Question

What will be the maximum speed of a car on a road turn of radius 30m if the coefficient of friction between the tyres and the road is 0.4? (Take \( g = 9.8 { m/s}^2 \))

• A. \( 10.84 { m/s} \)
• B. \( 9.84 { m/s} \)
• C. \( 8.84 { m/s} \)
• D. \( 6.84 { m/s} \)

Hint:

For circular motion:
- Maximum speed on a turn depends on friction: \( v_{\max} = \sqrt{\mu g R} \).
- Higher friction allows greater speed without skidding.

Answer 1

The Correct Option is A

Step 1: Identify the forces acting on the car
- The car experiences centripetal force due to friction when moving in a circular path.
- The maximum frictional force available for turning without skidding is given by:
\[ F_{{friction}} = \mu mg \]
Step 2: Use Newton’s second law for circular motion
- The required centripetal force is given by:
\[ F_c = \frac{m v^2}{R} \]
- Equating the friction force and centripetal force:
\[ \mu mg = \frac{m v^2}{R} \]
Step 3: Solve for \( v_{\max} \)
- Cancel \( m \) from both sides:
\[ \mu g = \frac{v^2}{R} \]
- Rearranging:
\[ v_{\max} = \sqrt{\mu g R} \]
Step 4: Substitute given values
\[ v_{\max} = \sqrt{(0.4) (9.8) (30)} \]
Step 5: Compute the value
\[ v_{\max} = \sqrt{117.6} \approx 10.84 { m/s} \]