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?

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)

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
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