Question

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?

Answer 1

Answer:

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

Explanation:

From the question we are told that

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

      The angular velocity after it slow down is  $w_2 = 12 \ rad/s$

      The time  for it to slow down is  $t = 18 \ s$

       The rotational inertia due to friction is  $I = 0.0004 \ kg \cdot m^2$

 Generally the change in the angular momentum is  mathematically represented as  

         $\Delta L = I *(w_1 -w_2)$

substituting values  

        $\Delta L = 0.0004 *(16 -12)$

       $\Delta L = 0.0016 \ kgm^2/s$