Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

Published by Pearson
ISBN 10: 0321740904
ISBN 13: 978-0-32174-090-8

Chapter 8 - Dynamics II: Motion in a Plane - Exercises and Problems - Page 214: 54

Answer

See the detailed answer below.

Work Step by Step

First, we need to draw the force diagram of the car, as shown in the two figures below. Now we need to apply Newton's second law; $$\sum F_z=F=m a_T $$ where $a_T$ is the tangential acceleration. Thus, $$a_T=\dfrac{F}{m}=\dfrac{1000}{1500}$$ $$a_T=\color{blue}{0.667}\;\rm m/s^2\tag 1$$ So, the tangential speed of the car after 10 seconds is given by $$v_T=v_0+at=0+a_Tt$$ Plugging from (1); $$v_T=\dfrac{F}{m}t=0.667\times 10$$ $$v=\bf 6.67\;\rm m/s$$ We know that the radial acceleration at some moment is given by $$a_r=\dfrac{v_T^2}{R}$$ So, the radial acceleration after 10 seconds is given by $$a_r=\dfrac{6.67^2}{25}$$ $$a_r=\color{blue}{1.78}\;\rm m/s^2\tag 2$$ Now we can find the car's acceleration by applying the Pythagorean theorem. $$a=\sqrt{a_T^2+a_r^2}$$ From (1) and (2); $$a=\sqrt{0.667^2+1.78^2}=\color{red}{\bf 1.90}\;\rm m/s^2$$ Its direction relative to the $r$-axis is given by $$\tan\theta=\dfrac{a_T}{a_r}$$ $$\theta=\tan^{-1}\left[\dfrac{a_T}{a_r}\right]=\tan^{-1}\left[\dfrac{0.667}{1.78}\right]$$ $$\theta=\color{red}{\bf 20.5^\circ}$$ ______________________________________________ b) The car will slide when the force between the tires and the road conquers the maximum static friction force between the tires and the road (concrete) which happens just after the static friction force between the tires and the road reaches its maximum value. To find when the happens, we need to find the speed of the car at that moment then we can find the time. Applying Newton's second law: $$\sum F_x=f_s=m a_x $$ where $f_s$ is the static friction force between the tires and the road and it is to the left at this point since at high speeds the car will be on the verge of sliding to the right. We know that the static friction force is given by $f_s=\mu_sF_n$ Thus, $$\mu_sF_n=ma_x\tag 1$$ Now we need to find the normal force, $$\sum F_y=F_n-mg=ma_y=m(0)=0$$ Thus, $$F_n=mg$$ Plugging into (1); $$\mu_s \color{red}{\bf\not} mg = \color{red}{\bf\not} ma_x $$ $$a_x=\mu_sg$$ Where $a_x$ is the radial acceleration which is given by $v^2/R$ $$\dfrac{v_T^2}{R}=\mu_sg$$ Solving for $v_T$; $$v_T=\sqrt{\mu_s gR}$$ Plugging the known and recalling the $\mu_s$ of rubber on concrete is 1. $$v_T=\sqrt{1\\times 9.8\times 25 }=\bf 15.65\;\rm m/s$$ Using the kinematic formula of velocity; $$v_T=v_0+a_Tt$$ Plugging the known; $$15.65=0+0.667 t$$ Hence, $$t=\dfrac{15.65}{0.667 }=\color{red}{\bf 23.5}\;\rm s$$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.