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Sav [38]
3 years ago
11

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 to

p is 0.0004kg.m2, by how much the angular momentum of the top changes?
Physics
1 answer:
Alex Ar [27]3 years ago
5 0

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

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