Answer:
The answers to the question are
(a) T = 2·π·r/v
(b) 3.3 % change in period of the under-inflated tire compared to the properly inflated tire
(c) Therefore the distance this car would have to travel in order for under-inflated tire to make one more complete rotation than the inflated tire is 57.685 meters.
Explanation:
(a) The period T = 2π/ω
The velocity v = ωr or ω = v/r
Therefore T = 2π/(v/r) = 2πr/v
T = 2·π·r/v
(b) Period of properly inflated tire with radius = 303 mm is 2π303/v
Period of under-inflated tire with radius = 293 mm is 2π293/v
Therefore we have percentage change in period of of the under-inflated tire compared to the properly inflated tire is given by
(2π303/v -2π293/v)/(2π303/v) = 2π10/v/(2π303/v) = 10/303 × 100 = 3.3 %
(c) The period of the under-inflated tire is 10/303 less than that of the inflated tire. Therefore for the under-inflated tire to make one complete turn more than the inflated tire, we have 1/(10/303) = 303/10 or 30.3 revolutions of either tire which is 30.3×2×π×303 = 57685.296 mm = 57.685 meters
Therefore the distance this car would have to travel in order for under-inflated tire to make one more complete rotation than the inflated tire is 57.685 meters
At 30.3 revolutions the distance covered by the under-inflated
= 55781.49 mm
Subtracting the two distances gives
1903.805 mm
The circumference of the inflated tire = 2×π×303 = 1903.805 mm
