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

Options

A. 0.3 rad `s^(-1)`
B. 5.3 rad `s^(-1)`
C. 10.3 rad `s^(-1)`
D. 8.3 rad `s^(-1)`

Answer

C

Explanation

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)`