00:01
Here in this question we are given t1 is equal to 1 ,600 plus 300 sine 2 theta newton meter and t2 is equal to 1 ,600 plus 170 sine theta newton meter.
00:24
The mass of rotating parts is equal to 240 kilogram.
00:29
Radius of gyration k is equal to 0 .5 meter maximum speed which is omega max is equal to 200 rpm or this is equal to 2 pi into 200 by 60 which is equal to 20 .94 radian per second square now we know that sine theta is equal to period is 2 pi and sine 2 theta is equal to here the period is pi therefore km will be equal to 2 pi and work done will be equal to integral t d theta or this is equal to t mean into theta or this is equal to t mean into theta or this is is equal to 1 ,600 into 2 pi.
01:33
Now in the first part we need to calculate the minimum speed of rotation.
01:42
So before that we need to calculate the work done.
01:47
Work done is equal to the equation is half into i into omega square max minus omega square minimum.
01:58
Here work done we already know which is 1 ,600 into 2.
02:02
2 pi into upon taking half to the other side it becomes 2 will be equal to here i is 240 into 0 .5 square that is m k square i is equal to moment of inertia will be equal to mass into radius of gyration it's a square this is the expression for i into here omega square max will be to omega maxis 20 .94 its square minus the minimum value of omega its square which is to be calculated.
02:37
Upon solving we will get the minimum speed of rotation is equal to 10 .10 radian per second.
02:46
So this is the value of minimum speed of rotation or converting it into rpm we get a omega minimum is equal to 10 into 60 divided by 2 pi and this turns out to be 96 .5 rpm...