Question

A truck of mass 2000 kg is moving along a circular path having a radius of curvature of 10 m. If the banking angle is 39°, then the maximum permissible speed of the truck is (Acceleration due to gravity = 10 ms\(^{-2}\), take tan 39° = 0.81):

• A. \( 14 \, \text{ms}^{-1} \)
• B. \( 5 \, \text{ms}^{-1} \)
• C. \( 18 \, \text{ms}^{-1} \)
• D. \( 9 \, \text{ms}^{-1} \)

Hint:

The banking angle is crucial for safe turning at higher speeds. By inclining the road, a component of the normal force contributes to the centripetal force, reducing the reliance on friction. If the speed exceeds this ideal value, friction will act downwards along the incline to provide the extra centripetal force needed, up to its maximum static value.

Answer 1

The Correct Option is D

Step 1: Understand the forces involved in motion on a banked curve.
When a vehicle travels on a banked road, the road is inclined at an angle \( \theta \) to the horizontal. This banking is designed to help the vehicle navigate the turn. The primary forces acting on the truck are its weight (\( mg \)) acting vertically downwards and the normal reaction force (\( N \)) exerted by the road, which acts perpendicular to the surface of the bank. For simplicity, to find the maximum permissible speed without relying on friction to prevent outward skidding, we first consider an ideal scenario where the frictional force between the tires and the road is negligible.
Step 2: Resolve the normal force into its horizontal and vertical components.
The normal force \( N \) can be resolved into two components:
A vertical component \( N \cos \theta \) acting upwards.
A horizontal component \( N \sin \theta \) acting towards the center of the circular path.

Step 3: Apply Newton's second law in the vertical and horizontal directions.
Vertical Equilibrium: Since there is no vertical acceleration, the net vertical force must be zero. The upward vertical component of the normal force balances the weight of the truck: \[ N \cos \theta = mg \] Horizontal Motion: The horizontal component of the normal force provides the necessary centripetal force \( F_c \) required for the circular motion of the truck. The centripetal force is given by \( F_c = \frac{mv^2}{r} \), where \( m \) is the mass, \( v \) is the speed, and \( r \) is the radius of the circular path. \[ N \sin \theta = \frac{mv^2}{r} \]
Step 4: Solve the system of equations for the maximum permissible speed \( v \).
To find \( v \), we can divide the equation for horizontal motion by the equation for vertical equilibrium: \[ \frac{N \sin \theta}{N \cos \theta} = \frac{\frac{mv^2}{r}}{mg} \] The normal force \( N \) and the mass \( m \) cancel out: \[ \tan \theta = \frac{v^2}{rg} \] Now, solve for \( v \): \[ v^2 = rg \tan \theta \] \[ v = \sqrt{rg \tan \theta} \]
Step 5: Substitute the given values to calculate the maximum permissible speed.
Given values are: Radius of curvature \( r = 10 \, \text{m} \)
Acceleration due to gravity \( g = 10 \, \text{ms}^{-2} \)
Banking angle \( \theta = 39^\circ \)
\( \tan 39^\circ = 0.81 \)
Substitute these values into the formula for \( v \): \[ v = \sqrt{(10 \, \text{m})(10 \, \text{ms}^{-2})(0.81)} \] \[ v = \sqrt{81 \, \text{m}^2\text{s}^{-2}} \] \[ v = 9 \, \text{ms}^{-1} \] Therefore, the maximum permissible speed of the truck on this banked curve, without relying on friction to prevent outward skidding, is \( 9 \, \text{ms}^{-1} \).