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Rotational Dynamics — Physics STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 SciencePhysicsRotational Dynamics4 Marks
Question
On a dry day, the maximum safe speed at which a car can be driven on a curved horizontal road without skidding is $7 \mathrm{~m} / \mathrm{s}$. When the road is wet, the frictional force between the tyres and road reduces by $25 \%$. How fast can the car safely take the turn on the wet road?
✓

Answer

Let subscripts 1 and 2 denote the values of a quantity under dry and wet conditions, respectively.
Data : $v_1=7 \mathrm{~m} / \mathrm{s}_2 \mathrm{f}_2=\mathrm{f}_1,-0.25 \mathrm{f}_1=0.75 \mathrm{f}_1$
On a dry horizontal curved road, the frictional force between the tyres and road is $f_1=\mu_1 m g$, where $\mathrm{m}$ is the mass of the car and $\mathrm{g}$ is the gravitational acceleration.
The maximum safe speed for taking a turn of radius $r$ on a dry horizontal curved road is
$
v_1=\sqrt{\mu_1 r g}=\sqrt{\frac{r}{m}} \sqrt{f_1}
$
If the road is wet, the corresponding quantities
are
$
f_2=\mu_2 m g \text { and } v_2=\sqrt{\frac{r}{m}} \sqrt{f_2}
$
Thus, for $m$ and $r$ remaining the same,
$
\begin{aligned}
\frac{v_1}{v_2} & =\sqrt{\frac{f_1}{f_2}} \\
\therefore v_2 & =\sqrt{\frac{f_2}{f_1}} \cdot v_1=\sqrt{\frac{0.75 f_1}{f_1}} \cdot(7) \\
& =7 \sqrt{0.75}=7 \times 0.866=6.062 \mathrm{~m} / \mathrm{s}
\end{aligned}
$

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