The Yo-Yo — Question 3

Section 11-4 · Physics pages 1616–1617

Question 3 Changing axle radius alters acceleration without changing body inertia Two yo-yos have the same mass M and inertia I = 4Mr2. Their axle radii are r and 2r. Each descends from rest on a fixed cord through the same height h. r 2r same M and I The two axle radii differ while body mass and rotational inertia remain the same. (a) Compare their downward accelerations. (b) Compare descent times and final speeds.
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Question 3 - Solution Changing axle radius alters acceleration without changing body inertia r 2r same M and I (a) Derivation and calculation For axle radius ra, the fixed-cord acceleration is a = g 1 + I/(Mr2a). Since the same inertia is I = 4Mr2, a1 = g 1 + 4 = g/5 , a2 = g 1 + 4r2/(2r)2 = g/2 . (b) Result and interpretation For equal drops from rest, ti = √ 2h/ai, vi = √ 2aih. Therefore t1 t2 = √a2 a1 = √5 2 , v2 v1 = √5 2 . Increasing axle radius reduces the effective rotational mass I/r2 a and speeds the descent. Physical check The larger axle has one quarter the effective rotational mass, despite unchanged physical inertia.

Question 3

Changing axle radius alters acceleration without changing body inertia

Two yo-yos have the same mass MM and inertia I=4Mr2I=4Mr^2. Their axle radii are rr and 2r2r. Each descends from rest on a fixed cord through the same height hh.

Diagram shown in the original SVG worksheet.

  1. Compare their downward accelerations.

  2. Compare descent times and final speeds.

Question 3 - Solution

Changing axle radius alters acceleration without changing body inertia

Diagram shown in the original SVG worksheet.

(a) Derivation and calculation

For axle radius rar_a, the fixed-cord acceleration is a=g1+I/(Mra2).a=\frac{g}{1+I/(Mr_a^2)}. Since the same inertia is I=4Mr2I=4Mr^2, a1=g1+4=g/5,a2=g1+4r2/(2r)2=g/2.\boxed{a_1=\frac{g}{1+4}=g/5},\qquad \boxed{a_2=\frac{g}{1+4r^2/(2r)^2}=g/2}.

(b) Result and interpretation

For equal drops from rest, ti=2h/ai,vi=2aih.t_i=\sqrt{2h/a_i},\qquad v_i=\sqrt{2a_ih}. Therefore t1t2=a2a1=52,v2v1=52.\boxed{\frac{t_1}{t_2}=\sqrt{\frac{a_2}{a_1}}=\sqrt{\frac52}},\qquad \boxed{\frac{v_2}{v_1}=\sqrt{\frac52}}. Increasing axle radius reduces the effective rotational mass I/ra2I/r_a^2 and speeds the descent.

Physical check

The larger axle has one quarter the effective rotational mass, despite unchanged physical inertia.