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?

jernplate8

jernplate8

Answered question

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?

Answer & Explanation

pierretteA

pierretteA

Skilled2021-04-20Added 102 answers

The problem is asking for centripetal acceleration, given by the equation:
ac=v2r, 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
ac=v2r=26.822.3048=2360.4ms2

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