A Rotating Toy

Classical Mechanics Level 3

A toy is made by attaching two thin discs at the ends of an axle of length l = 1.00 l=1.00 cm. Due to a manufacturing defect, one product has a pair of discs with slightly different radii: r 1 = 9.95 r_{1}=9.95 cm and r 2 = 10.05 r_{2}=10.05 cm. When this toy is placed on a horizontal ground and then pushed, it rolls without slip. In the process of this motion, the wheel rotates about its axle, advances on a circular path, and also rotates about a vertical axis.

If the center of the axle moves with speed v c = 10.0 v_{c}=10.0 cm/s, find the angular velocity of the rotation of the wheel about the vertical axis (in rad/s).


The answer is 0.1.

Rayyan Shahid by Rayyan Shahid , 20, India

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1 solution

Steven Chase
Feb 21, 2018

Let R R be the radius of the inner circle around the vertical axis.

Three equations:

1) Inner circle tangential speed = smaller wheel tangential speed
2) Outer circle tangential speed = bigger wheel tangential speed
3) Average wheel tangential speed = axle center speed

R ω 0 = r 1 ω ( R + l ) ω 0 = r 2 ω r 1 ω + r 2 ω = 2 v c R \, \omega_0 = r_1 \, \omega \\ (R+ l) \, \omega_0 = r_2 \, \omega \\ r_1 \, \omega + r_2 \, \omega = 2 v_c

Re-arranging and solving:

ω = 2 v c r 1 + r 2 l ω 0 = ( r 2 − r 1 ) ω = 2 v c ( r 2 − r 1 ) r 1 + r 2 ω 0 = 2 v c l r 2 − r 1 r 1 + r 2 = 20 1 0.1 20 = 0.1 \omega = \frac{2 v_c}{r_1 + r_2} \\ l \, \omega_0 = (r_2 - r_1) \omega = \frac{2 v_c (r_2 - r_1)}{r_1 + r_2} \\ \omega_0 = \frac{2 v_c}{l}\frac{r_2 - r_1}{r_1 + r_2} = \frac{20}{1} \frac{0.1}{20} = 0.1

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