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Found in: Page 335

Physics For Scientists & Engineers

Book edition 9th Edition
Author(s) Raymond A. Serway, John W. Jewett
Pages 1624 pages
ISBN 9781133947271

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 $N=54.4 rev$.

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

See the step by step solution

Step 1: Defining the final angular speed formula

Hence, the final angular speed formula is, $\omega =v/R$, where $\omega$ 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

$v=22.0 m/s,\phantom{\rule{0ex}{0ex}}t=9.00 s,$

D = 58.0 cm

Hence, the final angular speed can be found by using, $\omega =\frac{v}{R}$.

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

$\begin{array}{rcl}\omega & =& \frac{v}{R}\\ & =& \frac{22.0}{0.58/2}\\ & =& 75.9 \text{rad}/\text{s}\end{array}$

Thus, to find the number of revolutions, use the formula, $N=\frac{\varphi }{2\pi }$.

Solve for $\varphi$.

$\begin{array}{rcl}\varphi & =& \frac{\omega +{\omega }_{0}}{2}×t\\ & =& \frac{75.9+0.00}{2}×9.00\\ & =& 341 \text{rad}\end{array}$

Substitute $\varphi =341$ in the number of revolutions formula.

$\begin{array}{rcl}N& =& \frac{\varphi }{2\pi }\\ & =& \frac{341}{2\pi }\\ & =& 54.4\end{array}$

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

$n=2\pi \omega$

Substitute $\omega =75.9 rad/s$.

$\begin{array}{rcl}n& =& 2\pi \omega \\ & =& 6.28×75.9 \text{rad}/\text{s}\\ & =& 477 \text{rev}/\text{s}\end{array}$

Hence, the final angular speed in revolution per second is $n=477 \text{rev}/\text{s}$.