Question

A van is moving with a speed of 108 km/hr on a level road where the coefficient of friction between the tyres and the road is 0.5. For the safe driving of the van, the minimum radius of curvature of the road shall be (Acceleration due to gravity, g=10 m/s²)

• A. 180 m
• B. 120 m
• C. 80 m
• D. 40 m

Answer 1

The Correct Option is A

The centripetal force required to keep the van moving in a curved path is given by: 
F_c = \(\frac {mv^2}{r}\)
The maximum frictional force that can be provided by the tires is given by: 
F_friction = μmg 
For safe driving, the maximum centripetal force must be less than or equal to the maximum frictional force. Therefore, we can equate the two expressions: 
\(\frac {mv^2}{r}\) = μmg 
\(\frac {v^2}{r}\) = μg 
r = \(\frac {v^2}{μg}\)
Now we can substitute the given values: 
v = 108 km/hr 
v = = \(\frac {108,000\  m}{3600\  s}\) 
v = 30 m/s 
μ = 0.5 
g = 10 m/s² 
r = \(\frac {(30 m/s)^2}{0.5 \times  10 \ m/s^2}\) 
r = \(\frac {900}{5}\) m 
r = 180 m 
Therefore, the minimum radius of curvature of the road for safe driving is 180 meters. 
So, the correct option is (A) 180 m.

Answer 2

Given:
- Speed of the van: \( v = 108 \text{ km/hr} \)
- Coefficient of friction: \( \mu = 0.5 \)
- Acceleration due to gravity: \( g = 10 \text{ m/s}^2 \)
Steps to solve:
1. Convert the speed from km/hr to m/s:
\[ v = 108 \text{ km/hr} \]
  \[ v = \left( 108 \times \frac{1000}{3600} \right) \text{ m/s} \]
  \[ v = 30 \text{ m/s} \]
2. Frictional force provides the necessary centripetal force for the van to safely navigate the curve.
 The maximum frictional force (\( F_{\text{friction}} \)) is given by:
 \[ F_{\text{friction}} = \mu \cdot m \cdot g \]
The centripetal force (\( F_{\text{centripetal}} \)) required to keep the van moving in a circle of radius \( r \) is:
  \[ F_{\text{centripetal}} = \frac{m \cdot v^2}{r} \]
3. Equating the frictional force to the centripetal force:
\[ \mu \cdot m \cdot g = \frac{m \cdot v^2}{r} \]
 Cancel \( m \) from both sides (assuming \( m \neq 0 \)):
  \[ \mu \cdot g = \frac{v^2}{r} \]
4. Solve for \( r \):
 \[ r = \frac{v^2}{\mu \cdot g} \]
 Substitute the known values:
  \[ r = \frac{(30 \text{ m/s})^2}{0.5 \cdot 10 \text{ m/s}^2} \]
  \[ r = \frac{900 \text{ m}^2/\text{s}^2}{5 \text{ m/s}^2} \]
  \[ r = 180 \text{ m} \]
 Final Result:The minimum radius of curvature of the road for safe driving of the van is:
\[ \boxed{180 \text{ m}} \]