A typical road bike wheel has a diameter of 70 cm including the tire.

A) In a time trial, when a cyclist is racing along at 12 m/s how fast is a point at the top of the tire moving?

B) How fast, in rpm, are the wheels spinning?

### 5 Answers

Part A. The velocity for a point on the top of a tire would be 2V (2*12m/s)=24m/s

Part B. is just converting 12m/s into rad/s. Angular velocity = V/r ((12m/s)/0.35m)=34.2857 rad/s. Now convert into revolutions by dividing by 2pi to get revs per second, and multiply by 60 seconds to get 327.4 rpm.

hey mastering physics gamers, the correct answers are

A) 24 m/s

B) 330 rpm

A) The point at the top of he tire will also cover 12 m/s in a circular route.

B) rpm of wheels = angular velocity in rad/s x (1 revolutions / 2π) in rev/s x (60 s / 1 min)

rpm of wheels = (v/r) x (1/2π) x (60)

rpm of wheels = (12 m/s)/(0.35 m) x (1/2π) x (60) = 327.4 rpm

A) In a time trial, when a cyclist is racing along at 12 m/s how fast (Vt) is a point at the top of the tire moving?

Vt= 12 m/sec

B) How fast, in rpm, are the wheels spinning?

ω = Vt/r = 12/0.35 rad/sec = 2*PI*n/60

n = 12*60/(0.7*PI) = 1028.57/PI rpm ( ≅ 327.40 )

12 m/s and 327.4 Rpm