A 30 kg child is on a merry go round. The radius of the merry go round is 2.9m.

Question

A 30 kg child is on a merry go round. The radius of the merry go round is 2.9m. The inertia of the merry go round is 450 kg*m^2. The child starts on a pony that is.55 m from the center of the merry go round but then sees a cool dinosaur that is 2.75 m from the center goes to it. (The inertia due to the child is I = mR^2) Find the total inertia while the child is at the pony What is the inertia when the child is at the dinosaur If the Merry go round has an angular velocity of 60 rpm when the child is at the port what would be the new angular acceleration in rad/s when the child is at the dinosaur

Explanation / Answer

Moment of Inertia of child when he is at pony = MR^2 R= distance from center to pony

A)Total Moment of Inertia =

I1= MI of Merry Go ROund + MI of child at Pony = 450 + 30 * (0.55)^2 = 459.075 kg m^2

Moment of Inertia of child when he is at Dinasour = MD^2 D= distance from center to Dinasour

B)Total Moment of Inertia

I2= MI of Merry Go ROund + MI of child at Dinasour = 450 + 30 * (2.75)^2 = 676.875 kg m^2

C)As no external force acts on the system there is a conservation of angular momentum
so Angular Momentum at Pony = Angular Momentum at Dinasour

I1w1 = I2w2

459.075 * 60 = 676.875 *w2
w2 = 40.69 rpm

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