A uniform square plate of mass 2.0 kg and edge 10 cm rotates about one of its diagonals under the action of a constant torque of 0.10 Nm. Calculate the angular momentum and the kinetic energy of the plate at the end of the fifth second after the start.

Answer

B

Explanation

`I=0.10N, a=10cm=0.1m,m=2kg`
Therefore `(ma^2)/12)xxa=0.1N-m`
`rarr alpha=60rad/s`

Therefore `omega=omega_0+at`
`rarr omega=60xx5=300rad/s
Therefore angular mometum
`=Iomega=(0.10/60)xx300`
`=0.50kg-m^2/s`
`and kinetic energy `=1/2x(0.10/60)xx300^2`
=27jule