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amolent3u 2022-01-12 Answered
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redhotdevil13l3
Answered 2022-01-13 Author has 6117 answers
We are given following data for a bicycle wheel:
\(\displaystyle{d}_{{{1}}}={0.64}{m}\)
\(\displaystyle{m}={1.8}{k}{g}\)
\(\displaystyle{F}_{{{1}}}={120}{N}\)
It is also given that chain sprocket:
\(\displaystyle{d}_{{{2}}}={0.09}{m}\)
\(\displaystyle\alpha={4.5}{\frac{{{r}{a}{d}}}{{{s}^{{{2}}}}}}\)
We can assume that the wheel is a hoop. Hoop moment of inertia is given by:
\(\displaystyle{I}={m}\cdot{{r}_{{{1}}}^{{{2}}}}={m}\cdot{\left({\frac{{{d}_{{{1}}}}}{{{2}}}}\right)}^{{{2}}}={1.8}\cdot{\left({\frac{{{0.64}}}{{{2}}}}\right)}^{{{2}}}={0.184}{k}{g}\ {m}^{{{2}}}\)
Torque equation:
\(\displaystyle{I}\cdot\alpha={F}_{{{2}}}\cdot{r}_{{{2}}}-{F}_{{{1}}}\cdot{r}_{{{1}}}\)
Solving it for force applied by a chain:
\(\displaystyle{F}_{{{2}}}={\frac{{{I}\cdot\alpha+{F}_{{{1}}}\cdot{r}_{{{1}}}}}{{{r}_{{{2}}}}}}={\frac{{{I}\cdot\alpha+{F}_{{{1}}}\cdot{\frac{{{d}_{{{1}}}}}{{{2}}}}}}{{{\frac{{{d}_{{{2}}}}}{{{2}}}}}}}={\frac{{{0.184}\cdot{4.5}+{120}\cdot{\frac{{{0.64}}}{{{2}}}}}}{{{\frac{{{0.09}}}{{{2}}}}}}}\)
\(\displaystyle={871.7}{N}\)
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Karen Robbins
Answered 2022-01-14 Author has 4657 answers

Explanation: The net torque of the sprocket is
\(\displaystyle\tau_{{\ne{t}}}=\tau_{{{a}{p}{p}{l}{i}{e}{d}}}-\tau_{{{r}{e}{s}{i}{s}{t}{i}{v}{e}}}={I}\alpha\) and further it is modified as \(\displaystyle{r}{F}-{R}{f}={I}\alpha\).
Now the moment of inertia of the bicycle wheel is \(\displaystyle{I}={M}{R}^{{{2}}}\) and by using these relations, the force that must applied by the chain is found.
The formula for the force must be applied by the chain is,
\(\displaystyle{F}={\frac{{{M}{R}^{{{2}}}\alpha+{R}{f}}}{{{r}}}}\)
- M is mass of the wheel.
- R is radius of the wheel.
- \(\displaystyle\alpha\) is angular acceleration of the wheel.
- f is resistive force.
- r is radius sprocket.
Substitute 1.80kg for M, 64.0cm/2 for R, 120 N for f, 9.00cm/2 for r, and \(\displaystyle{4.50}{r}{a}\frac{{d}}{{s}^{{{2}}}}\ {f}{\quad\text{or}\quad}\ \alpha\) to find F.
\(\displaystyle{F}={\frac{{{\left[{\left({1.80}{k}{g}\right)}{\left({64.0}{c}\frac{{m}}{{2}}\right)}^{{{2}}}{\left({\frac{{{0.01}{m}}}{{{1}{c}{m}}}}\right)}^{{{2}}}\right]}{\left({4.50}{r}{a}\frac{{d}}{{s}^{{{2}}}}\right)}+{\left({64.0}{c}\frac{{m}}{{2}}\right)}{\left({\frac{{{0.01}{m}}}{{{1}{c}{m}}}}\right)}{\left({120}{N}\right)}}}{{{\left({9.00}{c}\frac{{m}}{{2}}\right)}{\left({\frac{{{0.01}{m}}}{{{1}{c}{m}}}}\right)}}}}\)
=872N
Thus, the force must be applied by a chain passing over a sprocket to give the wheel an acceleration of \(\displaystyle{4.50}\frac{{rad}}{{s}^{{{2}}}}\) is 872N.

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