00:01
Given that diameter of hole d is equal to 25 mm, thickness of hole p is equal to 19 mm.
00:10
Now area of carried hole pi dt is equal to pi multiply 25 multiply 19 mm square.
00:25
Here for p is equal to so multiply pi dt.
00:32
Now this is equal to 400 newton upon mm square multiply pi multiply 25 multiply 19 meter square.
00:47
Now this is now average energy used average energy used is equal to 1 divided by 2 multiply p multiply p.
01:01
Now e is equal to 1 divided by 2 multiply 400 multiply pi multiply 25 multiply 19 multiply 19 divided by 100 joule.
01:18
Now solve this we will get e is equal to e is equal to 1805 pi joule pi joule.
01:28
Now energy offered by flywheel is equal to pi omega square cs.
01:37
Now delta e max is equal to 0 .9 mr square 0 .9 mr square multiply 2 pi n divide by 60 square cs.
01:59
Now delta e max delta e max is equal to 0 .95 m multiply 0 .5 square multiply 2 pi multiply 200 divided by 6030 divided by 60 square now multiply 0 .1.
02:22
Now 0 .9 multiply m multiply 0 .25 multiply 400 pi square divided by 9 multiply 0 .1 this is equal to 1805 pi.
02:41
Now m is equal to m is equal to m is equal to 1805 pi multiply 2 multiply 9 divided by 0 .9 multiply 0 .25 multiply 400 multiply pi multiply 0 .1.
03:11
Now solve this we will get m is equal to 574 .4 kg.
03:21
Now this is approximate to 569 kg for 1 % error is allowed...