Question

While riding a multispeed bicycle, the rider can select the radius of the rear sprocket that is fixed to the rear axle. The front sprocket of a bicycle has radius 12.0 cm. If the angular speed of the front sprocket is 0.600 rev/s, what is the radius of the rear sprocket for which the tangential speed of a point on the rim of the rear wheel will be 5.00 m/s? The rear wheel has radius 0.310 m.

Answer 1

Answer:

2.8 cm

Explanation:

radius of front sprocket (R1) = 12 cm = 0.12 m

angular speed of the front sprocket (ω1) = 0.6 rev/s = 3.77 rad/s

radius of the rear wheel (R°2) = 0.31 m

tangential speed of the rear wheel (V°2) = 5 m/s

(take note that the tangential speed of the rear wheel is the same as that of the rear sprocket, V°2 = V2 (rear sprocket))

speed at any point on the front sprocket = speed at any point on the rear sprocket

V1 = V2 ....equation 1

• tangential speed of the front sprocket (V1) = R1 x ω1

        V1 = 0.12 x 3.77 = 0.45 m/s

• tangential speed of the rear sprocket (V2) = R1 x ω2

        take note that the speed of the rear wheel is the same as the speed of                

        the rear sprocket, which means ω2 is the same for both the rear  

        wheel and the rear sprocket. This also means we can use the

        parameters for the rear wheel (V°2 and R2°) to find ω2

         ω2 = $\frac{V°2}{R2°}$

         ω2 =  $\frac{5}{0.31}$ = 16.1 rad/s

        therefore V2 = R1 x ω2 = 16.1 x R1

recall that  V1 = V2 ....equation 1

now we can put the values of V1 and V2 into equation 1

0.45 = 16.1 x R1

R1 = 0.45 / 16.1 = 0.028 m = 2.8 cm