Question

An 18-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 pedaling rate from 62 rpm to 94 rpm in 12 s .

What is the tangential acceleration of the pedal?

What length of chain passes over the top of the sprocket during this interval?

Answer 1

Given that :

initial angular velocity, _i = 62 rpm = 6.49 rad/s

final angular velocity, _f = 94 rpm = 9.84 rad/s

time taken by cyclist, t = 12 sec

Using equation of rotational motion 1, we have

_f = _i + t

= [(9.84 rad/s) - (6.49 rad/s)] / (12 s)

= 0.279 rad/s²

(a) The tangential acceleration of the pedal will be given as :

we know that,    a_t = r = (0.279 rad/s²) (0.18 m)

a_t = 0.0502 m/s²

(b) Using equation of rotational motion 2, we have

= _i t + (1/2) t²

= (6.49 rad/s) (12 s) + (0.5) (0.279 rad/s²) (12 s)²

= (77.8 rad) + (20.088 rad)

= 97.8 rad

During this interval, the length of chain passes over the top of sprocket will be given as :

we know that,   L = r = (0.125 m) (97.8 rad)

L = 12.2 m