Circular Motion Turning on Road without Friction Questions in English

(A) When a railway track is banked (the outside rail is raised),the normal reaction force exerted by the rails on the wheels is tilted at an angle to the vertical.
This normal reaction force can be resolved into two components: a vertical component that balances the weight of the train,and a horizontal component directed towards the center of the curve.
This horizontal inward component provides the necessary centripetal force required for the train to move along the curved path.
Therefore,the correct option is $A$.

(A) For a vehicle to negotiate a curved road of radius $R$ safely without relying on friction,the road is banked at an angle $\theta$.
From the dynamics of circular motion,the condition for safe banking is given by $\tan \theta = \frac{v^2}{Rg}$.
From the geometry of the banked road,where $h$ is the height of the outer edge and $b$ is the width of the road,we have $\tan \theta = \frac{h}{b}$ (assuming $\theta$ is small).
Equating the two expressions for $\tan \theta$:
$\frac{h}{b} = \frac{v^2}{Rg}$
Therefore,the height $h$ is given by $h = \frac{v^2 b}{Rg}$.

(B) Given: Radius $r = 50 \ m$,width $l = 10 \ m$,height $h = 1.5 \ m$,and acceleration due to gravity $g = 10 \ m/s^2$.
For a banked road without friction,the condition for safe turning is given by $\tan \theta = \frac{v^2}{rg}$.
From the geometry of the banked road,$\tan \theta = \frac{h}{l}$ (assuming small angle approximation).
Equating the two,we get $\frac{v^2}{rg} = \frac{h}{l}$.
Solving for $v$: $v = \sqrt{\frac{hrg}{l}}$.
Substituting the values: $v = \sqrt{\frac{1.5 \times 50 \times 10}{10}} = \sqrt{75} \approx 8.66 \ m/s$.
Rounding to the nearest given option,$v = 8.6 \ m/s$.

(C) The formula for banking of roads is given by $\tan \theta = \frac{v^2}{rg}$.
Since the angle of banking $\theta$ is constant,we have $r \propto v^2$.
Let the initial speed be $v_1 = v$ and the final speed be $v_2 = v + 0.1v = 1.1v$.
The initial radius is $r_1 = 20 \ m$.
Using the proportionality $r_2 / r_1 = (v_2 / v_1)^2$,we get:
$r_2 / 20 = (1.1v / v)^2 = (1.1)^2 = 1.21$.
Therefore,$r_2 = 1.21 \times 20 = 24.2 \ m$.

(C) Given: Mass $m = 1000\, kg$,Radius $R = 90\, m$,Banking angle $\theta = 45^\circ$,Acceleration due to gravity $g = 10\, m\,s^{-2}$.
For a car on a frictionless banked road,the condition for safe turning is given by the formula: $\tan \theta = \frac{v^2}{Rg}$.
Rearranging for velocity $v$: $v = \sqrt{Rg \tan \theta}$.
Substituting the values: $v = \sqrt{90 \times 10 \times \tan 45^\circ}$.
Since $\tan 45^\circ = 1$,we get: $v = \sqrt{900 \times 1} = 30\, m\,s^{-1}$.

(A) Banking of roads is done to provide the necessary centripetal force for safe turning of vehicles on curved paths.
When a road is banked at an angle $\theta$,the normal reaction $N$ has a horizontal component $N \sin \theta$ which acts towards the center of the circular path.
This component provides the required centripetal force,$F_c = \frac{mv^2}{r}$,which prevents the vehicle from skidding or falling outwards due to inertia.

(C) The condition for a car to experience no friction on a banked road is given by the formula: $\tan \theta = \frac{v^2}{rg}$.
Given: $\theta = 30^o$,$r = 10\sqrt{3} \ m$,and taking $g = 10 \ m/s^2$.
Substituting the values: $\tan 30^o = \frac{v^2}{(10\sqrt{3})(10)}$.
Since $\tan 30^o = \frac{1}{\sqrt{3}}$,we have: $\frac{1}{\sqrt{3}} = \frac{v^2}{100\sqrt{3}}$.
Solving for $v^2$: $v^2 = \frac{100\sqrt{3}}{\sqrt{3}} = 100$.
Thus,$v = 10 \ m/s$.
To convert the velocity into $km/hr$,multiply by $\frac{18}{5}$: $v = 10 \times \frac{18}{5} = 36 \ km/hr$.

(A) For a vehicle moving on a banked road at the safe speed (where no friction is required),the forces acting on the vehicle are the gravitational force $(mg)$ acting downwards and the normal force $(N)$ acting perpendicular to the banked surface.
The vertical component of the normal force balances the weight: $N \cos \theta = mg$.
The horizontal component of the normal force provides the necessary centripetal force: $N \sin \theta = m a_c$,where $a_c$ is the centripetal acceleration.
Dividing the two equations: $\frac{N \sin \theta}{N \cos \theta} = \frac{m a_c}{mg}$.
This simplifies to: $\tan \theta = \frac{a_c}{g}$.
Therefore,the centripetal acceleration is $a_c = g \tan \theta$.
Given the banking angle $\theta = 45^o$,we have $a_c = g \tan(45^o) = g(1) = g$.

(A) Let $m$ be the mass of the train,$R = 1000 \ m$ be the radius of the curve,and $\theta$ be the angle of banking.
In the frame of the train,the forces acting along the inclined plane are the component of gravity $mg \sin \theta$,the centrifugal force $\frac{mv^2}{R} \cos \theta$,and the lateral force $F$ exerted by the rails.
For speed $v_1 = 10 \ ms^{-1}$,the lateral force $F_1$ on the inner rail is given by:
$F_1 + \frac{mv_1^2}{R} \cos \theta = mg \sin \theta \implies F_1 = mg \sin \theta - \frac{mv_1^2}{R} \cos \theta$
For speed $v_2 = 20 \ ms^{-1}$,the lateral force $F_2$ on the outer rail is given by:
$\frac{mv_2^2}{R} \cos \theta = mg \sin \theta + F_2 \implies F_2 = \frac{mv_2^2}{R} \cos \theta - mg \sin \theta$
Given $F_1 = F_2$,we have:
$mg \sin \theta - \frac{mv_1^2}{R} \cos \theta = \frac{mv_2^2}{R} \cos \theta - mg \sin \theta$
$2mg \sin \theta = \frac{m}{R} (v_1^2 + v_2^2) \cos \theta$
$2g \tan \theta = \frac{v_1^2 + v_2^2}{R} = \frac{10^2 + 20^2}{1000} = \frac{100 + 400}{1000} = \frac{500}{1000} = 0.5$
Thus,$\tan \theta = \frac{0.5}{2g} = \frac{1}{4g}$.
Therefore,$4g \tan \theta = 4g \times \frac{1}{4g} = 1$.

(B) For a car moving on a banked road without friction,the condition for safe turning is given by the formula: $\tan \theta = \frac{v^2}{Rg}$.
Here,the banking angle $\theta = 45^\circ$,the radius of the curve $R = 90\,m$,and the acceleration due to gravity $g = 10\,m/s^2$.
Substituting these values into the formula:
$\tan 45^\circ = \frac{v^2}{90 \times 10}$.
Since $\tan 45^\circ = 1$,we have:
$1 = \frac{v^2}{900}$.
$v^2 = 900$.
$v = \sqrt{900} = 30\,m/s$.
Therefore,the maximum speed of the car is $30\,m/s$.

(D) The angle of banking $\theta$ is given by $\tan \theta = \frac{h}{w}$,where $h = 1.5\, m$ is the height difference and $w = 10\, m$ is the width of the road.
$\tan \theta = \frac{1.5}{10} = 0.15$.
For a banked road,the optimum velocity $v$ is given by the formula $v = \sqrt{rg \tan \theta}$,where $r = 50\, m$ is the radius of curvature and $g = 9.8\, m/s^2$ is the acceleration due to gravity.
Substituting the values: $v = \sqrt{50 \times 9.8 \times 0.15}$.
$v = \sqrt{490 \times 0.15} = \sqrt{73.5} \approx 8.57\, m/s$.
Rounding to the nearest option,the most suitable velocity is $8.5\, m/s$.

(D) For a car on a frictionless banked road,the condition for safe turning is given by the formula: $\tan \theta = \frac{v^2}{rg}$.
Here,the radius $r = 90 \, m$,the banking angle $\theta = 45^\circ$,and taking acceleration due to gravity $g = 10 \, m \, s^{-2}$.
Rearranging the formula for velocity $v$,we get: $v = \sqrt{rg \tan \theta}$.
Substituting the values: $v = \sqrt{90 \times 10 \times \tan 45^\circ}$.
Since $\tan 45^\circ = 1$,we have $v = \sqrt{900 \times 1} = 30 \, m \, s^{-1}$.

(C) The $Assertion$ is true because when a vehicle moves at an optimum speed $v = \sqrt{rg \tan \theta}$ on a banked road,the horizontal component of the normal reaction $(N \sin \theta)$ is sufficient to provide the necessary centripetal force $(mv^2/r)$.
In this specific case,the frictional force is not required to maintain the circular path.
The $Reason$ is incorrect because,on a banked road,the vehicle does not inherently tend to remain inwards without skidding; rather,the banking angle is designed specifically to balance the forces at a particular speed to prevent skidding.

(N/A) The velocity of a vehicle on a banked circular road is given by the general formula:
$v = \left[ rg \left( \frac{\mu_s + \tan \theta}{1 - \mu_s \tan \theta} \right) \right]^{1/2}$
For a smooth surface,the coefficient of static friction $\mu_s = 0$.
Substituting $\mu_s = 0$ into the formula:
$v_{\max} = \left[ rg \left( \frac{0 + \tan \theta}{1 - 0 \cdot \tan \theta} \right) \right]^{1/2}$
$v_{\max} = \left[ rg \tan \theta \right]^{1/2}$
Therefore,the formula for the maximum permissible speed on a smooth banked track is:
$v_{\max} = \sqrt{rg \tan \theta}$
At this speed,the horizontal component of the normal force provides the necessary centripetal force,and no frictional force is required.

(N/A) The velocity of a vehicle on a banked circular road is given by the general formula:
$v = \left[ r g \left( \frac{\mu_{s} + \tan \theta}{1 - \mu_{s} \tan \theta} \right) \right]^{\frac{1}{2}}$
For the optimum speed,we assume the road surface is smooth,meaning the coefficient of static friction $\mu_{s} = 0$.
Substituting $\mu_{s} = 0$ into the equation:
$v_{0} = \left[ r g \left( \frac{0 + \tan \theta}{1 - 0 \cdot \tan \theta} \right) \right]^{\frac{1}{2}}$
$v_{0} = \sqrt{r g \tan \theta}$
At this speed,the horizontal component of the normal force provides the necessary centripetal force,and no frictional force is required. Driving at this speed on a banked road minimizes wear and tear on the tyres. This specific velocity $v_{0}$ is known as the optimum speed.

(B) For a vehicle moving on a level circular path of radius $r$,the necessary centripetal force is provided by the static friction $f_s$ between the tires and the road.
$f_s = \frac{mv^2}{r}$
The maximum value of static friction is $f_{s,max} = \mu_s N = \mu_s mg$,where $\mu_s$ is the coefficient of static friction and $m$ is the mass of the vehicle.
For safe turning,the required centripetal force must be less than or equal to the maximum static friction:
$\frac{mv^2}{r} \leq \mu_s mg$
Dividing both sides by $m$,we get:
$\frac{v^2}{r} \leq \mu_s g$
$v^2 \leq \mu_s rg$
$v \leq \sqrt{\mu_s rg}$
Since the mass $m$ cancels out from both sides of the inequality,the maximum safe speed $v_{max} = \sqrt{\mu_s rg}$ is independent of the mass of the vehicle.

(N/A) Optimum speed is the speed at which a vehicle can negotiate a banked curve without relying on friction between the tires and the road surface. At this speed,the necessary centripetal force is provided entirely by the horizontal component of the normal force.
The equation for optimum speed $(v_0)$ on a banked road with banking angle $(\theta)$ and radius of curvature $(r)$ is given by:
$v_0 = \sqrt{rg \tan \theta}$
where:
$r$ = radius of the curve
$g$ = acceleration due to gravity
$\theta$ = angle of banking

(N/A) When a vehicle moves on a curved road,it undergoes circular motion. In circular motion,a centrifugal force acts on the vehicle,tending to push it outward. To prevent the vehicle from skidding or overturning,a centripetal force is required to keep it on the curved path. By banking the road (raising the outer edge higher than the inner edge),a component of the normal force provides the necessary centripetal force,reducing the reliance on friction.

(A) For a vehicle moving on a banked road without friction,the centripetal force required for circular motion is provided by the horizontal component of the normal force.
Let $N$ be the normal force and $m$ be the mass of the vehicle.
The vertical component of the normal force balances the weight: $N \cos \theta = mg$.
The horizontal component of the normal force provides the centripetal force: $N \sin \theta = \frac{mv^2}{r}$.
Dividing the two equations: $\frac{N \sin \theta}{N \cos \theta} = \frac{mv^2/r}{mg}$.
This simplifies to $\tan \theta = \frac{v^2}{rg}$.
Therefore,the maximum safe speed is $v = \sqrt{rg \tan \theta}$.

(C) For a vehicle to move in a circular path of radius $r$ on a flat road, the necessary centripetal force is provided by the static friction between the tires and the road surface.
Since the road is smooth (frictionless), the coefficient of friction $\mu = 0$.
The maximum safe speed $v$ on a flat circular road is given by $v = \sqrt{\mu rg}$.
Substituting $\mu = 0$, we get $v = \sqrt{0 \cdot rg} = 0$.
Therefore, it is impossible for a vehicle to maintain a circular path on a perfectly smooth flat road at any non-zero speed.

(B) On a banked road,the normal force $N$ exerted by the road on the vehicle acts perpendicular to the surface of the road.
This normal force can be resolved into two rectangular components:
$1$. The vertical component $N \cos \theta$,which balances the weight of the vehicle $(mg)$.
$2$. The horizontal component $N \sin \theta$,which points towards the center of the circular path.
Therefore,the horizontal component of the normal force provides the necessary centripetal force for the vehicle to turn safely without relying on friction.

(B) For a smooth banked road,the angle of banking $\theta$ is given by the relation: $\tan \theta = \frac{v^2}{Rg}$.
Given: $v = 10\ m/s$,$R = 150\ m$,and taking $g = 10\ m/s^2$.
Substituting the values: $\tan \theta = \frac{10^2}{150 \times 10} = \frac{100}{1500} = \frac{1}{15}$.
From the geometry of the banked road,if $H$ is the height of the outer edge and $w$ is the width of the road,then $\tan \theta = \frac{H}{w}$.
Given $w = 7.5\ m$,we have $\frac{H}{7.5} = \frac{1}{15}$.
Solving for $H$: $H = \frac{7.5}{15} = 0.5\ m$.

(A) For a banked track,the angle of banking $\theta$ is given by $\tan \theta = \frac{v^2}{rg}$.
From the geometry of the track,where $h$ is the elevation and $d$ is the distance between the rails,we have $\tan \theta = \frac{h}{d}$.
Equating the two expressions for $\tan \theta$:
$\frac{v^2}{rg} = \frac{h}{d}$
Rearranging the equation to solve for the radius of curvature $r$:
$r = \frac{v^2 d}{gh}$

(D) For a banked road, the optimum velocity $v$ is given by the formula $\tan \alpha = \frac{v^2}{Rg}$.
Here, $\alpha$ is the banking angle, $R$ is the radius of curvature, and $g$ is the acceleration due to gravity.
From the geometry of the road, $\tan \alpha = \frac{h}{w}$, where $h$ is the height difference between the outer and inner edges and $w$ is the width of the road.
Equating the two expressions for $\tan \alpha$: $\frac{v^2}{Rg} = \frac{h}{w}$.
Rearranging for $v$: $v = \sqrt{\frac{Rgh}{w}}$.
Given: $R = 50 \text{ m}$, $h = 1.5 \text{ m}$, $w = 10 \text{ m}$, and $g = 9.8 \text{ m/s}^2$.
Substituting the values: $v = \sqrt{\frac{50 \times 9.8 \times 1.5}{10}} = \sqrt{5 \times 9.8 \times 1.5} = \sqrt{73.5} \approx 8.57 \text{ m/s}$.
Comparing this with the given options, the most suitable velocity is $8.5 \text{ m/s}$.

(A) For a banked track,the angle of banking $\theta$ is given by $\tan \theta = \frac{v^2}{rg}$.
From the geometry of the track,$\sin \theta = \frac{h}{\ell}$.
Since the angle of banking $\theta$ is small,we can approximate $\tan \theta \approx \sin \theta$.
Therefore,$\frac{v^2}{rg} = \frac{h}{\ell}$.
Solving for $v$,we get $v^2 = rg \left(\frac{h}{\ell}\right)$.
Thus,the safety speed limit is $v = \sqrt{rg \left(\frac{h}{\ell}\right)}$.

(A) In a conical pendulum,the bob moves in a horizontal circle. The forces acting on the bob are:
$1$. The tension $T$ in the string,acting along the string towards the point of suspension.
$2$. The weight $mg$ of the bob,acting vertically downwards.
Resolving the tension $T$ into two rectangular components:
- The vertical component $T \cos \theta$ balances the weight of the bob $(T \cos \theta = mg)$.
- The horizontal component $T \sin \theta$ provides the necessary centripetal force required for circular motion $(T \sin \theta = \frac{mv^2}{r})$.
In the rotating frame of reference (non-inertial frame),the centrifugal force acts outwards,which is balanced by the horizontal component of the tension. Therefore,the component of tension that balances the centrifugal force is $T \sin \theta$.

(B) The radius of the circular path is $R = 50 \ m$.
The velocity of the lorry is $V = 20 \ ms^{-1}$.
The acceleration due to gravity is $g = 10 \ ms^{-2}$.
The formula for the banking angle $\theta$ of a road is given by $\tan \theta = \frac{V^2}{Rg}$.
Substituting the given values into the formula:
$\tan \theta = \frac{(20)^2}{50 \times 10} = \frac{400}{500} = \frac{4}{5}$.
Therefore,the banking angle is $\theta = \tan^{-1} \left( \frac{4}{5} \right)$.

(D) Given:
Mass of the truck, $M = 2000 \,kg$
Radius of curvature, $R = 10 \,m$
Banking angle, $\theta = 39^{\circ}$
Acceleration due to gravity, $g = 10 \,ms^{-2}$
Value of $\tan 39^{\circ} = 0.81$
The formula for the maximum permissible speed $v$ on a banked road (without friction) is given by:
$v = \sqrt{Rg \tan \theta}$
Substituting the given values into the formula:
$v = \sqrt{10 \times 10 \times 0.81}$
$v = \sqrt{100 \times 0.81}$
$v = \sqrt{81}$
$v = 9 \,ms^{-1}$
Thus, the maximum permissible speed of the truck is $9 \,ms^{-1}$.

(C) Given, radius of curvature, $r = 250 \,m$.
Maximum velocity of train, $v = 90 \,km/h$.
Converting velocity to $m/s$: $v = 90 \times \frac{5}{18} = 25 \,m/s$.
Let $\theta$ be the angle of banking.
The formula for the angle of banking is $\tan \theta = \frac{v^2}{rg}$.
Substituting the values: $\tan \theta = \frac{(25)^2}{250 \times 10} = \frac{625}{2500} = \frac{1}{4}$.
Therefore, $\theta = \tan^{-1}\left(\frac{1}{4}\right)$.

(B) Given,$v_1 = 14 \ m/s$ and $v_2 = 28 \ m/s$.
For a banked road,the angle of banking $\theta$ is related to the velocity $v$ and radius $r$ by the formula: $\tan \theta = \frac{v^2}{rg}$.
Rearranging for radius,we get $r = \frac{v^2}{g \tan \theta}$.
Since the ramp is similar,the banking angle $\theta$ remains constant.
Thus,$r \propto v^2$.
Therefore,$\frac{r_2}{r_1} = \left( \frac{v_2}{v_1} \right)^2 = \left( \frac{28}{14} \right)^2 = (2)^2 = 4$.
This implies $r_2 = 4r_1$.
Hence,the radius should be increased by a factor of $4$.