Question

This problem refers to the bicycle generator considered in the previous problem. It is driven by a \({\rm{1}}{\rm{.60 cm}}\) diameter wheel that rolls on the outside rim of the bicycle tire. (a) What is the velocity of the bicycle if the generator’s angular velocity is \({\rm{1875 rad/s}}\)? (b) What is the maximum emf of the generator when the bicycle moves at \({\rm{10}}{\rm{.0 m/s}}\), noting that it was \({\rm{18}}{\rm{.0 V}}\) under the original conditions? (c) If the sophisticated generator can vary its own magnetic field, what field strength will it need at \({\rm{5}}{\rm{.00 m/s}}\) to produce a \({\rm{9}}{\rm{.00 V}}\) maximum emf?

Short answer

a) Velocity of the bicycle is \({\rm{v}} = {\rm{15\;m/s}}\).

b) The emf of the generator is maximum when \({{\rm{v}}_{\rm{2}}} = {\rm{8}}{\rm{.3\;m/s}}\).

c) Field strength produced by the emf is \({\rm{B}} = {\rm{0}}{\rm{.96\;T}}\)

Step-by-step solution

Calculation the speed of the bicycle.

(a)

The speed of the bicycle will be equal to the perimeter of the wheel that rotates the generator times the frequency (not angular frequency). That is,

\({\rm{v}} = {\rm{Pf}}\)

Since we are given the diameter, the perimeter will be\({\rm{P}} = {\rm{\pi d}}\)and we also should remember that the frequency can be given by the angular frequency as\({\rm{f}} = \frac{{\rm{\omega }}}{{{\rm{2\pi }}}}\).

This means that our expression of the result is

\({\rm{v}} = \frac{{{\rm{\omega d}}}}{{\rm{2}}}\)

Numerically we will have

\(\begin{aligned}{}{\rm{v}} &= \frac{{{\rm{1875}}\;{\rm{rad/s}} \times {\rm{0}}{\rm{.016}}\;{\rm{m}}}}{{\rm{2}}}\\ &= {\rm{15\;m/s}}\end{aligned}\)

Therefore, the value of the speed of the bicycle is \({\rm{v}} = {\rm{15\;m/s}}\).

Calculation the maximum emf of the generator.

(b)

Let us remember that the emf is directly proportional to the speed of the bicycle.

This allows us to simply write the following relation

\(\begin{aligned}{l}\frac{{{{\rm{v}}_{\rm{1}}}}}{{{{\rm{v}}_{\rm{2}}}}} &= \frac{{{{\rm{\varepsilon }}_{\rm{1}}}}}{{{{\rm{\varepsilon }}_{\rm{2}}}}}\\{{\rm{\varepsilon }}_{\rm{2}}} &= \left( {\frac{{{{\rm{v}}_{\rm{2}}}}}{{{{\rm{v}}_{\rm{1}}}}}} \right){{\rm{\varepsilon }}_{\rm{1}}}\end{aligned}\)

Applying numerically, we can find this speed to be

\(\begin{aligned}{}{{\rm{\varepsilon }}_{\rm{2}}} &= \left( {\frac{{{\rm{10}}\;{\rm{m/s}}}}{{{\rm{15 m/s}}}}} \right) \times {\rm{18}}{\rm{.0}}\;{\rm{V}}\\ &= {\rm{12}}{\rm{.0\;V}}\end{aligned}\)

Therefore, the emf of the generator is \({\rm{12}}{\rm{.0\;V}}\).

Calculation for the field strength produced by emf.

c)

With the given conditions, the speed is\(\frac{{\rm{1}}}{{\rm{3}}}\)of the initial speed, while the electromotive force needs to be\(\frac{{\rm{1}}}{{\rm{2}}}\)of the initial.

This means that the field will have to be\(\frac{{\rm{3}}}{{\rm{2}}}\)of the initial field, yielding

\(\begin{aligned}{}{\rm{B}} &= \frac{{\rm{3}}}{{\rm{2}}} \times {\rm{0}}{\rm{.64}}\;{\rm{T}}\\ &= {\rm{0}}{\rm{.96\;T}}\end{aligned}\)

Therefore, the field strength produced by emf is \({\rm{B}} = {\rm{0}}{\rm{.96\;T}}\)