The minimum velocity (in ms^-1) with which a car driver must traverse a flat curve of radius 150 m a

Q: Why do we use static friction and not kinetic friction? The car is moving!

Even though the car is moving forward, the tires are not sliding *sideways* (radially) across the road. The contact patch of the tire is momentarily stationary relative to the road in the radial direction. As long as there is no radial slipping, the friction acting towards the center is static friction.

Q: Why does the mass of the car not matter? Shouldn't a heavier car be harder to turn?

It's a fascinating balance! A heavier car does indeed require a larger centripetal force to turn (since $F_c \propto m$). However, a heavier car also presses down harder on the road, which increases the normal reaction and proportionally increases the maximum available friction ($f_{\text{max}} \propto m$). Because both the required force and the available force scale linearly with mass, the mass cancels out entirely.

Q: What happens if the car goes faster than $30\text{ m/s}$?

If the velocity exceeds $30\text{ m/s}$, the required centripetal force ($\frac{mv^2}{r}$) will become greater than the maximum static friction ($\mu mg$) the road can provide. The friction will 'break', transitioning to weaker kinetic friction, and the car will skid outwards, failing to follow the curve.

Q: Why does the question ask for 'minimum' velocity? Isn't $30\text{ m/s}$ the maximum?

You are absolutely correct. The phrasing in the original exam question is slightly flawed. It should ask for the 'maximum' safe velocity. Any speed below $30\text{ m/s}$ is safe. However, in the context of such problems, it implies finding the threshold or boundary velocity.

Q: Would putting wider tires on the car increase this maximum speed?

Surprisingly, according to the standard Coulomb model of friction ($f = \mu N$), the maximum frictional force is independent of the contact area. So, theoretically, wider tires wouldn't change the limit. In reality, tire dynamics are more complex, and wider tires can offer better heat dissipation and slight mechanical keying advantages, but the primary physics remains governed by $\mu$ and $N$.

Question

The minimum velocity (in $	ext{ms}^{-1}$) with which a car driver must traverse a flat curve of radius $150	ext{ m}$ and coefficient of friction $0.6$ to avoid skidding is

Options:

(A) 60

(B) 30

(C) 15

(D) 25

Sigma Percentile · Accepted Answer

### The Physics of a Flat Turn

Imagine you are driving a car and approaching a sharp, flat circular turn. As you steer the wheel, your car gracefully follows the curve. But what invisible hand is pulling your car towards the center of the turn, preventing it from continuing in a straight line and crashing off the road?

That invisible hand is **static friction**. When the tires try to slide outward due to inertia, the rough surface of the road grips them, providing a force directed exactly towards the center of the circular path. This is the required **centripetal force**.

### Balancing the Forces

To understand the limits of this turn, we must look at all the forces acting on the car. Vertically, the car is in equilibrium. The Earth pulls it down with its weight, $mg$, and the road pushes back up with an equal and opposite normal reaction, $N$. Therefore, we can write:

$$N = mg$$

The maximum grip the road can offer before the tires start to slide is the maximum static frictional force, denoted by $f_{	ext{max}}$. According to the laws of friction, this is proportional to the normal reaction:

$$f_{	ext{max}} = \mu N = \mu mg$$

### Calculating the Safe Speed

For the car to successfully navigate the turn of radius $r$ at a velocity $v$, it demands a centripetal force equal to $\frac{mv^2}{r}$. This demand must be met by the available friction. Thus, for a safe, non-skidding turn, the required force must be less than or equal to the maximum available friction:

$$\frac{mv^2}{r} \le \mu mg$$

Here lies a beautiful piece of physics: **the mass $m$ cancels out from both sides!**

$$v^2 \le \mu rg$$

This profound result tells us that the maximum safe speed on a flat curve is entirely independent of the vehicle's mass. A massive eighteen-wheeler truck and a lightweight sports car both share the exact same speed limit on this curve, provided their tires have the same coefficient of friction with the road.

Let's calculate this maximum velocity, $v_{	ext{max}}$, using the values given in our problem. We have a coefficient of friction $\mu = 0.6$, a curve radius $r = 150	ext{ m}$, and we'll use the standard approximation for gravity, $g = 10	ext{ m/s}^2$.

$$v_{	ext{max}} = \sqrt{0.6 	imes 150 	imes 10}$$

$$v_{	ext{max}} = \sqrt{900}$$

$$v_{	ext{max}} = 30	ext{ m/s}$$

Therefore, the driver must keep the speed at or below $30	ext{ m/s}$ to avoid skidding outward. *(Note: The original question asks for the 'minimum' velocity to avoid skidding, which is a slight phrasing error common in exams; physically, it is the 'maximum' safe velocity before skidding occurs.)*

Animated Solution for Physics - Laws of Motion: The minimum velocity (in $ms^{- 1}$ ) with which a car driver must traverse a flat curve of radius $150 m$ and coefficient of friction $0.6$ to avoid skidding is

Select Answer:

Visualized Solution

$Visualizing the Turn$

$The Centripetal Force$

$Balancing Vertical Forces$

$Condition for No Skidding$

$The Safe Speed Limit$

$Substituting Values$

$Final Calculation$

$The Way Forward$

The Sigma Insight: Dynamics of Circular Motion

The Physics of a Flat Turn

Balancing the Forces

Calculating the Safe Speed

Similar Questions

The normal reaction $N$ for a vehicle of $800 kg$ mass, negotiating a turn on a $3 0^{\circ}$ banked road at maximum possible speed without skidding is $\dots \times 1 0^{3} kg-m/s^{2}$ . [Take, $μ_{s} = 0.2$ ]

A car is moving in a circular horizontal track of radius $10 m$ with a constant speed of $10 m/s$ . A plumb bob is suspended from the roof of the car by a light rigid rod. The angle made by the rod with the vertical is (Take $g = 10 m/s^{2}$ )

A block of $200 g$ mass moves with a uniform speed in a horizontal circular groove, with vertical side walls of radius $20 cm$ . If the block takes $40 s$ to complete one round, the normal force by the side walls of the groove is

A particle of mass $m$ is suspended from a ceiling through a string of length $L$ . The particle moves in a horizontal circle of radius $r$ such that $r = \frac{L}{2}$ . The speed of particle will be

A particle is moving with uniform speed along the circumference of a circle of radius $R$ under the action of a central fictitious force $F$ which is inversely proportional to $R^{3}$ . Its time period of revolution will be given by

Select Answer:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Visualizing the Turn

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The Centripetal Force

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Balancing Vertical Forces

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Condition for No Skidding

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The Safe Speed Limit

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Substituting Values

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final Calculation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The Way Forward