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Answer :
Final answer:
The new angular speed when the man walks to a point 1m from the center of the merry-go-round is approximately 0.174 rad/s.
Explanation:
To find the new angular speed when the man walks to a point 1m from the center, we can use the principle of conservation of angular momentum. The initial angular momentum of the merry-go-round is equal to the final angular momentum after the man walks to the new position.
The initial moment of inertia (I1) of the merry-go-round with the man at a distance of 2.9m from the axis of rotation is given by:
I1 = (1/2) * m1 * r1^2
Where m1 is the mass of the merry-go-round and r1 is the initial distance of the man from the axis of rotation.
The final moment of inertia (I2) of the merry-go-round with the man at a distance of 1m from the center is given by:
I2 = I_merry-go-round + m2 * r2^2
Where I_merry-go-round is the moment of inertia of the merry-go-round, m2 is the mass of the man, and r2 is the new distance of the man from the axis of rotation.
Since the initial angular momentum is equal to the final angular momentum, we can write:
I1 * w1 = I2 * w2
Where w1 is the initial angular speed (0.42 rev/s) and w2 is the final angular speed (unknown).
Solving for w2, we get:
w2 = (I1 * w1) / I2
Substituting the values, we get:
w2 = ((1/2) * m1 * r1^2 * w1) / (I_merry-go-round + m2 * r2^2)
Calculating the expression, we have:
w2 = ((1/2) * 68kg * (2.9m)^2 * 0.42 rev/s) / (68kg * (2.9m)^2 + 93kg * (1m)^2)
w2 ≈ 0.174 rad/s
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