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Physics For Scientists & Engineers
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Short Answer

Question: A car accelerates uniformly from rest and reaches a speed of 22.0m/s in 9.00s. Assuming the diameter of a tire is 58.0cm, (a) find the number of revolutions the tire makes during this motion, assuming that no slipping occurs. (b) What is the final angular speed of a tire in revolutions per second?

a. The number of revolutions the tire makes during the car accelerates uniformly, assuming that no slipping occurs is N=54.4rev.

b. The final angular speed of a tire in revolutions per second is n=477rev/s.

See the step by step solution

Step by Step Solution

Step 1: Defining the final angular speed formula

Hence, the final angular speed formula is, ω=v/R, where ω is the angular speed, v is the velocity, R is the radius.

Step 2: Calculating the final angular speed to determine the number of revolutions

(a)

Consider the given values

v=22.0m/s,t=9.00s,

D = 58.0 cm

Hence, the final angular speed can be found by using, ω=vR.

Substitute.v=22.0m/s,R=0.58m

ω=vR=22.00.58/2=75.9rad/s

Thus, to find the number of revolutions, use the formula, N=ϕ2π.

Solve for ϕ.

ϕ=ω+ω02×t=75.9+0.002×9.00=341rad

Substitute ϕ=341 in the number of revolutions formula.

N=ϕ2π=3412π=54.4

Therefore, the number of revolutions the tire makes during the acceleration of the car is 54.4.

Step 3: Calculating the final angular speed in revolution per second

(b)

Thus, the formula to find the final angular speed in revolution per second is

n=2πω

Substitute ω=75.9rad/s.

n=2πω=6.28×75.9rad/s=477rev/s

Hence, the final angular speed in revolution per second is n=477rev/s.

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