Question

A tire 2.00ft in diameter is placed on a balancing machine,where it is spun so that its tread is moving at a constant speed of60.0 mi/h. A small stone is stuck in the tread of the tire. What isthe acceleration of the stone as the tire is being balanced?

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asked 2021-04-18
A tire 2.00ft in diameter is placed on a balancing machine,where it is spun so that its tread is moving at a constant speed of 60.0 mi/h. A small stone is stuck in the tread of the tire. What isthe acceleration of the stone as the tire is being balanced?

Answers (1)

2021-04-20
The problem is asking for centripetal acceleration, given by the equation:
\(\displaystyle{a}_{{{c}}}=\frac{{v}^{{{2}}}}{{r}}\), wherev is velocity and r is radius.
First we need to be in the correct units:
1 mi =1609.344 m
1 hr = 3600 s
60mph = (60)(1609.344)/3600 = 26.82m/s
r = .5(2) = 1 ft
1 ft = 0.3048m
\(\displaystyle{a}_{{{c}}}=\frac{{v}^{{{2}}}}{{r}}=\frac{{26.82}^{{{2}}}}{{.3048}}={2360.4}\frac{{m}}{{s}^{{{2}}}}\)
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