A spinning top initially spins at 16rad/s but slows down to 12rad/s in 18s, due to friction. If the rotational inertia of the top is 0.0004kg.m2, by how much the angular momentum of the top changes?

Respuesta :

Answer:

The  change  in angular momentum is [tex]\Delta L = 0.0016 \ kgm^2/s[/tex]

Explanation:

From the question we are told that

      The initial angular velocity of the spinning top is  [tex]w_1 = 16 \ rad/s[/tex]

      The angular velocity after it slow down is  [tex]w_2 = 12 \ rad/s[/tex]

      The time  for it to slow down is  [tex]t = 18 \ s[/tex]

       The rotational inertia due to friction is  [tex]I = 0.0004 \ kg \cdot m^2[/tex]

 Generally the change in the angular momentum is  mathematically represented as  

         [tex]\Delta L = I *(w_1 -w_2)[/tex]

substituting values  

        [tex]\Delta L = 0.0004 *(16 -12)[/tex]

       [tex]\Delta L = 0.0016 \ kgm^2/s[/tex]