RBD BYU 6

The car moves along a circular path of radius \( R = 20 \) m with a linear speed \( v_c = 36 \) km/h \( = 10 \) m/s. The angular velocity of the car about the center of curvature is:

$$ \Omega_{car} = \frac{v_c}{R} = \frac{10}{20} = 0.5 \text{ rad/s} $$

The outer wheel moves on a radius \( R_{out} = R + l/2 \) and the inner wheel on \( R_{in} = R – l/2 \), where \( l = 1.5 \) m. The difference in their linear speeds is:

$$ \Delta v = v_{out} – v_{in} = \Omega_{car} R_{out} – \Omega_{car} R_{in} $$ $$ \Delta v = \Omega_{car} (R_{out} – R_{in}) = \Omega_{car} \cdot l $$ $$ \Delta v = 0.5 \times 1.5 = 0.75 \text{ m/s} $$

Assuming rolling without slipping, the linear speed of a wheel is \( v = \omega r \), where \( r = 25 \) cm \( = 0.25 \) m is the radius of the wheel itself. The difference in angular velocity is:

$$ \Delta \omega = \frac{\Delta v}{r} = \frac{0.75}{0.25} = 3 \text{ rad/s} $$