Find the maximum velocity for overturn for a car moved on a circular track of radius 100m . The coefficient of friction between the road and tyre is 0.2

To find the maximum safe velocity for a car moving on a circular track without skidding or overturning, we can analyze the forces acting on the car.

When a car moves on a circular track of radius $r$, the necessary centripetal force is provided by the frictional force between the tyres and the road. For the car to not slip or overturn, the centripetal force must be less than or equal to the maximum static frictional force.

The maximum frictional force is given by:
$F_{\text{friction}}=\mu N=\mu mg$
where:
• $\mu$ is the coefficient of friction,
• $m$ is the mass of the car,
• $g$ is the acceleration due to gravity ($g=9.8{\text{m/s}}^2$).

The centripetal force required for circular motion is:
$F_{\text{centripetal}}=\frac{m{v}^2}{r}$
where $v$ is the velocity of the car.

To prevent the car from skidding or overturning, we equate the centripetal force to the maximum frictional force:
$\frac{m{v}^2}{r}=\mu mg$

Solving for the maximum velocity $v$:
$v^2=\mu gr$
$v=\sqrt{\mu gr}$

Given parameters in the question:
• Radius, $r=100\text{m}$
• Coefficient of friction, $\mu =0.2$
• Acceleration due to gravity, $g=9.8{\text{m/s}}^2$

Substituting these values into the equation:
$v=\sqrt{0.2\times 9.8\times 100}$
$v=\sqrt{2\times 9.8\times 10}$
$v=\sqrt{196}$
$v=14\text{m/s}$

Thus, the maximum velocity of the car to avoid overturning is 14 m / s.