A uniform rod of mass `m=2kg` and length `l=0.5m` is sliding along two mutually perpendicular smooth walls with the two ends P and Q having velocities `U_(P)=4m//s` and `v_(Q)=3m//s` as shown then

Options

A. The angular velocity of rod, `omega=10rad//s` counter clockwise
B. The angular veloicty of rod `omega=5.0rad//s` counter clockwie
C. The velocity of centre of mass of rod `v_(cm)=2.5m//s`
D. The total kinetic energy of rod, `K=(25)/(3)"joule"`

Explanation

`x^(2)+y^(2)=l^(2)=(0.25)m^(2)`
`yomega=4` ..(i)
`xomega=3` ..(ii)
Squaring and adding we get
`(x^(2)+y^(2))omega^(2)=25`
or `(0.25)omega^(2)=25`
`thereforeomega=10rad//s`
`OC=(l)/(2)=0.25m`
`K_("total")=(1)/(2)mv_(cm)^(2)+(1)/(2)I_(cm)omega^(2)`
`=(1)/(2)xx2xx(0.25)^(2)+(1)/(2)((2xx0.25)/(12))(10)^(2)`
`=(25)/(3)J`