Ask Question
13 September, 20:30

Three children are riding on the edge of a merry-go-round that is 100 kg, has a 1.60-m radius, and is spinning at 20.0 rpm. The children have masses of 22.0, 28.0, and 33.0 kg. If the child who has a mass of 28.0 kg moves to the center of the merry-go-round, what is the new angular velocity in rpm?

+5
Answers (1)
  1. 13 September, 21:40
    0
    25.33 rpm

    Explanation:

    M = 100 kg

    m1 = 22 kg

    m2 = 28 kg

    m3 = 33 kg

    r = 1.60 m

    f = 20 rpm

    Let the new angular speed in rpm is f'.

    According to the law of conservation of angular momentum, when no external torque is applied, then the angular momentum of the system remains constant.

    Initial angular momentum = final angular momentum

    (1/2 x M x r^2 + m1 x r^2 + m2 x r^2 + m3 x r^2) x ω =

    (1/2 x M x r^2 + m1 x r^2 + m3 x r^2) x ω'

    (1/2 M + m1 + m2 + m3) x 2 x π x f = (1/2 M + m1 + m3) x 2 x π x f'

    (1/2 x 100 + 22 + 28 + 33) x 20 = (1/2 x 100 + 22 + 33) x f'

    2660 = 105 x f'

    f' = 25.33 rpm
Know the Answer?
Not Sure About the Answer?
Get an answer to your question ✅ “Three children are riding on the edge of a merry-go-round that is 100 kg, has a 1.60-m radius, and is spinning at 20.0 rpm. The children ...” in 📙 Physics if there is no answer or all answers are wrong, use a search bar and try to find the answer among similar questions.
Search for Other Answers