1.

In the above question angular velocity of the system after the particle sticks to the cylinder

Answer»

0.3 rad `s^(-1)`
5.3 rad `s^(-1)`
10.3 rad `s^(-1)`
8.3 rad `s^(-1)`

Solution :Here according to the LAW of conservation of angular MOMENTUM, we get
`L+mvR=(I+MR^(2))omega`
`THEREFORE omega=(L+mvR)/((I+mR^(2)))`
`=(0.12+0.5xx5xx0.2)/(((1)/(2)MR^(2)+MR^(2)))`
`=(0.12+0.5)/((1)/(2)xx2+0.5)xx0.4`
`=(0.62)/(1.5xx.04)=(0.62)/(0.06)`
`=10.33" rad "s^(-1)`


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