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3 years ago

5

A student decides to give his bicycle a tune up. He flips it upside down (so there's no friction with the ground) and applies a

force of 34 N over 0.6 seconds to the pedal, which has a length of 16.5 cm. If the back wheel has a radius of 33.0 cm and moment of inertia of 1200 kg cm^2, what is the tangential velocity of the rim of the back wheel in m/s

1 answer:

aleksklad [387]3 years ago

5 0

Answer:

$V=9.2565m/s$

Explanation:

From the question we are told that:

Force $F = 34 N$

Time $t = 0.6 s$

Length of pedal $l_p=16.5cm \approx0.165m$

Radius of wheel $r = 33 cm = 0.33 m$

Moment of inertia, $I = 1200 kgcm2 = 0.12 kg.m2$

Generally the equation for Torque on pedal $\mu$ is mathematically given by

$\mu=F*L\\\mu=34*0.165$

$\mu=5.61N.m$

Generally the equation for angular acceleration $\alpha$ is mathematically given by

$\alpha=\frac{\mu}{l}$

$\alpha=\frac{5.61}{0.12}$

$\alpha=46.75$

Therefore Angular speed is \omega

$\omega=\alpha*t$

$\omega=(46.75)*(0.6)$

$\omega=28.05rad/s$

Generally the equation for Tangential velocity V is mathematically given by

$V=r\omega$

$V=(0.33)(28.05)$

$V=9.2565m/s$

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2 years ago

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3 years ago

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<em>Solving for distance, we would use the second equation of motion;</em>

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<em>Substituting the values into the equation;</em>

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<u>For the second phase, time T2 = 5 secs;</u>

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$Speed = \frac{distance}{time}$

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$Distance, S = speed * time$

<em>Substituting into the above equation;</em>

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