Statement I: A cyclist is moving on an unbanked road with a speed of 7 km/h and takes a sharp circul

Q: Why do we use static friction instead of kinetic friction if the car is moving?

Even though the car is moving forward, the tires are not sliding *sideways* across the road. The contact patch of the tire is momentarily stationary relative to the road in the radial direction, so static friction applies.

Q: Why is there a *minimum* speed on a banked road?

If the road is banked steeply and you drive too slowly, gravity will pull you down the incline. You need a minimum speed so that the 'outward' centrifugal tendency is strong enough to prevent you from sliding down.

Q: The question gave $g = 9.8 \text{ m/s}^2$, but the solution used $g = 10 \text{ m/s}^2$. Why?

In many competitive exams, $g = 10 \text{ m/s}^2$ is used to simplify calculations unless the exact precision of $9.8$ changes the final conclusion. Here, both $9.8$ and $10$ yield a $v_{\text{max}}$ greater than the cyclist's speed, so the approximation is safe.

Q: What happens if the cyclist exceeds $v_{\text{max}}$ on the banked road?

If the speed exceeds $v_{\text{max}}$, the required centripetal force becomes greater than what the horizontal component of the normal force and maximum static friction can provide. The cyclist will skid outwards up the incline.

Q: How does the banking angle $\theta$ affect the safe speed?

Increasing the banking angle increases the horizontal component of the normal force, which provides more centripetal force. This significantly raises the maximum safe speed, which is why high-speed racetracks have steeply banked curves.

Question

Statement I A cyclist is moving on an unbanked road with a speed of $7 	ext{ kmh}^{-1}$ and takes a sharp circular turn along a path of radius of 2 m without reducing the spee(d) The static friction coefficient is 0.2. The cyclist will not slip and pass the curve ($g = 9.8 	ext{ m} / 	ext{s}^2$)
Statement II If the road is banked at an angle of $45^\circ$, cyclist can cross the curve of 2 m radius with the speed of $18.5 	ext{ kmh}^{-1}$ without slipping.
In the light of the above statements, choose the correct answer from the options given below.

Options:

(A) Statement I is false and statement II is true.

(B) Statement I is true and statement II is false.

(C) Both statement I and statement II are false.

(D) Both statement I and statement II are true.

Sigma Percentile · Accepted Answer

### The Physics of Taking a Turn: Flat vs. Banked Roads

Taking a sharp turn on a bicycle or in a car is an everyday experience, yet it is governed by a beautiful interplay of invisible forces. When you turn, your body naturally wants to continue moving in a straight line due to inertia. To force you into a circular path, a net inward force—the **centripetal force**—is required. Let's explore how this force is generated on different types of roads.

### The Unbanked Turn: Relying on Friction

Imagine you are taking a turn on a perfectly flat, unbanked road. In this scenario, the only horizontal force capable of pulling you toward the center of the curve is the **static friction** between your tires and the road.

For you to safely navigate the turn without skidding outward, the required centripetal force must be less than or equal to the maximum available static friction:

$$ \frac{mv^2}{R} \le \mu_s mg $$

By canceling the mass $m$, we find the maximum safe speed:

$$ v_{	ext{max}} = \sqrt{\mu_s g R} $$

In Statement I of our problem, we are given $\mu = 0.2$, $R = 2 	ext{ m}$, and we can use $g = 10 	ext{ m/s}^2$ for simplicity. Plugging these in, we get $v_{	ext{max}} = \sqrt{0.2 	imes 10 	imes 2} = 2 	ext{ m/s}$. The cyclist's speed is $7 	ext{ km/h}$, which converts to approximately $1.94 	ext{ m/s}$. Since $1.94 < 2$, the cyclist is well within the safe limit and will not slip. Thus, Statement I is true.

### The Banked Turn: A Safer Path

Now, imagine the curved edge of a professional racing track. The road is tilted or **banked** at an angle $	heta$. This banking is a game-changer. Now, the normal force exerted by the road is no longer purely vertical. It tilts inward, meaning a component of the normal force ($N \sin	heta$) actively helps push you toward the center of the circle.

Because gravity and the normal force are doing some of the heavy lifting, you don't have to rely entirely on friction. This allows for much higher safe speeds.

### Calculating the Safe Speed Range

On a banked road with friction, there isn't just a maximum speed; there is a **safe speed range**. If you go too fast, you'll slide up the incline. If you go too slow, gravity will pull you down the incline. The formulas for these limits are:

$$ v_{	ext{max}} = \sqrt{\frac{gR(\mu + 	an	heta)}{1 - \mu 	an	heta}} $$

$$ v_{	ext{min}} = \sqrt{\frac{gR(	an	heta - \mu)}{1 + \mu 	an	heta}} $$

In Statement II, the road is banked at $	heta = 45^\circ$. Substituting $	an 45^\circ = 1$, $\mu = 0.2$, $R = 2 	ext{ m}$, and $g = 10 	ext{ m/s}^2$, we calculate:

$$ v_{	ext{max}} = \sqrt{\frac{20(1.2)}{0.8}} = \sqrt{30} \approx 5.47 	ext{ m/s} $$

$$ v_{	ext{min}} = \sqrt{\frac{20(0.8)}{1.2}} = \sqrt{13.33} \approx 3.65 	ext{ m/s} $$

The cyclist's speed is $18.5 	ext{ km/h}$, which converts to $5.13 	ext{ m/s}$. Since $5.13 	ext{ m/s}$ falls perfectly between $3.65 	ext{ m/s}$ and $5.47 	ext{ m/s}$, the cyclist is in the safe zone and will not slip. Therefore, Statement II is also true.

Both statements hold up to the rigorous laws of physics, making option (d) the correct choice.

Select Answer:

Visualized Solution

$Statement I: Unbanked Road$

$v_{max} on Unbanked Road$

$Calculating v_{max}$

$Checking Safe Speed$

$Statement II: Banked Road$

$Safe Speed Limits on Banked Road$

$Calculating v_{max}$

$Calculating v_{min}$

$Checking Safe Speed$

$Final Conclusion$

The Sigma Insight: Dynamics of Circular Motion

The Physics of Taking a Turn

The Unbanked Turn

The Banked Turn

Calculating the Safe Speed Range

Similar Questions

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

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 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

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

Select Answer:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Statement I: Unbanked Road

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } vmax​ on Unbanked Road

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

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Statement II: Banked Road

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Safe Speed Limits on Banked Road

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

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

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

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

The Sigma Insight: Dynamics of Circular Motion

The Physics of Taking a Turn

The Unbanked Turn

The Banked Turn

Calculating the Safe Speed Range

Similar Questions

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

$Statement I: Unbanked Road$

$v_{max} on Unbanked Road$

$Calculating v_{max}$

$Checking Safe Speed$

$Statement II: Banked Road$

$Safe Speed Limits on Banked Road$

$Calculating v_{max}$

$Calculating v_{min}$

$Checking Safe Speed$

$Final Conclusion$

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