Question

A small child doing the hokey-pokey makes one complete rotation 2.7 seconds. If her elbow is located 21 cm from her axis of rotation, determine the linear speed of her elbow. If the child extends her arms, so that her moment of inertia increases to 0.838 kgm2, what will her new angular speed be?

Answer 1

Solution:

The initial angular velocity _i= 2/2.7 = 2.33 rad/s

a) The linear speed of her elbow,

v_i = r_i = 0.21*2.33

= 0.488 m/s

= 0.5 m/s

b)Let the initial moment of inertia be I_i = mr² = 0.2095 kg-m²

so, The final moment of inertia i₂ = m[2r]² = 0.838

4mr² = 0.838

mr² = 0.2095.

Using conservation of angular momentum,

I₁₁ = I₂₂

₂ = [0.2095/0.838]*2.33

= 0.5825 rad/s.

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