Problem 8 - Entrance Test

A child of mass m is standing at the edge of a merry-go-round (a uniform disk) of mass M and radius R, which is initially rotating with an angular speed ω₀. The child then walks radially inward until they are at a distance R/2 from the center. Assume the child can be treated as a point mass. What is the change in the total rotational kinetic energy of the system during this process?

A. A) (3m R² ω₀² (M + 2m)) / (4 (2M + m))B. B) (3m R² ω₀² (M + 2m)) / (2 (2M + m))C. C) (m R² ω₀² (M + 2m)) / (4 (2M + m))D. D) (3m R² ω₀² (M + 2m)) / (8 (2M + m))E. E) 0, because angular momentum is conserved.

Correct: A

This problem involves the conservation of angular momentum but not the conservation of rotational kinetic energy because the child does work by walking inward (internal forces do work). 1. **Calculate Initial Moment of Inertia (I_initial):** The merry-go-round is a uniform disk: I_MGR = (1/2)MR². The child is a point mass at the edge (r=R): I_child_initial = mR². I_initial = I_MGR + I_child_initial = (1/2)MR² + mR² = ((1/2)M + m)R². 2. **Calculate Final Moment of Inertia (I_final):** The child walks to a distance R/2 from the center: I_child_final = m(R/2)² = mR²/4. I_final = I_MGR + I_child_final = (1/2)MR² + mR²/4 = ((1/2)M + m/4)R². 3. **Apply Conservation of Angular Momentum (L_initial = L_final):** L_initial = I_initial * ω₀ L_final = I_final * ω_final I_initial * ω₀ = I_final * ω_final ((1/2)M + m)R² * ω₀ = ((1/2)M + m/4)R² * ω_final ω_final = ω₀ * [ ((1/2)M + m) / ((1/2)M + m/4) ] 4. **Calculate Initial Rotational Kinetic Energy (KE_initial):** KE_initial = (1/2)I_initial * ω₀² = (1/2) * ((1/2)M + m)R² * ω₀². 5. **Calculate Final Rotational Kinetic Energy (KE_final):** KE_final = (1/2)I_final * ω_final² = (1/2) * ((1/2)M + m/4)R² * (ω₀ * [ ((1/2)M + m) / ((1/2)M + m/4) ])² KE_final = (1/2) * ((1/2)M + m/4)R² * ω₀² * [ ((1/2)M + m)² / ((1/2)M + m/4)² ] KE_final = (1/2)R²ω₀² * [ ((1/2)M + m)² / ((1/2)M + m/4) ] 6. **Calculate the Change in Rotational Kinetic Energy (ΔKE):** ΔKE = KE_final - KE_initial ΔKE = (1/2)R²ω₀² * [ ( ((1/2)M + m)² / ((1/2)M + m/4) ) - ((1/2)M + m) ] Let A = (1/2)M + m and B = (1/2)M + m/4. ΔKE = (1/2)R²ω₀² * [ A²/B - A ] ΔKE = (1/2)R²ω₀² * A * [ A/B - 1 ] ΔKE = (1/2)R²ω₀² * A * [ (A - B) / B ] Now, substitute A and B back: A = (M + 2m)/2 B = (2M + m)/4 A - B = (M + 2m)/2 - (2M + m)/4 = (2M + 4m)/4 - (2M + m)/4 = (3m)/4 ΔKE = (1/2)R²ω₀² * ( (M + 2m)/2 ) * ( (3m/4) / ( (2M + m)/4 ) ) ΔKE = (1/2)R²ω₀² * ( (M + 2m)/2 ) * ( 3m / (2M + m) ) ΔKE = (3m R² ω₀² (M + 2m)) / (4 * (2M + m)) The correct choice is A. Note that since ΔKE > 0, the kinetic energy of the system increases. This is because the child does positive work by walking inward against the 'centrifugal' force.