Answer to Question #226632 in Mechanical Engineering for Papi Chulo

Question #226632

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

1
Expert's answer
2021-08-20T01:36:42-0400

"l = 2\u03c0r =4\\ m\\\\\nr = 2\/\u03c0 \\ m\n\n\\\\d = 300\\ mm\\\\\nr'= d\/2 =150\\ mm= 0.150\\ m\\\\\n\\omega_o = 80\\ rev\/min = 80 \u00d72\u03c0\/60 = 8\u03c0\/3 \\ rad\/s\\\\\n\\omega_1 = 240\\ rev\/min = 240\u00d72\u03c0\/60 = 8\u03c0 \\ rad\/s\\\\ t = 40s"


1.

"\\omega_1 = \\omega_o+ \\alpha t\\\\\n8\u03c0= 8\u03c0\/3 + \\alpha (40)\\\\\n\\alpha= \\dfrac{8\u03c0-8\u03c0\/3}{40} =2\u03c0\/15\\ rad\/s\u00b2 \\\\\n= 0.42\\ rad\/s\u00b2"


"a =r\\ \\alpha\\\\\na = \\dfrac2\u03c0\u00d7 \\dfrac{2\u03c0}{15} = \\dfrac{4}{15} =0.27\\ ms^{-2}"


2.

"\\theta = w_ot+ 1\/2\\alpha t\u00b2\n\\\\ \\theta =8\u03c0(40)\/3 + 2\u03c0\u00d7(40)\u00b2\/30\\\\\n\\theta = 320\u03c0\/3 + 320\u03c0\/3\\\\\n\\theta = 640\u03c0\/3\\ rad\\\\\n\\theta =640\u03c0\/(3\u00d72\u03c0)=320\/3\\ rev = 106.67 \\text{ revolutions}"


3.

"\\text{number of times = number of revolutions}\\\\\n\\text{number of revolutions}=n\\\\\nn = \\omega_1t_1 = 240\\ rev\/min\u00d7 12\/60 min= 48\\text{ revolutions}"

"\\therefore" 48 times





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