Aplayground merry-go-round has a radius of 3.0 m and a rotational inertia of 600 kg. m2. it is initially spinning at 0.80 rad/s when a 20-kg child crawls from the center to the rim. when the child reaches the rim the angular velocity of the merry-go-round is:

The solution for this problem is:

Compute first for the initial Inertia:

Initial inertia = 600 + (20 * (0 ^ 2)) = 600 kg m^2 since the radius of the child is 0 originally 

Then look for the final inertia:

Final Inertia = 600 + (20 * (3^2)) = 780 kg m^2 

Now use conservation of angular momentum:

(I final) (W final) = (I initial) (w initial) = 780w = 600(0.80)

w = (600*0.80)/780 

So, final angular velocity:

W = 0.6154 rad/s

~0.62 rad/s

answer: at 24 february, 2014 the population gets twice of its initial population.

explanation:

since we have given that

the population for alpha city, t years after january 1, 2004 is given as

first we find out the initial population, i.e. t=0,

so, our quadratic equation becomes,

according to question, we have said that if the population has twice its initial population, then it becomes

so, time taken to reach this above mentioned population is given by

now, we know that time can't be negative so, we take t=9.149 years.

hence, at 24 february, 2014 the population gets twice of its initial population.

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