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Laws of Motion — Physics STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 SciencePhysicsLaws of Motion5 Marks
Question
A racing car travels on a track (without banking) ABCDEFA ABC is a circular arc of radius 2 R. CD and FA are straight paths of length R and DEF is a circular arc of radius $R = 100 m$. The co-effecient of friction on the road is $\mu = 0.1.$ The maximum speed of the car is $50m s^{–1}$. Find the minimum time for completing one round.
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Answer

The main concept used: The centripetal force to keep the car in circular motion is provided by frictional force inward to centre O.Explanation:
  1. Time taken from $\text{A}\rightarrow\text{B}\rightarrow\text{C}$
$\text{s}_1=$ length of path $=\frac{3}{4}2\pi(2\text{R})=\frac{3}{4}\times4\pi\times100=300\pi\text{m}$
$\text{v}_1=$ the maximum speed of car along the circular path
$\sqrt{\mu\text{r}\text{g}}=\sqrt{0.1\times2\text{R}\times\text{g}}$
$\text{v}_1=\sqrt{0.1\times2\times100\times10}=\sqrt{200}=10\times\sqrt{2\text{m/ s}}=14.14\text{m/}\sec$
$\therefore\ \text{t}_1=\frac{\text{s}_1}{\text{v}_1}=\frac{300\pi}{14.14}=\frac{300\times3.14}{14.14}=66.62\text{s}$
  1. Time from $\text{c}\rightarrow\text{D}$ and $\text{F}\rightarrow\text{A}$
$\text{s}_2=\text{CD}+\text{FA}=\text{R+R}=100+100=200\text{m}$
As path CD and FA are in straight so car will travel with it’s maximum speed $=\text{v}_2=50\text{m/ s}$
$\therefore\ \text{t}_2=\frac{\text{s}_2}{\text{v}_2}=\frac{200}{50}=4\sec$
  1. Time for path $\text{D}\rightarrow\text{E}\rightarrow\text{F}$ is
$\text{s}_3=\frac{1}{4}2\pi\text{R}=\frac{1}{4}\times2\pi\times100=50\pi$
$\text{v}_3=\sqrt{\mu}\text{r}\text{g}=\sqrt{0.1\times\text{R}\times\text{g}}=\sqrt{0.1\times100\times10}=\text{m/ s}$
$\text{t}_3=\frac{\text{s}_3}{\text{v}_3}=\frac{50\pi}{10}=5\sec=5\times3.14=154.70\text{s}$
Total time taken by car $\text{t}_1+\text{t}_2+\text{t}_3$
$\therefore\text{t}=66.62+4+15.70=86.32\sec.$

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