A uniform rod of mass `M` and length `L` is free to rotate in `X-Z` plane i.e. `vecF=(3hati+2hatj+6hatk)N` is acting on the rod at `(L//2, 0,0)` in the situtation shown in figure. The angular acceleration of the rod is (Take `M=6 kg` and `L=4m`)

Options

A. `-3/2hatj+1/2hatk`
B. `-3/2hatj`
C. `1/2k`
D. `4j`

Answer

B

Explanation

`I=(ML^(2))/12=(6xx4xx4)/12=8kgm^(2)`
From `vectau=vecrxxvecLF=[2hatixx(3hati+2hatj+6hatk)]=4hatk-12hatj`
`tau_=Ialphaimplies-12j=8alphaimpliesalpha=-3/2jrad//s^(2)`