Circular motion with Friction Questions in English

(A) For a body of mass $m$ moving in a horizontal circular path of radius $r$ with speed $v$,the centripetal force required is $F_c = \frac{mv^2}{r}$.
This centripetal force is provided by the static friction force $f_s$ acting between the body and the surface.
The maximum value of static friction is $f_{s,max} = \mu N = \mu mg$.
For the body to move in a circular path without skidding,the required centripetal force must be less than or equal to the maximum available static friction.
Therefore,the condition for no skidding is $\frac{mv^2}{r} \leq \mu mg$.

(D) For a car to safely negotiate a turn on a flat road,the centripetal force is provided by the static friction force.
$F_c = F_f$
$\frac{mv^2}{r} = \mu mg$
$\frac{v^2}{r} = \mu g$
Rearranging for the radius $r$:
$r = \frac{v^2}{\mu g}$
Given $v = 10\,m/s$,$\mu = 0.5$,and taking $g = 10\,m/s^2$:
$r = \frac{10^2}{0.5 \times 10} = \frac{100}{5} = 20\,m$.
Thus,the minimum radius of the turn is $20\,m$.

(B) For a car moving on a level circular road,the centripetal force required for turning is provided by the static friction between the tyres and the road.
The condition for the car not to skid is that the required centripetal force must be less than or equal to the maximum static frictional force.
$F_{c} \leq f_{s,max}$
$\frac{mv^{2}}{R} \leq \mu mg$
$v^{2} \leq \mu gR$
$v_{max} = \sqrt{\mu gR}$
Given: $\mu = 0.2,$ $R = 450\,m,$ and taking $g = 10\,m/s^{2}.$
$v_{max} = \sqrt{0.2 \times 10 \times 450}$
$v_{max} = \sqrt{2 \times 450}$
$v_{max} = \sqrt{900}$
$v_{max} = 30\,m/s.$

(B) For a motorcyclist to negotiate a curve of radius $r$ with speed $v$ safely,the centripetal force required is provided by the static frictional force between the tires and the road.
The required centripetal force is given by $F_c = \frac{mv^2}{r}$.
The maximum available frictional force is $f_{max} = \mu N = \mu mg$,where $\mu$ is the coefficient of friction and $N = mg$ is the normal force.
For safe negotiation,the frictional force must be at least equal to the required centripetal force:
$f_{max} \geq F_c$
$\mu mg \geq \frac{mv^2}{r}$
$\mu \geq \frac{v^2}{gr}$
Therefore,the minimum value of the coefficient of friction is $\mu_{min} = \frac{v^2}{gr}$.

(C) The coin placed on the rotating table slips when the centripetal force $mr\omega^2$ is greater than or equal to the maximum static frictional force $f_{max} = \mu mg$.
The condition for the coin to just slip is given by:
$mr\omega^2 = \mu mg$
This simplifies to:
$r = \frac{\mu g}{\omega^2}$
From this relation,we can see that $r \propto \frac{1}{\omega^2}$.
Let the initial distance be $r_1 = 4r$ at angular velocity $\omega_1 = \omega$.
Let the new distance be $r_2$ at angular velocity $\omega_2 = 2\omega$.
Using the proportionality $r_1 \omega_1^2 = r_2 \omega_2^2$:
$(4r)(\omega)^2 = r_2 (2\omega)^2$
$4r \cdot \omega^2 = r_2 \cdot 4\omega^2$
$r_2 = r$
Thus,the coin will just slip when it is at a distance $r$ from the centre.

(B) To avoid skidding on a flat circular curve,the centripetal force required is provided by the static friction force.
$f_{L} \geq \frac{mv^{2}}{r}$
For maximum speed $(v_{\max})$,the limiting friction is equal to the centripetal force:
$f_{L} = \frac{mv_{\max}^{2}}{r}$
Since $f_{L} = \mu mg$,we have:
$\mu mg = \frac{mv_{\max}^{2}}{r}$
$v_{\max} = \sqrt{\mu rg}$
Given $\mu = 0.6$,$r = 150\, m$,and taking $g = 10\, m/s^2$:
$v_{\max} = \sqrt{0.6 \times 150 \times 10}$
$v_{\max} = \sqrt{900}$
$v_{\max} = 30\, m/s$.

(C) For the block to remain stationary,the upward frictional force must balance the downward gravitational force.
$f_s = mg$
The normal force $N$ provided by the wall acts as the centripetal force for the block's circular motion:
$N = mr\omega^2$
The maximum static frictional force is given by:
$f_{L} = \mu N = \mu mr\omega^2$
For the block to stay in equilibrium,the frictional force must be at least equal to the weight:
$f_{L} \geq mg$
$\mu mr\omega^2 \geq mg$
Solving for $\omega$:
$\omega^2 \geq \frac{g}{\mu r}$
$\omega \geq \sqrt{\frac{g}{\mu r}}$
Substituting the given values ($g = 10\; m/s^2$,$\mu = 0.1$,$r = 1\; m$):
$\omega_{\min} = \sqrt{\frac{10}{0.1 \times 1}} = \sqrt{\frac{10}{0.1}} = \sqrt{100} = 10\; rad/s$

(A) On an unbanked road,the frictional force provides the necessary centripetal force for circular motion. The condition for the cyclist not to slip is $v^{2} \leq \mu_{s} R g$.
Given: $v = 18 \; km/h = 18 \times \frac{5}{18} = 5 \; m/s$,$R = 3 \; m$,$\mu_{s} = 0.1$,and $g = 9.8 \; m/s^{2}$.
Calculating the maximum allowed speed squared: $\mu_{s} R g = 0.1 \times 3 \times 9.8 = 2.94 \; m^{2}/s^{2}$.
Calculating the actual speed squared: $v^{2} = (5)^{2} = 25 \; m^{2}/s^{2}$.
Since $v^{2} > \mu_{s} R g$ $(25 > 2.94)$,the frictional force is insufficient to provide the required centripetal force. Therefore,the cyclist will slip.

(A) On a banked road,the horizontal component of the normal force and the frictional force contribute to provide the centripetal force required to keep the car moving on a circular turn without slipping.
$(a)$ At the optimum speed,the horizontal component of the normal reaction is sufficient to provide the necessary centripetal force,and the frictional force is not required. The optimum speed $v_{o}$ is given by the formula:
$v_{o} = \sqrt{R g \tan \theta}$
Given $R = 300 \; m$,$\theta = 15^{\circ}$,and $g = 9.8 \; m/s^{2}$:
$v_{o} = \sqrt{300 \times 9.8 \times \tan(15^{\circ})} = \sqrt{2940 \times 0.2679} \approx \sqrt{787.6} \approx 28.1 \; m/s$.
$(b)$ The maximum permissible speed $v_{\max}$ to avoid slipping is given by the formula:
$v_{\max} = \sqrt{R g \left( \frac{\mu_{s} + \tan \theta}{1 - \mu_{s} \tan \theta} \right)}$
Given $\mu_{s} = 0.2$:
$v_{\max} = \sqrt{300 \times 9.8 \times \left( \frac{0.2 + 0.2679}{1 - (0.2 \times 0.2679)} \right)} = \sqrt{2940 \times \left( \frac{0.4679}{0.9464} \right)} \approx \sqrt{2940 \times 0.4944} \approx \sqrt{1453.5} \approx 38.1 \; m/s$.

(A) Mass of the stone,$m = 0.25 \; kg$.
Radius of the circle,$r = 1.5 \; m$.
Frequency of revolution,$n = \frac{40}{60} = \frac{2}{3} \; rps$.
Angular velocity,$\omega = 2 \pi n = 2 \times 3.1416 \times \frac{2}{3} \approx 4.189 \; rad/s$.
The centripetal force is provided by the tension $T$ in the string: $T = m r \omega^2$.
$T = 0.25 \times 1.5 \times (4.189)^2 \approx 6.58 \; N$.
For the maximum speed,$T_{\max} = 200 \; N$.
Using $T_{\max} = \frac{m v_{\max}^2}{r}$,we get $v_{\max} = \sqrt{\frac{T_{\max} \times r}{m}}$.
$v_{\max} = \sqrt{\frac{200 \times 1.5}{0.25}} = \sqrt{1200} \approx 34.64 \; m/s$.

(D) Radius of the circular track,$r = 30 \; m$.
Speed of the train,$v = 54 \; km/h = 54 \times \frac{5}{18} \; m/s = 15 \; m/s$.
Mass of the train,$m = 10^{6} \; kg$.
The centripetal force is provided by the lateral thrust exerted by the rails on the wheels of the train. According to Newton's third law of motion,the wheels exert an equal and opposite force on the rails,which causes wear and tear.
The angle of banking $\theta$ required to prevent wear and tear is given by the relation:
$\tan \theta = \frac{v^{2}}{rg}$
Substituting the values:
$\tan \theta = \frac{(15)^{2}}{30 \times 9.8} = \frac{225}{294} \approx 0.765$
$\theta = \tan^{-1}(0.765) \approx 37.4^{\circ}$.
(Note: Using $g = 10 \; m/s^{2}$,$\tan \theta = \frac{225}{300} = 0.75$,so $\theta = \tan^{-1}(0.75) \approx 36.87^{\circ}$).

(B) For a coin to revolve with the disc,the frictional force must be greater than or equal to the required centripetal force: $\mu mg \geq mr\omega^2$,or $\mu g \geq r\omega^2$.
Given:
Frequency $\nu = 33 \frac{1}{3} \; rev/min = \frac{100}{3 \times 60} = \frac{5}{9} \; rev/s$.
Angular velocity $\omega = 2\pi\nu = 2 \times \pi \times \frac{5}{9} = \frac{10\pi}{9} \approx 3.49 \; rad/s$.
Maximum acceleration provided by friction $a_{max} = \mu g = 0.15 \times 10 = 1.5 \; m/s^2$.
For the coin at $r_1 = 4 \; cm = 0.04 \; m$:
Required centripetal acceleration $a_1 = r_1\omega^2 = 0.04 \times (3.49)^2 \approx 0.49 \; m/s^2$.
Since $a_1 < a_{max}$ $(0.49 < 1.5)$,the coin at $4 \; cm$ will revolve with the disc.
For the coin at $r_2 = 14 \; cm = 0.14 \; m$:
Required centripetal acceleration $a_2 = r_2\omega^2 = 0.14 \times (3.49)^2 \approx 1.70 \; m/s^2$.
Since $a_2 > a_{max}$ $(1.70 > 1.5)$,the coin at $14 \; cm$ will slip.

(C) Mass of the man,$m = 70 \; kg$.
Radius of the drum,$r = 3 \; m$.
Coefficient of friction,$\mu = 0.15$.
Acceleration due to gravity,$g = 10 \; m/s^2$.
For the man to remain stuck to the wall,the frictional force $f$ must balance the weight of the man,$mg$.
$f = \mu F_N = mg$,where $F_N$ is the normal force provided by the wall.
The normal force provides the necessary centripetal force: $F_N = mr\omega^2$.
Substituting $F_N$ into the friction equation: $\mu(mr\omega^2) = mg$.
Simplifying for angular velocity $\omega$: $\omega^2 = \frac{g}{\mu r}$.
$\omega = \sqrt{\frac{g}{\mu r}} = \sqrt{\frac{10}{0.15 \times 3}} = \sqrt{\frac{10}{0.45}} = \sqrt{22.22} \approx 4.71 \; rad/s$.

(N/A) Consider a vehicle of mass $m$ moving on a level curved road of radius $R$. The forces acting on the vehicle are:
$(1)$ The gravitational force $(mg)$ acting downwards.
$(2)$ The normal reaction force $(N)$ acting upwards from the road surface. Since there is no vertical motion,$N = mg$.
$(3)$ The static frictional force $(f_s)$ acting towards the center of the circular path,which provides the necessary centripetal force.
For safe turning,the required centripetal force must be provided by the static friction:
$\frac{mv^2}{R} \leq f_s$
Since the maximum value of static friction is $f_{s,max} = \mu_s N = \mu_s mg$,we have:
$\frac{mv_{max}^2}{R} = \mu_s mg$
Solving for $v_{max}$:
$v_{max}^2 = \mu_s Rg$
$v_{max} = \sqrt{\mu_s Rg}$
Where $\mu_s$ is the coefficient of static friction between the tyres and the road surface.

(N/A) For a vehicle of mass $m$ moving on a banked road with angle $\theta$ and radius $R$,the forces acting are the normal force $N$,the gravitational force $mg$,and the maximum static friction $f_s = \mu_s N$.
Resolving forces vertically:
$N \cos \theta = mg + f_s \sin \theta$
$N \cos \theta = mg + \mu_s N \sin \theta$
$N(\cos \theta - \mu_s \sin \theta) = mg$ --- $(1)$
Resolving forces horizontally (providing centripetal force):
$N \sin \theta + f_s \cos \theta = \frac{mv_{max}^2}{R}$
$N \sin \theta + \mu_s N \cos \theta = \frac{mv_{max}^2}{R}$
$N(\sin \theta + \mu_s \cos \theta) = \frac{mv_{max}^2}{R}$ --- $(2)$
Dividing $(2)$ by $(1)$:
$\frac{\sin \theta + \mu_s \cos \theta}{\cos \theta - \mu_s \sin \theta} = \frac{v_{max}^2}{Rg}$
Dividing numerator and denominator by $\cos \theta$:
$\frac{\tan \theta + \mu_s}{1 - \mu_s \tan \theta} = \frac{v_{max}^2}{Rg}$
Thus,$v_{max} = \sqrt{Rg \left( \frac{\mu_s + \tan \theta}{1 - \mu_s \tan \theta} \right)}$.

(B) When a vehicle moves on a level circular path,it requires a centripetal force to maintain its circular motion.
This necessary centripetal force is provided by the static frictional force acting between the tires of the vehicle and the surface of the road.
The frictional force acts towards the center of the circular path,thus satisfying the requirement for centripetal force $(F_c = \frac{mv^2}{r})$.
If the required centripetal force exceeds the maximum available static friction $(f_{s,max} = \mu_s N)$,the vehicle will skid.

(B) The maximum safe speed on a level circular road is given by $v_{max} = \sqrt{\mu R g}$,where $\mu$ is the coefficient of friction,$R$ is the radius,and $g$ is the acceleration due to gravity.
On a banked road (circular road with slope $\theta$),the maximum speed is given by $v_{max} = \sqrt{Rg \left( \frac{\mu + \tan \theta}{1 - \mu \tan \theta} \right)}$.
Since $\tan \theta > 0$,the term $\left( \frac{\mu + \tan \theta}{1 - \mu \tan \theta} \right)$ is greater than $\mu$.
Therefore,a circular road with a slope (banked road) allows for a higher maximum speed compared to a level circular road.

(C) When a vehicle takes a turn on a level circular track,it tends to skid outwards due to inertia.
To prevent this,the force of static friction between the tires and the road surface acts towards the center of the circular path.
This static friction provides the necessary centripetal force required for circular motion.
The condition for safe turning is $f_s \leq \mu_s N$,where $f_s$ is the static friction,$\mu_s$ is the coefficient of static friction,and $N$ is the normal reaction.
Thus,the correct option is $C$.

(A) For a vehicle moving on a flat curved road,the centripetal force required for circular motion is provided by the static friction force between the tires and the road.
Let $m$ be the mass of the vehicle,$v$ be its velocity,$r$ be the radius of the curve,and $\mu_s$ be the coefficient of static friction.
The required centripetal force is $F_c = \frac{mv^2}{r}$.
The maximum available static friction force is $f_{s,max} = \mu_s N = \mu_s mg$.
For safe turning,the centripetal force must be less than or equal to the maximum static friction force: $\frac{mv^2}{r} \leq \mu_s mg$.
Solving for $v$,we get $v^2 \leq \mu_s rg$,which gives the maximum safe speed as $v_{max} = \sqrt{\mu_s rg}$.

(A) The maximum safe speed on a flat curved road is given by $v_{max} = \sqrt{\mu_s rg}$,where $\mu_s$ is the coefficient of static friction.
On a banked road with banking angle $\theta$,the maximum safe speed is given by $v_{max} = \sqrt{rg \tan \theta}$ (ignoring friction).
Even with friction,banking provides an additional centripetal force component,allowing for a higher speed compared to a flat road where the vehicle relies solely on friction.
Therefore,the maximum safe speed on a banked curved road is higher than on a flat curved road.

(B) When a vehicle moves at the optimum speed (also known as the safe speed) on a banked road,the necessary centripetal force is provided entirely by the horizontal component of the normal reaction force.
In this condition,there is no requirement for frictional force between the tires and the road surface to provide the centripetal force.
Since the frictional force is zero,the wear and tear on the tires is minimum.

(A) The optimum speed $v_0$ on a banked road is given by $v_0 = \sqrt{rg \tan \theta}$.
When the vehicle moves at a speed $v < v_0$,it has a tendency to slide down the incline due to the component of gravity being greater than the required centripetal force provided by the normal reaction.
To prevent this downward sliding,the static frictional force acts upwards along the incline.

(B) The maximum safe speed $v$ of a vehicle on a horizontal curved road is given by the formula $v = \sqrt{\mu r g}$.
Given:
Coefficient of friction $\mu = 0.25$
Radius of the curve $r = 20 \ m$
Acceleration due to gravity $g = 9.8 \ m/s^2$
Substituting the values:
$v = \sqrt{0.25 \times 20 \times 9.8}$
$v = \sqrt{5 \times 9.8}$
$v = \sqrt{49}$
$v = 7 \ m/s$.
Thus,the safe speed of the vehicle is $7 \ m/s$.

(B) In the rotating frame of reference,the forces acting on the bead are the gravitational force $mg$ (downwards),the centrifugal force $m x \omega^2$ (outwards),and the normal reaction $N$ from the wire.
For the bead to remain stationary at point $P(a, b)$,the net force along the tangent to the parabola must be zero.
The slope of the parabola $y = 4Cx^2$ is given by $\frac{dy}{dx} = 8Cx$.
At point $P(a, b)$,the slope is $\tan \theta = 8Ca$,where $\theta$ is the angle the tangent makes with the horizontal.
Resolving forces along the tangent,we have $m x \omega^2 \cos \theta = mg \sin \theta$.
This simplifies to $x \omega^2 = g \tan \theta$.
Substituting $x = a$ and $\tan \theta = 8Ca$,we get $a \omega^2 = g(8Ca)$.
Thus,$\omega^2 = 8gC$,which gives $\omega = \sqrt{8gC} = 2\sqrt{2gC}$.

(C) For the block to remain stationary on the rotating disc,the centripetal force must be provided by the static friction force.
Let $m$ be the mass of the block,$\omega$ be the angular frequency,and $x$ be the distance from the axis.
The condition for no sliding is: $f_s \leq \mu_s N$.
Since $N = mg$ and $f_s = m\omega^2 x$,we have $m\omega^2 x \leq \mu_s mg$.
Thus,$x \leq \frac{\mu_s g}{\omega^2}$.
Given $\mu_s = 0.4$,$g = 10\, m/s^2$,and $\omega = 10\, rad/s$:
$x = \frac{0.4 \times 10}{10^2} = \frac{4}{100}\, m$.
Converting to centimeters: $x = 0.04\, m \times 100 = 4\, cm$.

(A) For a car moving in a circular path on a flat track,the centripetal force is provided by the static friction force $f_{s}$.
The maximum static friction force is given by $f_{s,max} = \mu_{s} N$,where $N$ is the normal force.
The centripetal force required is $F_{c} = \frac{mv^{2}}{R}$.
Thus,$\mu_{s} N = \frac{mv^{2}}{R}$,which gives $N = \frac{mv^{2}}{\mu_{s} R}$.
The normal force $N$ on the car is the sum of its weight $mg$ and the downward negative lift force $F_{L}$ (aerodynamic downforce).
Therefore,$N = mg + F_{L}$.
Substituting the expression for $N$,we get $\frac{mv^{2}}{\mu_{s} R} = mg + F_{L}$.
Solving for $F_{L}$,we get $F_{L} = \frac{mv^{2}}{\mu_{s} R} - mg = m \left(\frac{v^{2}}{\mu_{s} R} - g\right)$.

(D) At maximum speed $V_{\max}$,the frictional force $f$ acts downwards along the incline and is limiting in nature,so $f = \mu N$.
Resolving forces in the vertical direction:
$N \cos 30^{\circ} - mg - f \sin 30^{\circ} = 0$
Substituting $f = \mu N$:
$N \cos 30^{\circ} - mg - \mu N \sin 30^{\circ} = 0$
$N (\cos 30^{\circ} - \mu \sin 30^{\circ}) = mg$
Given $m = 800 \, kg$,$g = 10 \, m/s^{2}$,$\cos 30^{\circ} = 0.87$,$\sin 30^{\circ} = 0.5$,and $\mu = 0.2$:
$N (0.87 - 0.2 \times 0.5) = 800 \times 10$
$N (0.87 - 0.1) = 8000$
$N (0.77) = 8000$
$N = \frac{8000}{0.77} \approx 10389.6 \, N \approx 10.4 \times 10^{3} \, N$.
Note: Based on the provided options and the calculation method in the prompt,the closest value is $10.2 \times 10^{3} \, N$.

(B) For the beaker to revolve with the disc without slipping,the necessary centripetal force must be provided by the static friction force.
The required centripetal force is $F_{c} = m \omega^{2} R$,where $m$ is the mass of the beaker.
The maximum available static friction force is $f_{s,max} = \mu N = \mu mg$,where $N = mg$ is the normal force.
For the beaker to move in a circular path with the disc,the static friction must be greater than or equal to the required centripetal force:
$f_{s} \leq f_{s,max}$
Substituting the expressions:
$m \omega^{2} R \leq \mu mg$
Dividing both sides by $m \omega^{2}$:
$R \leq \frac{\mu g}{\omega^{2}}$

(A) The maximum speed $v$ of a car on a level curved road is given by the formula $v = \sqrt{\mu Rg}$,where $\mu$ is the coefficient of friction,$R$ is the radius of the curve,and $g$ is the acceleration due to gravity.
Since $\mu$ and $g$ are constant,we have $v \propto \sqrt{R}$.
Therefore,the ratio of the speeds is $\frac{v_2}{v_1} = \sqrt{\frac{R_2}{R_1}}$.
Given $v_1 = 30 \, m/s$,$R_1 = 75 \, m$,and $R_2 = 48 \, m$.
Substituting these values: $\frac{v_2}{30} = \sqrt{\frac{48}{75}}$.
Simplifying the fraction inside the square root: $\frac{48}{75} = \frac{16}{25}$.
So,$\frac{v_2}{30} = \sqrt{\frac{16}{25}} = \frac{4}{5}$.
$v_2 = 30 \times \frac{4}{5} = 6 \times 4 = 24 \, m/s$.

(C) The plank is resting on the Earth's surface,which is rotating about its axis with angular velocity $\omega$. The plank undergoes circular motion with radius $r = r_e \cos 45^{\circ}$.
The centripetal force required for this circular motion is $F_c = m \omega^2 r = m \omega^2 (r_e \cos 45^{\circ})$.
This centripetal force acts towards the axis of rotation. The horizontal ground at latitude $45^{\circ}$ is inclined at an angle of $45^{\circ}$ with the axis of rotation. The component of the centripetal force acting parallel to the ground provides the necessary frictional force to keep the plank at rest relative to the ground.
$f = F_c \sin 45^{\circ} = (m \omega^2 r_e \cos 45^{\circ}) \sin 45^{\circ}$
Since $\sin 45^{\circ} = \cos 45^{\circ} = \frac{1}{\sqrt{2}}$,we have:
$f = m \omega^2 r_e \left(\frac{1}{\sqrt{2}}\right) \left(\frac{1}{\sqrt{2}}\right) = \frac{m r_e \omega^2}{2}$

(C) On a banked road with friction,the forces acting on the vehicle are the normal reaction $R$,the frictional force $f = \mu R$,the weight $mg$,and the centripetal force $\frac{mv^2}{r}$.
Resolving forces in the horizontal and vertical directions:
Horizontal: $R \sin \theta + f \cos \theta = \frac{mv^2}{r}$
Vertical: $R \cos \theta - f \sin \theta = mg$
Dividing the horizontal equation by the vertical equation:
$\frac{R \sin \theta + \mu R \cos \theta}{R \cos \theta - \mu R \sin \theta} = \frac{mv^2/r}{mg}$
$\frac{\sin \theta + \mu \cos \theta}{\cos \theta - \mu \sin \theta} = \frac{v^2}{rg}$
Dividing numerator and denominator by $\cos \theta$:
$\frac{\tan \theta + \mu}{1 - \mu \tan \theta} = \frac{v^2}{rg}$
Cross-multiplying:
$rg(\tan \theta + \mu) = v^2(1 - \mu \tan \theta)$
$rg \tan \theta + \mu rg = v^2 - \mu v^2 \tan \theta$
$\tan \theta(rg + \mu v^2) = v^2 - \mu rg$
$\tan \theta = \frac{v^2 - \mu rg}{rg + \mu v^2}$

(B) For the block not to slip,the required centripetal force must be provided by the static friction force.
The condition for the block not to slip is:
$f_{s} \leq \mu N$
Here,the centripetal force is $F_{c} = m \omega^2 x$.
The normal force $N$ on the horizontal surface is $N = mg$.
Therefore,the maximum static friction force is $f_{s,max} = \mu mg$.
Equating the centripetal force to the maximum static friction force:
$m \omega^2 x = \mu mg$
Dividing both sides by $mx$:
$\omega^2 = \frac{\mu g}{x}$
Taking the square root of both sides,we get the maximum angular speed:
$\omega = \sqrt{\frac{\mu g}{x}}$

(C) For a cyclist taking a turn on a circular road,the angle of inclination $\theta$ with the vertical is given by the formula: $\tan \theta = \frac{v^2}{rg}$.
Given values are: speed $v = 14 \sqrt{3} \, m/s$,radius $r = 20 \sqrt{3} \, m$,and acceleration due to gravity $g \approx 10 \, m/s^2$.
Substituting these values into the formula:
$\tan \theta = \frac{(14 \sqrt{3})^2}{20 \sqrt{3} \times 10} = \frac{196 \times 3}{200 \sqrt{3}} = \frac{588}{200 \sqrt{3}} = \frac{2.94}{\sqrt{3}} \approx \sqrt{3}$.
Since $\tan \theta = \sqrt{3}$,we find that $\theta = 60^{\circ}$.

(B) For a car moving on a horizontal circular track,the centripetal force is provided by the static friction between the tyres and the road.
For safe turning without skidding,the condition is: $F_c \leq f_{max}$.
$\frac{mv^2}{r} \leq \mu mg$.
Thus,the maximum speed $v_{max} = \sqrt{\mu rg}$.
Given: $\mu = 0.45$,$r = 0.2 \, km = 200 \, m$,and $g = 10 \, m/s^2$.
Substituting the values:
$v_{max} = \sqrt{0.45 \times 200 \times 10}$.
$v_{max} = \sqrt{0.45 \times 2000} = \sqrt{900}$.
$v_{max} = 30 \, m/s$.
Therefore,the correct option is $B$.

(A) The boy is in circular motion on the platform. The necessary centripetal force is provided by the static friction force between the boy and the platform.
For the boy to just slip,the maximum static friction must equal the required centripetal force:
$f_{max} = F_c$
$\mu N = m \omega^2 r$
Since the platform is horizontal,the normal force $N = mg$.
Substituting this into the equation:
$\mu mg = m \omega^2 r$
$\mu = \frac{\omega^2 r}{g}$
Given values: $\omega = 1 \, rad/s$,$r = 5 \, m$,and $g = 10 \, m/s^2$.
$\mu = \frac{(1)^2 \times 5}{10} = \frac{5}{10} = 0.5$
Therefore,the coefficient of friction is $0.5$.

(D) The maximum speed $v_{\text{max}}$ for a car on a horizontal circular road is given by the condition that the centripetal force is provided by the static friction: $\frac{mv^2}{r} \leq \mu mg$.
Thus,$v^2 \leq \mu rg$.
Given: radius $r = 0.1 \, km = 100 \, m$,coefficient of friction $\mu = 0.4$,and $g = 10 \, m/s^2$.
Calculating the maximum speed: $v_{\text{max}} = \sqrt{\mu rg} = \sqrt{0.4 \times 100 \times 10} = \sqrt{400} = 20 \, m/s$.
Since the car can move at any speed less than or equal to the maximum speed without skidding,the speed can be any value $v \leq 20 \, m/s$.
Therefore,$5 \, m/s$,$10 \, m/s$,and $20 \, m/s$ are all possible speeds.
Hence,the correct option is $D$.

(B) Step $1$: Identify the forces acting on the block. The forces are the gravitational force $mg$ acting downwards and the normal reaction force $N$ acting perpendicular to the surface of the bowl towards the center of the sphere.
Step $2$: Resolve the forces into vertical and horizontal components. The angle between the normal force $N$ and the vertical axis is $\theta$.
Vertical equilibrium: $N \cos \theta = mg$ (Equation $1$)
Horizontal component provides the centripetal force: $N \sin \theta = m \omega^2 R$,where $R$ is the radius of the circular path of the block. From the geometry of the bowl,$R = r \sin \theta$.
So,$N \sin \theta = m \omega^2 (r \sin \theta)$ (Equation $2$)
Step $3$: Solve the equations. Divide Equation $2$ by Equation $1$:
$\frac{N \sin \theta}{N \cos \theta} = \frac{m \omega^2 r \sin \theta}{mg}$
$\tan \theta = \frac{\omega^2 r \sin \theta}{g}$
$\frac{\sin \theta}{\cos \theta} = \frac{\omega^2 r \sin \theta}{g}$
$\frac{1}{\cos \theta} = \frac{\omega^2 r}{g}$
$\omega^2 = \frac{g}{r \cos \theta}$
$\omega = \sqrt{\frac{g}{r \cos \theta}}$

(C) For a car moving on a horizontal curved road,the centripetal force required for turning is provided by the static friction between the tyres and the road.
The condition for safe turning without skidding is given by $f_s \leq \mu N$,where $N = mg$.
The maximum speed $v_{\max}$ is achieved when the frictional force is at its maximum value:
$\frac{mv_{\max}^2}{r} = \mu mg$
Simplifying for $v_{\max}$:
$v_{\max} = \sqrt{\mu rg}$
Given values are $\mu = 0.34$,$r = 50\,m$,and $g = 10\,ms^{-2}$.
Substituting these values:
$v_{\max} = \sqrt{0.34 \times 50 \times 10}$
$v_{\max} = \sqrt{0.34 \times 500}$
$v_{\max} = \sqrt{170}$
$v_{\max} \approx 13.038\,ms^{-1}$
Therefore,the approximate maximum speed is $13\,ms^{-1}$.

(A) For a stone rotating in a horizontal circle (conical pendulum),the forces acting on the stone are the tension $T$ in the string and the gravitational force $mg$.
Resolving the tension $T$ into components:
Vertical component: $T \cos \theta = mg$ $(1)$
Horizontal component providing centripetal force: $T \sin \theta = \frac{mv^2}{r}$,where $r = l \sin \theta$ is the radius of the circular path.
So,$T \sin \theta = \frac{mv^2}{l \sin \theta}$ $(2)$
From $(1)$,$\cos \theta = \frac{mg}{T} = \frac{1 \times 10}{400} = 0.025$.
Since $\sin^2 \theta + \cos^2 \theta = 1$,we have $\sin^2 \theta = 1 - (0.025)^2 = 1 - 0.000625 = 0.999375$.
From $(2)$,$v^2 = \frac{T l \sin^2 \theta}{m} = \frac{400 \times 1 \times 0.999375}{1} = 399.75$.
Taking the square root,$v = \sqrt{399.75} \approx 19.99 \approx 20\,ms^{-1}$.

(D) The condition for the coin to just slip on a rotating table is that the required centripetal force is provided by the maximum static friction force.
$f_{s,max} = m \omega^2 R$
Since $f_{s,max} = \mu mg$,we have:
$\mu mg = m \omega^2 R$
$R = \frac{\mu g}{\omega^2}$
This shows that the distance $R$ is inversely proportional to the square of the angular velocity,i.e.,$R \propto \frac{1}{\omega^2}$.
Given initial conditions: $R_1 = 1\,cm$ and $\omega_1 = \omega$.
New conditions: $\omega_2 = \frac{\omega}{2}$ and $R_2 = ?$.
Using the proportionality: $\frac{R_2}{R_1} = \left( \frac{\omega_1}{\omega_2} \right)^2$
$\frac{R_2}{1} = \left( \frac{\omega}{\omega/2} \right)^2 = (2)^2 = 4$
$R_2 = 4\,cm$.
Therefore,the coin will just slip when placed at a distance of $4\,cm$ from the center.

(C) For the coin to remain on the disc without slipping,the centripetal force required for circular motion must be provided by the static frictional force.
The normal force acting on the coin is $N = mg$.
The maximum static frictional force available is $f_{max} = \mu N = \mu mg$.
The required centripetal force for the coin moving in a circle of radius $r$ with angular velocity $\omega$ is $F_c = m \omega^2 r$.
For the coin not to slip,the centripetal force must be less than or equal to the maximum static friction:
$m \omega^2 r \leq \mu mg$
Solving for $\omega$:
$\omega^2 \leq \frac{\mu g}{r}$
Therefore,the maximum angular velocity is $\omega_{max} = \sqrt{\frac{\mu g}{r}}$.

(B) Given: Mass $m = 800 \,kg$, Radius $r = 300 \,m$, Angle of banking $\theta = 30^{\circ}$, Coefficient of static friction $\mu = 0.2$, Acceleration due to gravity $g = 10 \,m/s^2$.
The formula for the maximum safe speed on a banked road with friction is given by:
$V_{\max} = \sqrt{rg \left[ \frac{\tan \theta + \mu}{1 - \mu \tan \theta} \right]}$
Substitute the values:
$V_{\max} = \sqrt{300 \times 10 \times \left[ \frac{\tan 30^{\circ} + 0.2}{1 - 0.2 \times \tan 30^{\circ}} \right]}$
Since $\tan 30^{\circ} = \frac{1}{\sqrt{3}} \approx \frac{1}{1.73} \approx 0.577$
$V_{\max} = \sqrt{3000 \times \left[ \frac{0.577 + 0.2}{1 - 0.2 \times 0.577} \right]}$
$V_{\max} = \sqrt{3000 \times \left[ \frac{0.777}{0.8846} \right]}$
$V_{\max} = \sqrt{3000 \times 0.8783} \approx \sqrt{2635} \approx 51.33 \,m/s$
Rounding to the nearest value, $V_{\max} \approx 51.4 \,m/s$.

(C) For a car moving at maximum speed $v_0$ on a banked road,the forces acting on it are the normal force $N$,gravity $mg$,and the maximum static friction $f = \mu N$ acting down the incline to prevent slipping outwards.
Resolving forces horizontally and vertically:
$N \sin \theta + f \cos \theta = \frac{m v_0^2}{r}$
$N \cos \theta - f \sin \theta = m g$
Substituting $f = \mu N$ into the equations:
$N(\sin \theta + \mu \cos \theta) = \frac{m v_0^2}{r}$
$N(\cos \theta - \mu \sin \theta) = m g$
Dividing the two equations:
$\frac{\sin \theta + \mu \cos \theta}{\cos \theta - \mu \sin \theta} = \frac{v_0^2}{r g}$
Cross-multiplying:
$r g \sin \theta + \mu r g \cos \theta = v_0^2 \cos \theta - \mu v_0^2 \sin \theta$
Rearranging terms to solve for $\mu$:
$\mu(r g \cos \theta + v_0^2 \sin \theta) = v_0^2 \cos \theta - r g \sin \theta$
Dividing by $\cos \theta$:
$\mu(r g + v_0^2 \tan \theta) = v_0^2 - r g \tan \theta$
$\mu = \frac{v_0^2 - r g \tan \theta}{r g + v_0^2 \tan \theta}$

(B) The maximum tension the string can withstand is $T_{\max} = mg = 50 \times 10 = 500 \ N$.
For a mass $m$ rotating in a horizontal circle of radius $r$ with angular velocity $\omega$,the tension $T$ is given by $T = m \omega^2 r$.
In this case,the length of the string $L = 10 \ m$ acts as the radius $r$ (assuming the string remains horizontal for the maximum speed case).
Substituting $\omega = 2 \pi n$,where $n$ is the frequency in revolutions per second (rps):
$T_{\max} = m (2 \pi n_{\max})^2 L$
$500 = 1 \times (2 \pi n_{\max})^2 \times 10$
$50 = (2 \pi n_{\max})^2$
$2 \pi n_{\max} = \sqrt{50}$
$n_{\max} = \frac{\sqrt{50}}{2 \pi} \text{ rps}$.

(B) For a motorcyclist moving in a horizontal circle inside a cylindrical wall,the forces acting on the motorcyclist are:
$1$. The weight $(mg)$ acting downwards.
$2$. The normal reaction $(N)$ from the wall acting towards the center of the circle,which provides the necessary centripetal force: $N = \frac{mv^2}{R}$.
$3$. The frictional force $(f)$ acting upwards to balance the weight: $f = mg$.
For the motorcyclist not to slip downwards,the frictional force must be less than or equal to the limiting friction: $f \leq \mu_{s} N$.
Substituting the values: $mg \leq \mu_{s} \left( \frac{mv^2}{R} \right)$.
$g \leq \frac{\mu_{s} v^2}{R}$.
$v^2 \geq \frac{Rg}{\mu_{s}}$.
Therefore,the minimum speed required is $v_{min} = \sqrt{\frac{Rg}{\mu_{s}}}$.

(B) The maximum safe speed $V$ of a vehicle on a circular road of radius $r$ is given by $V = \sqrt{\mu rg}$,where $\mu$ is the coefficient of friction.
For the dry road: $V = \sqrt{\mu rg}$ $(i)$
For the wet road,let the new coefficient of friction be $\mu^{\prime}$. The new speed is $\frac{V}{2} = \sqrt{\mu^{\prime} rg}$ (ii)
Dividing equation $(i)$ by equation (ii):
$\frac{V}{V/2} = \frac{\sqrt{\mu rg}}{\sqrt{\mu^{\prime} rg}}$
$2 = \sqrt{\frac{\mu}{\mu^{\prime}}}$
Squaring both sides:
$4 = \frac{\mu}{\mu^{\prime}}$
$\therefore \mu^{\prime} = \frac{\mu}{4}$

(D) For a motorcyclist moving in a horizontal circle inside a cylindrical wall,the forces acting on the motorcyclist are:
$1$. The weight $(mg)$ acting downwards.
$2$. The normal force $(N)$ acting horizontally towards the center,which provides the necessary centripetal force: $N = \frac{mv^2}{r}$.
$3$. The frictional force $(f)$ acting upwards,which balances the weight to prevent skidding: $f = mg$.
For the motorcyclist not to skid,the frictional force must be less than or equal to the limiting friction: $f \le \mu N$.
Substituting the values: $mg \le \mu \left(\frac{mv^2}{r}\right)$.
$g \le \frac{\mu v^2}{r}$.
$v^2 \ge \frac{rg}{\mu}$.
Therefore,the minimum speed is $v_{min} = \sqrt{\frac{rg}{\mu}}$.

(C) The child starts from rest with constant tangential acceleration $a$. After time $t$,the tangential velocity is $v = at$.
The radial (centripetal) acceleration is $a_r = \frac{v^2}{r} = \frac{(at)^2}{r} = \frac{a^2 t^2}{r}$.
The total acceleration $a_{net}$ experienced by the child is the vector sum of tangential and radial accelerations: $a_{net} = \sqrt{a_t^2 + a_r^2} = \sqrt{a^2 + \left(\frac{a^2 t^2}{r}\right)^2} = \sqrt{a^2 + \frac{a^4 t^4}{r^2}}$.
Slipping starts when the required frictional force equals the limiting friction,i.e.,$F_{net} = m a_{net} = \mu m g$.
Thus,$\mu g = \sqrt{a^2 + \frac{a^4 t^4}{r^2}} = \sqrt{\frac{a^2 r^2 + a^4 t^4}{r^2}} = \frac{1}{r} \sqrt{a^2 r^2 + a^4 t^4}$.
Therefore,$\mu = \frac{[a^4 t^4 + a^2 r^2]^{1/2}}{r g}$.

(C) The maximum safe speed $v$ on a banked road is given by the formula $v = \sqrt{rg(\tan \theta + \mu) / (1 - \mu \tan \theta)}$.
Assuming the friction coefficient $\mu$ and banking angle $\theta$ remain constant,the speed $v$ is directly proportional to the square root of the radius $r$,i.e.,$v \propto \sqrt{r}$.
Let the initial radius be $r_1 = 20 \ m$ and initial speed be $v_1$. The new speed is $v_2 = v_1 + 0.20v_1 = 1.2v_1$.
Since $v \propto \sqrt{r}$,we have $v_2 / v_1 = \sqrt{r_2 / r_1}$.
Substituting the values: $1.2 = \sqrt{r_2 / 20}$.
Squaring both sides: $1.44 = r_2 / 20$.
$r_2 = 1.44 \times 20 = 28.8 \ m$.
The increase in the radius of curvature is $\Delta r = r_2 - r_1 = 28.8 \ m - 20 \ m = 8.8 \ m$.

(A) For a car to safely negotiate a curved road of radius $R$ with velocity $V$,the road is banked at an angle $\theta$.
From the theory of banking of roads,the condition for safe turning is $\tan \theta = \frac{V^2}{Rg}$.
For a small angle $\theta$,$\tan \theta \approx \sin \theta = \frac{h}{b}$,where $h$ is the height of the outer edge and $b$ is the width of the road.
Equating the two expressions for $\tan \theta$,we get $\frac{h}{b} = \frac{V^2}{Rg}$.
Therefore,the value of $h$ is $h = \frac{V^2 b}{Rg}$.