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

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

Step-by-step 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`