Answer:
The linear speed of the bike is 19.242 miles per hour.
Step-by-step explanation:
If sliding between the bottom of the wheel and ground can be neglected, the motion of the wheel can be well described by rolling, which is a superposition of coplanar pure rotation and translation, The speed of the bike occurs at the center of the wheel, where resulting instantaneous motion is pure translation parallel to ground orientation. The magnitude of the speed of bike ([tex]v_{B}[/tex]), measured in inches per second, is:
[tex]v_{B} = R\cdot \omega[/tex]
Where:
[tex]R[/tex] - Radius, measured in inches.
[tex]\omega[/tex] - Angular speed, measured in radians per second.
Now, the angular speed must be converted from revolutions per minute into radians per second:
[tex]\omega = \left(154\,\frac{rev}{min} \right)\cdot \left(2\pi\,\frac{rad}{rev} \right)\cdot \left(\frac{1}{60}\,\frac{min}{s} \right)[/tex]
[tex]\omega \approx 16.127\,\frac{rad}{s}[/tex]
The speed of the bike is: ([tex]R = 21\,in[/tex] and [tex]\omega \approx 16.127\,\frac{rad}{s}[/tex])
[tex]v_{B} = (21\,in)\cdot \left(16.127\,\frac{rad}{s} \right)[/tex]
[tex]v_{B} = 338.667\,\frac{in}{s}[/tex]
Lastly, the outcome is converted into miles per hour:
[tex]v_{B} = (338.667\,\frac{in}{s} )\cdot \left(3600\,\frac{s}{h} \right)\cdot \left(\frac{1}{63360}\,\frac{mi}{in} \right)[/tex]
[tex]v_{B} = 19.242\,\frac{mi}{h}[/tex]
The linear speed of the bike is 19.242 miles per hour.
