Question

An 19-cm-long bicycle crank arm, with a pedal at one end, is attached to a 25-cm-diameter sprocket, the toothed disk around which the chain moves. A cyclist riding this bike increases her pedallng rate from 64 rpm to 92 rpm in 10 s Part A What is the tangential acceleration of the pedal? Express your answer to two significant figures and include the approprlate units. Value Units at Request Answer Submit Part B What length of chain passes over the top of the sprocket during this interval? Express your answer to two signlfhcant figures and Include the approprlate units. Value Units 1= Bequest Answer Submit

Answer 1

Part A -

Initial angular speed, w1 = 64 rpm

Concert this in rad/s -

w1 = (2π/60)*64 rad/s

Final angular speed, w2 = 92 rpm = (2π/60)*92 rad/s

So, change in the speed,

dw = (2π/60)*92 - (2π/60)*64

= 9.632 - 6.701 = 2.931 rad/sec

Time required to change this speed, dt = 10 s

So, angular acceleration, α = dw/dt

= 2.931 / 10 = 0.2931 rad/sec²

Length of the bicycle crank arm, Rp = 19 cm = 0.19 m

Hence, tangential acceleration of the pedal,

Ap = α*Rp = 0.2931 * 0.19 = 0.0557 m/s² (Answer)

Part B -

Now, angular displacement during the acceleration,

Θ = d(w²)/(2α)

= (9.632² - 6.701²)/(2*0.2931)

= (47.872) / 0.5862 = 81.66 rad.

Diameter of the sprocket = 25 cm = 0.25 m

Hence, radius of the sprocket, Rc = 0.25/2 = 0.125 m

Therefore, length of the chain passes over the top of the sprocket during this interval,

Lc = Rc*Θ

= 0.125 * 81.66 = 10.21 m (Answer)