00:01
Here in this given problem initial spinning rate of the ice skater that is f1 is equal to 6 .8 revolutions per second.
00:30
His initial moment of inertia rotational inertia that is i1 is equal to 0 .24 kilogram into meter square.
00:47
In the first part of the problem we have to find his angular momentum which is found using the expression l is equal to i into omega.
01:01
So that will be i1 omega 1 omega that is the angular velocity and this is given 2 pi f.
01:10
So that is 2 pi multiplied by f1.
01:14
So plugging in the known values for i1 this is 0 .24 multiplied by 2 times of 3 .14 multiplied by f1 which is 6 .8 and this angular momentum is calculated to be equal to 10 .25 kilogram meter square per second.
01:38
Answer for the first part of this given problem here.
01:44
Now in the second part of the problem new rate of spinning when skater outstretched his hands.
02:04
Now the rate of spinning is reduced to 1 .25 revolutions per second.
02:14
But as there is no external torque acting on him so angular momentum will be conserved l remain conserved.
02:55
So using the same expression l is equal to i2 omega 2 or we can say this is i2 multiplied by 2 pi f2.
03:09
So i2 means new moment of inertia of the person of the skater when he outstretched his hands...