Ask Question
28 August, 21:16

A 4.2-m-diameter merry-go-round is rotating freely with an angular velocity of 0.80 rad? s. Its total moment of inertia is 1360 kg? m2. Four people standing on the ground, each of mass 65 kg, suddenly step onto the edge of the merry-go-round. (a) What is the angular velocity of the merry-go-round now? (b) What if the people were on it initially and then jumped off in a radial direction (relative to the merry-go-round)

+2
Answers (1)
  1. 29 August, 01:11
    0
    A) 0.44 rad/s

    B) 3.091 rad/s

    Explanation:

    Initial angular momentum must be equal to final angular momentum.

    W1 = 0.80 rad/s

    I = 1360 kgm^2

    Initial moment Iw = 0.8 x 1360 = 1088 rad-kg-m^2/s

    For second case

    I = (65 x 4) 4.2 + 1360 = 2452 kg-m2-rad/s

    New momentum = 2452 x W2

    Where w2 = final angular momentum

    = 2452W2

    Equating both moment,

    1088 = 2452W2

    W2 = 0.44 rad/s0

    If they were on it before,

    Initial momentum

    I = 1360 - (60 x 4) 4.2

    = 352 rad-kg-m^2/s

    352W1 = 1088

    W1 = 3.091 rad/s
Know the Answer?
Not Sure About the Answer?
Get an answer to your question ✅ “A 4.2-m-diameter merry-go-round is rotating freely with an angular velocity of 0.80 rad? s. Its total moment of inertia is 1360 kg? m2. ...” 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