Question

A car of mass $800\text{kg}$ moves on a horizontal circular track of radius $40\text{m}$. If the coefficient of static friction (${\mu }_s$) is $0.5$, then the maximum velocity with which the car can move will be

• A. $15\text{m/s}$
• B. $20\text{m/s}$
• C. $25\text{m/s}$
• D. $14\text{m/s}$

Answer 1

The correct option is D $14\text{m/s}$

For this case, the road is an unbanked horizontal circular track, hence friction is the only source of centripetal force ($F_c$).

$F_c=\frac{m{v}^2}{r}$..........($1$)

Vertical forces on the body are balanced, since $a_y=0$
i.e $N=mg$........($2$)

For executing circular motion successfully:
$F_c\le f_{max}$..............($3$)

From Eq.($1$) and ($3$):
$\frac{m{v}^2}{r}\le f_{max}$
or $\frac{m{v}^2}{r}\le {\mu }_{s}mg$
$\therefore v\le \sqrt{{\mu }_{s}gr}$

$\Rightarrow v_{\text{max}}=\sqrt{{\mu }_{s}gr}$
$=\sqrt{0.5\times 9.8\times 40}=14\text{m/s}$