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KE= W v v rev s m s v π (1 m) rev s m s KE= m s KE=3646.270515 J =3.646270515 KJ

The document is a math problem solving the kinetic energy needed to punch a 350kg flywheel when its speed varies from 200rpm to 180rpm with a 1m diameter. It provides the formula, substitutions, and shows the calculation that the kinetic energy is 3.63KJ.

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162 views1 page

KE= W v v rev s m s v π (1 m) rev s m s KE= m s KE=3646.270515 J =3.646270515 KJ

The document is a math problem solving the kinetic energy needed to punch a 350kg flywheel when its speed varies from 200rpm to 180rpm with a 1m diameter. It provides the formula, substitutions, and shows the calculation that the kinetic energy is 3.63KJ.

Uploaded by

GodisGood AlltheTime
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
Available Formats
Download as DOCX, PDF, TXT or read online on Scribd
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Perez, Mhon Aldrin H.

ME – 5202

Seatwork No.7

Problem No. 35

During punching process of 350 kg flywheel, the speed varies from 200 rpm to
180 rpm with 1 m mean diameter. Determine the kinetic energy needed.

A. 3.63 KJ B. 6.28 KJ C. 4.51 KJ D. 5.62 KJ

Given:

m = 350 kg

N1 = 200 rpm

N2 = 180 rpm

Dm = 1m

Required:

Kinetic energy

Solution:

Wf 2 2
KE= ( v −v )
2g 2 1

v1 =π ( 1 m ) ( 200 rev
60 s ) =10.47197551
m
s

v 2=π ( 1 m ) ( 180 rev


60 s ) =9.424777961
m
s

350 kg m
KE= (10.471975512−9.4247779612 )
2 s

KE=3646.270515 J =3.646270515 KJ ans. A.

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