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 60 rpm to 90 rpm in 10 s.
a. What is the tangential acceleration of the pedal?
b. What length of chain passes over the top of the sprocket during this interval?

Answer 1

Answer:

Part a)

$a = 0.056 m/s^2$

Part b)

$L = 7.85 m$

Explanation:

Part a)

Angular speed of the pedal is changing from 60 rpm to 90 rpm in 10 s

so the angular acceleration is given as

$\alpha = \frac{\omega_2 - \omega_1}{\Delta t}$

so we will have

$\alpha = \frac{2\pi(\frac{90}{60}) - 2\pi(\frac{60}{60})}{10}$

$\alpha = 0.314 rad/s^2$

now the tangential acceleration of the pedal is given as

$a = r \alpha$

$a = 0.18 \times 0.314$

$a = 0.056 m/s^2$

Part b)

Total angular displacement made by the sprocket in the interval of 10 s is given as

$\theta = \frac{\omega_f + \omega_i}{2} t$

$\theta = \frac{2\pi (\frac{90}{60}) + 2\pi (\frac{60}{60})}{2}(10)$

$\theta = 78.5 radian$

now length of the chain passing over it is given as

$L = R\theta$

$L = 0.10 \times 78.5$

$L = 7.85 m$

Answer 2

The tangential acceleration of the pedal is 0.0283 m/s².

The length of chain passes over the top of the sprocket during this interval is 7.85 m.

Angular speed of the pedal in radian per second

The angular speed of the pedal in radian per second is calculated as follows;

$\omega_i = 60\frac{rev}{\min} \times \frac{2\pi \ rad}{1 \ rev} \times \frac{1 \min}{60 \ s} = 2\pi \ rad/s = 6.283 \ rad/s\\\\\omega _f = 90\frac{rev}{\min} \times \frac{2\pi \ rad}{1 \ rev} \times \frac{1 \min}{60 \ s} =9.425 \ rad/s$

Angular acceleration of the pedal

The angular acceleration of the pedal is calculated as follows;

$\alpha = \frac{\Delta \omega}{t} = \frac{\omega _f - \omega _i }{t} \\\\\alpha = \frac{9.425 - 6.283}{10} \\\\\alpha = 0.314 \ rad/s^2$

Tangential acceleration of the pedal

The tangential acceleration of the pedal is calculated as follows;

a = αR

where;

• R is the radius of the pedal = 9 cm

a = 0.314 x 0.09

a = 0.0283 m/s²

Displacement of the sprocket

$\theta =( \frac{\omega _f + \omega _i}{2} ) \times t\\\\\theta =( \frac{9.425 + 6.283}{2} )(10)\\\\\theta = 78.54 \ rad$

Length of the chain passing through

L = Rθ

L = 0.1 x 78.54

L = 7.85 m

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