A torque of 160 N.m is applied to a body capable of rotating about a given axis. If the body starts from rest and acquires kinetic energy of 8000 J in 5 seconds, find (i) its moment of inertia about the given axis( ii) its angular momentum at the end of 5 seconds.

Explanation

Data : ` t = 160 N.m , omega_(i) = 0 , E_(1) = 0, E_(f) = 8 xx 10^(3) J , t = 5s`
` t = (DeltaL)/(Deltat) = (L_(r)-L_(i))/(Deltat)`
Since the body starts from rest, its initial angular momentum `,L_(i) = 0`
The final angular momentum.
` L_(f) = t Deltat = (160) (5) = 800 kg. m^(2)//s`
THe final rotational kinetic energy = ` E_(f) = 1/2 L_(f)omega_(f)`
` omega_(f) = (2E_(f))/L_(f) = (2xx (8xx10^(3)))/800 = 20 ` rad/s
The moment of inertia of the body,
` I = (I_(f))/(omega_(f)) = 800/20 = 40 kg.m^(2)`