Please I want some to do answer it clearly and step by step A 30-kg child runs a
ID: 1555457 • Letter: P
Question
Please I want some to do answer it clearly and step by step
A 30-kg child runs around the edge of a playground merry-go-round. The merry-go-round has a mass of 200 kg and a radius of 2.0 m. Consider it to be a uniform solid disk for which I = 1/2 mr^2. The child is observed to be running, relative to the Earth, at a speed of 3.0 m/s while the merry-go-round rotates in the opposite direction making one revolution every 12 Sec. (a) If the child stops running, what is the angular speed of the child plus merry- go-round combination? (b) How much work is done as the child stops running?Explanation / Answer
a) Using law of conservation of angular momentum
Initial Angular momentum = Final angular momentum
initial angular velocity of the meery go round is w1= 2*3.142/12 = 0.523 rad/sec
Initial angular momentum is Li =m*v*r + (I1*w1) = (30*3*2)-(0.5*200*2^2*0.523) =-29.2 Kg m^2/sec
Final angular momentum is Lf= (I2*w2) = ((m*r^2)+I1)*w2 = ((30*2^2)+(0.5*200*2^2))*w2
29.2 = ((30*2^2)+(0.5*200*2^2))*w2
w2 = 0.0562 rad/sec
b) work done is W = Kf-Ki
Kf = (0.5*I*w^2) = 0.5*((30*2^2)+(0.5*200*2^2))*0.0562^2
Kf = 0.821 J
Ki = 0.5*I*w^2 + 0.5*m*v^2 = (0.5*0.5*200*2^2*(2*3.142/12)^2) + (0.5*30*3^2) = 189.84 J
W = Kf-Ki = 189.84-0.821 = 189.019 J