Question

An 20-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 64 rpm to 95 rpm in 12 seconds .-What is the tangential acceleration of the pedal?-What length of chain passes over the top of the sprocket during this interval?

Answer 1

Answer:

0.033815 m/s²

12.48810875 m

Explanation:

t = Time taken = 12 s

$\omega_f$ = Final angular velocity = 95 rpm

$\omega_i$ = Initial angular velocity = 64 rpm

$\alpha$ = Angular acceleration

$\theta$ = Angle of rotation

$\omega_f=\omega_i+\alpha t\\\Rightarrow \alpha=\frac{\omega_f-\omega_i}{t}\\\Rightarrow \alpha=\frac{95\times \frac{2\pi}{60}-64\times \frac{2\pi}{60}}{12}\\\Rightarrow \alpha=0.27052\ rad/s^2$

Tangential acceleration is given by

$a_t=\alpha r\\\Rightarrow a_t=0.27052\times 0.125\\\Rightarrow a_t=0.033815\ m/s^2$

The tangential acceleration of the pedal is 0.033815 m/s²

$\omega_f^2-\omega_i^2=2\alpha \theta\\\Rightarrow \theta=\frac{\omega_f^2-\omega_i^2^2}{2\alpha}\\\Rightarrow \theta=\frac{\left(95\times\frac{2\pi }{60}\right)^2-\left(64\times\frac{2\pi}{60}\right)^2}{2\times 0.27052}\\\Rightarrow \theta=99.90487\ rad$

Linear displacement would be $99.90487\times \frac{1}{2\pi}\times 2\pi\times 0.125=12.48810875\ m$

Length of chain that passes in the interval is 12.48810875 m