A runner of mass m = 59 kg and running at 2.6 m/s runs as shown and jumps on the

Question

A runner of mass m = 59 kg and running at 2.6 m/s runs as shown and jumps on the rim of a playground merry-go-round which has a moment of inertia of 395 kg m^2 and a radius of 2 meters. Assuming the merry-go-round is initially at rest, what is its final angular velocity to 3 decimal places? A 1-kg mass hangs by a string from a disk with radius 13.8 cm which has a rotational inertia of 5 Times 10^2 kg m^2. After it falls a distance of 0.8 meters, how fast is it going to the nearest hundredth of a m/s?

Explanation / Answer

1)    Here, Initial angular momentum = Final angular momentum

=> m * v * r = I * w

=> 59 * 2.6 * 2 = (395 + 59 * 22) * w

=> w   =   0.486 rad/sec

=> Final angular velocity = 0.486 rad/sec

2) Here, Torque = I * angular acceleration

=>    1 * 9.8 * 0.138 = 5 * 10-2 * angular acceleration

=> angular acceleration = 27.048 rad/sec2

=> linear acceleration = 0.138 * 27.048 = 3.732 m/sec2

=>    fast it is going = sqrt(2 * 3.732 * 0.8)

                             =   2.443 m/sec

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