Question

An 18-cm-long bicycle crank arm, with a pedal at one end, is attached to a 20-cm-diameter sprocket, the toothed disk around which the chain moves. A cyclist riding this bike increases her pedaling rate from 60rpm to 90 rpm in 10s .

a. What is the tangential acceleration of the pedal?

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

Short answer

a. the tangential acceleration of the pedal is$5.652\times {10}^{-2}\text{m}/{\text{sec}}^2$

b. the length of chain passes over the top of the sprocket during this interval is$7.85\text{m}$

Step-by-step solution

Part (a)

The initial and final angular velocity, 60 and 90 rpm respectively, the time period for this change of velocity, 10 seconds, and a

The radius of the crank arm,$r=18\text{cm}$

$\begin{array}{l}\alpha =\frac{d\omega }{dt}=\frac{(90-60)\text{rpm}}{10\text{sec}}=\frac{30\text{rpm}}{10\text{sec}}=\frac{\left(\frac{30}{60\text{sec}}\times \frac{\text{rotation}}{\text{min}}\times \frac{2\pi \text{rad}}{\text{rotation}}\right)}{10\text{sec}}\\ \alpha =0.314\text{rad}/{\text{s}}^2\end{array}$

Tangential acceleration

$\begin{array}{l}{a}_t=r\alpha \\ a_t=\left(18\times {10}^{-2}\text{m}\right)\times \left(0.314\frac{\text{rad}}{{\text{s}}^2}\right)\\ a_t=5.652\times {10}^{-2}\text{m}/{\text{sec}}^2\end{array}$

Part (b)

The initial and final angular velocity

$\begin{array}{l}60\text{\hspace{0.17em}rpm}=60\text{\hspace{0.17em}rpm}\cdot \frac{\frac{2\pi \text{\hspace{0.17em}\hspace{0.17em}rad}}{60\text{\hspace{0.17em}s}}}{1\text{\hspace{0.17em}\hspace{0.17em}rpm}}=6.28\frac{\text{rad}}{\text{s}}\\ \text{and\hspace{0.17em}}\\ 90\text{\hspace{0.17em}rpm\hspace{0.17em}=\hspace{0.17em}}90\text{\hspace{0.17em}rpm}\cdot \frac{\frac{2\pi \text{\hspace{0.17em}rad}}{60\text{\hspace{0.17em}s}}}{1\text{\hspace{0.17em}\hspace{0.17em}rpm}}=9.42\frac{\text{rad}}{\text{s}}\end{array}$

Angle turned by the bicycle crank in

$\begin{array}{l}\theta ={\omega }_{i}t+\frac{1}{2}\alpha t^2\\ \theta =\left(2\pi \right)\left(10\text{sec}\right)+\frac{1}{2}\left(0.314\text{rad}/{\text{sec}}^2\right){\left(10\text{sec}\right)}^2\\ \theta =78.5\end{array}$

If the length of chain passes over the top of sprocket

$\begin{array}{l}S=\theta r\\ S=78.5\times \left(0.10\text{m}\right)\\ S=7.85\text{m}\end{array}$