Question 45: Work, Energy and Power & Circular Motion

Question

A car of $$800 \mathrm{~kg}$$ is taking turn on a banked road of radius $$300 \mathrm{~m}$$ and angle of banking $$30^{\circ}$$. If coefficient of static friction is 0.2 then the maximum speed with which car can negotiate the turn safely: $$(\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2, \sqrt{3}=1.73)$$

Accepted Answer

When a car takes a turn on a banked road, the forces involved are the gravitational force, the normal force from the surface, and frictional force (if any). The net force provides the necessary centripetal force for the circular motion. The angle of banking and static friction contribute to the maximum speed the car can achieve without slipping.
The forces acting on the car are as follows:
1. The normal force ($$N$$) acts perpendicular to the surface of the road.
2. Gravitational force ($$mg$$) acts downward.
3. Frictional force ($$f$$), which can provide additional centripetal force if needed. It acts parallel to the surface of the road, towards the center of the circle.
The normal force and the gravitational force components can be resolved into two directions: perpendicular and parallel to the road surface. The maximum speed is achieved when all available forces (normal, frictional) are utilized to provide the necessary centripetal force ($$F_c$$) without slipping.
The centripetal force required for circular motion is given by:
$F_c = \frac{mv^2}{r}$
Where:
$$m$$ = mass of the car = $$800 kg$$
$$v$$ = speed of the car
$$r$$ = radius of the turn = $$300 m$$
On a banked curve, the maximum velocity can be calculated using the formula:
$v = \sqrt{rg(	an	heta + \mu)\Big/\big(1-\mu	an	heta)}$
Where:
$$	heta$$ = angle of banking = $$30^{\circ}$$
$$\mu$$ = coefficient of static friction = $$0.2$$
$$g$$ = acceleration due to gravity = $$10 m/s^2$$
First, calculate the tangent of the angle:
$	an30^{\circ} = \frac{1}{\sqrt{3}} = \frac{1}{1.73}$
Substitute all the values into the formula:
$v = \sqrt{300 	imes 10 	imes \left(\frac{1}{1.73} + 0.2ight)\Big/\big(1 - 0.2 	imes \frac{1}{1.73}\big)}$
$v = \sqrt{3000 	imes \left(\frac{1}{1.73} + 0.2ight)\Big/\big(1 - \frac{0.2}{1.73}\big)}$
$v = \sqrt{3000 	imes \frac{1 + 1.73 	imes 0.2}{1.73 - 0.2}}$
$v = \sqrt{3000 	imes \frac{1 + 0.346}{1.73 - 0.2}}$
$v = \sqrt{3000 	imes \frac{1.346}{1.53}}$
$v = \sqrt{3000 	imes 0.8797}$
$v = \sqrt{2639.1} \approx 51.37 \, m/s$
Hence, the closest option to the calculated maximum speed without slipping is:
Option A: 51.4 m/s

Answer

51.4 m/s

Answer

102.8 m/s

Answer

70.4 m/s

Answer

264 m/s

Work, Energy and Power & Circular Motion

A car of $800 \mathrm{~kg} is taking turn on a banked road of radius 300 \mathrm{~m} and angle of banking 30^{\circ}. If coefficient of static friction is 0.2 then the maximum speed with which car can negotiate the turn safely: (\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2, \sqrt{3}=1.73)

Choose the correct answer: