in 40 seconds. If there is no belt slip, calculate:

7.1 the angular acceleration of the pulley and the linear acceleration of the belt.

7.2 the number of revolutions made by the pulley during its acceleration.

7.3 the number of times a point on the belt will pass over the pulley in 12 seconds.

Also clearly show the values of the following:

ω1

ω2

α

a

θ

ω2

θ

s

times

Answer 1

$l = 2πr =4\ m\\ r = 2/π \ m \\d = 300\ mm\\ r'= d/2 =150\ mm= 0.150\ m\\ \omega_o = 80\ rev/min = 80 ×2π/60 = 8π/3 \ rad/s\\ \omega_1 = 240\ rev/min = 240×2π/60 = 8π \ rad/s\\ t = 40s$

1.

$\omega_1 = \omega_o+ \alpha t\\ 8π= 8π/3 + \alpha (40)\\ \alpha= \dfrac{8π-8π/3}{40} =2π/15\ rad/s² \\ = 0.42\ rad/s²$

$a =r\ \alpha\\ a = \dfrac2π× \dfrac{2π}{15} = \dfrac{4}{15} =0.27\ ms^{-2}$

2.

$\theta = w_ot+ 1/2\alpha t² \\ \theta =8π(40)/3 + 2π×(40)²/30\\ \theta = 320π/3 + 320π/3\\ \theta = 640π/3\ rad\\ \theta =640π/(3×2π)=320/3\ rev = 106.67 \text{ revolutions}$

3.

$\text{number of times = number of revolutions}\\ \text{number of revolutions}=n\\ n = \omega_1t_1 = 240\ rev/min× 12/60 min= 48\text{ revolutions}$

$\therefore$ 48 times