Answer to Question #186450 in Mechanical Engineering for Deepanshi singh

Question #186450

A flywheel rotates with a constant retardation due to breaking from t=0to t=10 seconds. it made 400 revolutions . At time t=8 sec , it's angular velocity was 45πrad/sec. Determine

1 . Value of constant ratardation.

2. Total time taken to come to rest.

3. Total revolution made till it comes to rest.


1
Expert's answer
2021-05-07T08:12:27-0400

"t_o = 0s\\\\\n\\omega_o= ?\\\\\nt_1 = 8s \\\\\n\\omega_1 = 40\u03c0 rad\/s\\\\\nt_2 = 10s\\\\\n\\omega_2 = ?"


Retardation is constant throughout,

"\\therefore \\alpha = \\dfrac{\\omega_1- \\omega_o}{t_1- t_o}=\\dfrac{\\omega_2- \\omega _1}{t_2- t_1}"


"\\dfrac{40\u03c0-\\omega_o}{8} = \\dfrac{\\omega_2 - 40\u03c0}{2}"


"\\dfrac{40\u03c0-\\omega_o}{4} = {\\omega_2 - 40\u03c0}"


"40\u03c0-\\omega_o= 4{\\omega_2 - 160\u03c0}\\\\\n4\\omega_2+ \\omega_o=200\u03c0---(i)"



"\\theta =\\dfrac{\\omega_o +\\omega_2}2\u00d7 t\\\\"


"400(2\u03c0) =\\dfrac{\\omega_o +\\omega_2}2\u00d7 10"


"800\u03c0 = \\dfrac{\\omega_o +\\omega_2}2\u00d7 10"


"160\u03c0= \\omega_o +\\omega_2---(ii)"


Comparing equation (i) with equation (ii),

"\\omega_o = 461\\ rad\/s, \\quad \\omega_2= 52\\ rad\/s"


"\\theta= \\omega_o t + 0.5\\alpha t\u00b2\\\\\n800\u03c0 =461(10) + 0.5\\alpha(10)^ 2"


"\\alpha = -42\\ rad\/s\u00b2"

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