Circular motion with Friction Questions in English

(B) The centripetal force required for a cyclist to turn around a curve is given by $F = \frac{mv^2}{r}$,where $m$ is the mass,$v$ is the velocity,and $r$ is the radius of the curve.
From this relation,we can see that the force $F$ is directly proportional to the square of the velocity,i.e.,$F \propto v^2$.
If the speed $v$ is doubled $(v' = 2v)$,the new force $F'$ becomes $F' \propto (2v)^2 = 4v^2$.
Therefore,$F' = 4F$.
This means the tendency to overturn becomes four times (quadrupled) the original value.

(B) When a car takes a turn on a horizontal road,the necessary centripetal force is provided by the static friction between the tyres and the road surface.
If the speed of the car is too high or the friction is insufficient,the required centripetal force $(F_c = \frac{mv^2}{r})$ cannot be maintained.
Consequently,the car fails to follow the circular path and is thrown out of the road due to the lack of sufficient centripetal force.

(A) When a car takes a turn,it experiences a centrifugal force acting outward at its center of gravity.
This force creates a torque that tends to rotate the car about the outer wheels.
As the speed of the car increases,the normal reaction on the inner wheels decreases.
When the centrifugal torque becomes sufficient to overcome the stabilizing torque provided by the car's weight,the normal reaction on the inner wheels becomes zero.
Therefore,the inner wheels lose contact with the ground first.

(B) The velocity of the motorcyclist is $v = 72\, km/h = 72 \times \frac{5}{18}\, m/s = 20\, m/s$.
Given the radius of curvature $r = 20\, m$ and acceleration due to gravity $g = 10\, m/s^2$.
To avoid skidding while taking a turn on a flat road,the angle of banking $\theta$ with the vertical is given by the formula $\tan \theta = \frac{v^2}{rg}$.
Substituting the values: $\tan \theta = \frac{20^2}{20 \times 10} = \frac{400}{200} = 2$.
Therefore,the angle is $\theta = \tan^{-1}(2)$.

(B) When a car takes a turn on a circular path,the centripetal force is provided by the friction between the tires and the road. This creates a torque that tends to tilt the car outward.
To maintain equilibrium,the normal reaction on the outer wheel increases while the normal reaction on the inner wheel decreases.
The reaction on the inner wheel is given by $R_1 = \frac{1}{2}M \left[ g - \frac{v^2 h}{ra} \right]$.
The reaction on the outer wheel is given by $R_2 = \frac{1}{2}M \left[ g + \frac{v^2 h}{ra} \right]$.
Here,$r$ is the radius of the circular path,$2a$ is the distance between the two wheels,$h$ is the height of the center of gravity of the car,$M$ is the mass,and $v$ is the velocity.
Comparing the two expressions,it is clear that $R_1 < R_2$.

(A) The formula for the maximum safe speed $v$ on a banked road with friction is given by:
$v = \sqrt{gr \left( \frac{\mu + \tan \theta}{1 - \mu \tan \theta} \right)}$
Given values are:
Radius $r = 1000 \, m$
Banking angle $\theta = 45^\circ$
Coefficient of friction $\mu = 0.5$
Acceleration due to gravity $g = 9.8 \, m/s^2$
Substituting these values into the formula:
$v^2 = 9.8 \times 1000 \times \left( \frac{0.5 + \tan 45^\circ}{1 - 0.5 \times \tan 45^\circ} \right)$
Since $\tan 45^\circ = 1$:
$v^2 = 9800 \times \left( \frac{0.5 + 1}{1 - 0.5 \times 1} \right)$
$v^2 = 9800 \times \left( \frac{1.5}{0.5} \right)$
$v^2 = 9800 \times 3 = 29400$
$v = \sqrt{29400} \approx 171.46 \, m/s \approx 172 \, m/s$.

(D) For a car moving on a circular track,the centripetal force is provided by the static friction between the tyres and the road.
To avoid skidding,the maximum centripetal force must be less than or equal to the maximum static friction force.
$F_c \leq f_{s, \max}$
$\frac{mv^2}{r} \leq \mu mg$
$v^2 \leq \mu rg$
$v_{\max} = \sqrt{\mu rg}$
Given: $\mu = 0.2$,$r = 100 \, m$,and $g = 9.8 \, m/s^2$.
$v_{\max} = \sqrt{0.2 \times 100 \times 9.8}$
$v_{\max} = \sqrt{20 \times 9.8}$
$v_{\max} = \sqrt{196}$
$v_{\max} = 14 \, m/s$.

(D) When a car of mass $M$ moves over a convex bridge of radius $r$ with a velocity $v$,the forces acting on the car are its weight $Mg$ acting downwards and the normal reaction $N$ from the bridge acting upwards.
The net centripetal force required for circular motion is provided by the difference between the weight and the normal reaction:
$Mg - N = \frac{Mv^2}{r}$
Therefore,the normal reaction $N$ exerted by the car on the bridge is:
$N = Mg - \frac{Mv^2}{r}$
Thus,the correct option is $D$.

(C) The maximum force of friction provides the necessary centripetal force for the car to move in a circular path.
The formula for centripetal force is $F_c = \frac{mv^2}{r}$.
Given:
Mass $m = 100 \, kg$
Velocity $v = 9 \, m/s$
Radius $r = 30 \, m$
Substituting the values:
$F_c = \frac{100 \times (9)^2}{30}$
$F_c = \frac{100 \times 81}{30}$
$F_c = \frac{8100}{30} = 270 \, N$.
Therefore,the maximum force of friction is $270 \, N$.

(A) To find the maximum speed of a car on a curved road,we use the formula for safe circular motion on a flat road provided by friction:
$v_{max} = \sqrt{\mu rg}$
Given:
Radius $(r)$ = $30\, m$
Coefficient of friction $(\mu)$ = $0.4$
Acceleration due to gravity $(g)$ = $9.8\, m/s^2$
Substituting the values:
$v_{max} = \sqrt{0.4 \times 30 \times 9.8}$
$v_{max} = \sqrt{12 \times 9.8}$
$v_{max} = \sqrt{117.6}$
$v_{max} \approx 10.84\, m/s$
Therefore,the correct option is $A$.

(C) For a cyclist taking a circular turn on a level road,the necessary centripetal force is provided by the static friction between the tyres and the road.
The condition for safe turning is given by the formula: $v^2 \leq \mu rg$,where $v$ is the speed,$\mu$ is the coefficient of friction,$r$ is the radius,and $g$ is the acceleration due to gravity.
To find the minimum coefficient of friction,we use: $\mu = \frac{v^2}{rg}$.
Given: $v = 4.9 \, m/s$,$r = 4 \, m$,and $g = 9.8 \, m/s^2$.
Substituting the values: $\mu = \frac{(4.9)^2}{4 \times 9.8} = \frac{24.01}{39.2} = 0.6125$.
Rounding to two decimal places,we get $\mu = 0.61$.

(B) To avoid skidding on a flat circular curve,the centripetal force required is provided by the static friction force.
$F_c = F_f$
$\frac{mv^2}{r} = \mu mg$
$v^2 = \mu rg$
$v = \sqrt{\mu rg}$
Given:
Radius $r = 150 \, m$
Coefficient of friction $\mu = 0.6$
Acceleration due to gravity $g = 10 \, m/s^2$
Substituting the values:
$v = \sqrt{0.6 \times 150 \times 10}$
$v = \sqrt{900}$
$v = 30 \, m/s$
Therefore,the maximum velocity is $30 \, m/s$.

(D) For a car moving on a flat circular curve,the centripetal force required for turning is provided by the static friction between the tyres and the road.
To avoid skidding,the centripetal force must be less than or equal to the maximum static friction force:
$\frac{mv^2}{r} \leq \mu mg$
$v^2 \leq \mu rg$
$v_{max} = \sqrt{\mu rg}$
Given:
$\mu = 0.25$
$r = 40 \,m$
$g = 10 \,ms^{-2}$
Substituting the values:
$v_{max} = \sqrt{0.25 \times 40 \times 10}$
$v_{max} = \sqrt{10 \times 10}$
$v_{max} = 10 \,ms^{-1}$

(B) For a car moving on a circular track,the centripetal force is provided by the frictional force.
$F_c = F_f$
$\frac{mv^2}{r} = \mu mg$
$v^2 = \mu rg$
$v = \sqrt{\mu rg}$
Given: $\mu = 0.5$,$r = 40 \, m$,and taking $g = 9.8 \, m/s^2$.
$v = \sqrt{0.5 \times 40 \times 9.8}$
$v = \sqrt{20 \times 9.8}$
$v = \sqrt{196}$
$v = 14 \, m/s$.

(A) Let the bead start slipping after time $t$.
For the bead to slip,the net force acting on it must exceed the maximum static frictional force.
The forces acting on the bead in the rotating frame are the centrifugal force $F_c = m\omega^2 L$ (acting radially outward) and the tangential force $F_t = m a_t = m \alpha L$ (acting perpendicular to the rod).
The normal force $N$ is zero as gravity is neglected and there are no other vertical forces.
Wait,the friction force $f$ must balance the resultant of these forces. The maximum static friction is $f_{max} = \mu N$. Since there is no vertical force,$N$ is effectively provided by the constraint of the rod. However,in this specific problem context,the friction is assumed to be $\mu m a_t$ or similar. Re-evaluating: The bead slips when the resultant force $F_{net} = \sqrt{(m\omega^2 L)^2 + (m\alpha L)^2} = \mu N$. Assuming $N = m a_t = m \alpha L$ is not standard. Actually,the friction must oppose the tendency to move. The bead will slip when the net force required to keep it in circular motion exceeds the maximum available friction. The force required is $F = \sqrt{(m\omega^2 L)^2 + (m\alpha L)^2}$. Given the options,the intended logic is $m\omega^2 L = \mu m \alpha L$,which implies $\omega^2 = \mu \alpha$. Since $\omega = \alpha t$,we get $(\alpha t)^2 = \mu \alpha$,so $t^2 = \mu / \alpha$,or $t = \sqrt{\mu / \alpha}$.

(D) For a car moving on a circular unbanked road,the necessary centripetal force is provided by the static frictional force between the tires and the road surface.
Let $m$ be the mass of the car,$v$ be its speed,and $R$ be the radius of the circular path.
The centripetal force required is $F_c = \frac{mv^2}{R}$.
The maximum static frictional force available is $f_{max} = \mu N = \mu mg$,where $N = mg$ is the normal force.
To avoid skidding,the centripetal force must be less than or equal to the maximum static friction:
$\frac{mv^2}{R} \leq \mu mg$
$v^2 \leq \mu Rg$
$v_{max} = \sqrt{\mu Rg}$.

(A) For a car turning on a flat circular 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 values: $v = 10\,m/s$,$\mu = 0.5$,and taking $g = 10\,m/s^2$.
$r = \frac{10^2}{0.5 \times 10} = \frac{100}{5} = 20\,m$.
Thus,the minimum radius is $20\,m$.

(C) For a car moving on an unbanked circular road,the centripetal force required for turning is provided by the static friction between the tyres and the road.
For the car not to slip,the maximum centripetal force must be less than or equal to the maximum static frictional force:
$\frac{mv^2}{r} \leq \mu mg$
$v^2 \leq \mu gr$
$v_{max} = \sqrt{\mu gr}$
Given: $\mu = 0.5$,$r = 40.0\, m$,and taking $g = 9.8\, m/s^2$:
$v_{max} = \sqrt{0.5 \times 9.8 \times 40.0}$
$v_{max} = \sqrt{196}$
$v_{max} = 14\, m/s$.

(B) To find the maximum speed of a car on a circular turn without skidding,we use the formula for the maximum safe speed on a flat circular road:
$v_{max} = \sqrt{\mu rg}$
Given:
Radius of the turn,$r = 30\,m$
Coefficient of friction,$\mu = 0.4$
Acceleration due to gravity,$g = 9.8\,m/s^2$
Substituting the values into the formula:
$v_{max} = \sqrt{0.4 \times 30 \times 9.8}$
$v_{max} = \sqrt{12 \times 9.8}$
$v_{max} = \sqrt{117.6}$
$v_{max} \approx 10.84\,m/s$
Therefore,the maximum speed is $10.84\,m/s$.

(A) To avoid skidding on a curved track,the centripetal force required is provided by the static friction between the tyres and the road.
For a motorcycle on a flat curved track,the condition for safe turning without skidding is given by $v \le \sqrt{\mu rg}$.
Given:
Radius $r = 500\,m$
Coefficient of friction $\mu = 0.5$
Acceleration due to gravity $g = 10\,m/s^2$
Substituting the values into the formula:
$v = \sqrt{0.5 \times 500 \times 10}$
$v = \sqrt{2500}$
$v = 50\,m/s$
Therefore,the maximum speed to avoid skidding is $50\,m/s$.

(C) For the coin to revolve with the record without slipping,the required centripetal force must be provided by the static friction force.
The condition for the coin to remain stationary relative to the record is:
$f_s \ge F_c$
where $f_s$ is the static friction force and $F_c$ is the centripetal force.
The maximum static friction force is $f_{s,max} = \mu N = \mu mg$.
The centripetal force required is $F_c = mr\omega^2$.
Therefore,the condition is:
$\mu mg \ge mr\omega^2$
Dividing both sides by $m\omega^2$ (assuming $m \neq 0$ and $\omega \neq 0$):
$r \le \frac{\mu g}{\omega^2}$

(D) The maximum safe speed $v$ of a vehicle on a circular 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.
From this formula,we can see that $v \propto \sqrt{\mu}$.
Given initial conditions: $v_1 = 10\;m/s$ and $\mu_1 = \mu$.
New conditions: $\mu_2 = \frac{\mu}{2}$.
Using the proportionality $v_2 = v_1 \sqrt{\frac{\mu_2}{\mu_1}}$:
$v_2 = 10 \times \sqrt{\frac{\mu/2}{\mu}} = 10 \times \sqrt{\frac{1}{2}} = \frac{10}{\sqrt{2}}$.
Rationalizing the denominator: $v_2 = \frac{10\sqrt{2}}{2} = 5\sqrt{2}\;m/s$.

(D) The maximum safe speed $v_{\max}$ on a flat circular road is given by the formula:
$v_{\max} = \sqrt{\mu \, r \, g}$
Given:
Radius $r = 40 \, m$
Coefficient of friction $\mu = 0.25$
Acceleration due to gravity $g = 10 \, m \, s^{-2}$
Substituting the values:
$v_{\max} = \sqrt{0.25 \times 40 \times 10}$
$v_{\max} = \sqrt{100}$
$v_{\max} = 10 \, m \, s^{-1}$
Therefore,the maximum safe speed is $10 \, m \, s^{-1}$.

(D) Given: Radius $r = 4 \; m$,Maximum speed $v = 4.9 \; m/s$,Acceleration due to gravity $g = 9.8 \; m/s^2$.
For a car moving on a flat circular road,the centripetal force is provided by the static friction force.
Therefore,$F_c = F_f \implies \frac{mv^2}{r} = \mu mg$.
Simplifying for the coefficient of friction $\mu$,we get $\mu = \frac{v^2}{rg}$.
Substituting the values: $\mu = \frac{4.9 \times 4.9}{4 \times 9.8}$.
Since $4.9 / 9.8 = 0.5$,we have $\mu = \frac{4.9 \times 0.5}{4} = \frac{2.45}{4} = 0.6125$.
Rounding to two decimal places,the coefficient of friction is $0.61$.

(B) Given: Speed $v = 60 \; km/hr = 60 \times \frac{5}{18} \; m/s = \frac{50}{3} \; m/s$.
Radius $r = 0.1 \; km = 100 \; m$.
Acceleration due to gravity $g = 9.8 \; m/s^2$.
For a bike to take a turn without slipping,the angle of banking $\theta$ with the vertical is given by the formula:
$\tan \theta = \frac{v^2}{rg}$.
Substituting the values:
$\tan \theta = \frac{(50/3)^2}{100 \times 9.8}$.
Therefore,$\theta = \tan^{-1} \left[ \frac{(50/3)^2}{100 \times 9.8} \right]$.

(C) Given: Height of center of mass $h = 2 \, m$,acceleration due to gravity $g = 10 \, m/s^2$,distance between tires $2a = 1.5 \, m$ (so $a = 0.75 \, m$),and radius of the turn $r = 120 \, m$.
The condition to avoid overturning while taking a turn is given by the formula $v_{\max} = \sqrt{\frac{g \cdot r \cdot a}{h}}$.
Substituting the values: $v_{\max} = \sqrt{\frac{10 \times 120 \times 0.75}{2}}$.
$v_{\max} = \sqrt{\frac{900}{2}} = \sqrt{450} \approx 21.2 \, m/s$.
Rounding to the nearest integer provided in the options,the correct speed is $21 \, m/s$.

(A) The maximum safe speed $v$ on a horizontal curved road is given by the formula $v = \sqrt{\mu rg}$.
Given:
Radius $r = 500\,m$
Coefficient of friction $\mu = 0.5$
Acceleration due to gravity $g = 10\,m/s^2$
Substituting the values:
$v = \sqrt{0.5 \times 500 \times 10}$
$v = \sqrt{2500}$
$v = 50\,m/s$.

(A) For vertical equilibrium on the road:
$N \cos\theta = mg + f \sin\theta$
$mg = N \cos\theta - f \sin\theta$ ... $(i)$
For safe turning,the centripetal force is provided by the horizontal components of the normal force and friction:
$N \sin\theta + f \cos\theta = \frac{mv^2}{R}$ ... $(ii)$
Dividing equation $(ii)$ by equation $(i)$:
$\frac{v^2}{Rg} = \frac{N \sin\theta + f \cos\theta}{N \cos\theta - f \sin\theta}$
At maximum velocity,the friction force $f$ reaches its limiting value,$f = \mu_s N$. Substituting this into the equation:
$\frac{v_{\max}^2}{Rg} = \frac{N \sin\theta + \mu_s N \cos\theta}{N \cos\theta - \mu_s N \sin\theta}$
Dividing the numerator and denominator by $N \cos\theta$:
$\frac{v_{\max}^2}{Rg} = \frac{\tan\theta + \mu_s}{1 - \mu_s \tan\theta}$
Therefore,the maximum safe velocity is:
$v_{\max} = \sqrt{gR \left( \frac{\mu_s + \tan\theta}{1 - \mu_s \tan\theta} \right)}$

(B) The maximum speed $v_{m}$ at which the cyclist will not skid is given by the formula $v_{m} = \sqrt{\mu r g}$.
Substituting the given values: $\mu = 0.2$,$r = 3 \; m$,and $g = 10 \; m \cdot s^{-2}$.
$v_{m} = \sqrt{0.2 \times 3 \times 10} = \sqrt{6} \approx 2.45 \; m \cdot s^{-1}$.
To convert this speed into $km \cdot h^{-1}$,we multiply by $\frac{18}{5}$:
$v_{m} = 2.45 \times 3.6 = 8.82 \; km \cdot h^{-1}$.
The cyclist will not skid if their speed is less than or equal to $8.82 \; km \cdot h^{-1}$.
Among the given options,$7.2 \; km \cdot h^{-1}$ is the only speed less than $8.82 \; km \cdot h^{-1}$.

(C) For the coin to revolve with the record without slipping,the required centripetal force must be provided by the static friction force.
The centripetal force required is $F_c = mr\omega^2$.
The maximum static friction force available is $f_{s,max} = \mu N = \mu mg$.
For the coin to stay in place,the friction force must be greater than or equal to the required centripetal force:
$f_{s,max} \ge F_c$
$\mu mg \ge mr\omega^2$
Dividing both sides by $m\omega^2$ (where $m$ is the mass of the coin),we get:
$r \le \frac{\mu g}{\omega^2}$

(D) The necessary centripetal force for circular motion is provided by the static friction between the road and the tyres.
The condition for the car not to skid is $f \leq \mu_s N$.
Since the track is level,the normal force $N = mg$.
The centripetal force required is $F_c = \frac{mv^2}{R}$.
Equating the maximum friction to the centripetal force: $\mu_s mg = \frac{mv_{max}^2}{R}$.
Solving for $v_{max}$:
$v_{max}^2 = \mu_s Rg$
$v_{max} = \sqrt{\mu_s Rg}$.

(D) The maximum velocity $v$ for a cyclist to take a turn on a circular track of radius $r$ with a coefficient of friction $\mu$ is given by the formula $v = \sqrt{\mu r g}$.
Given:
Radius $r = 100 \ m$
Coefficient of friction $\mu = 0.2$
Acceleration due to gravity $g = 10 \ m/s^2$
Substituting the values:
$v = \sqrt{0.2 \times 100 \times 10}$
$v = \sqrt{20 \times 10}$
$v = \sqrt{200}$
$v = 10\sqrt{2} \approx 14.14 \ m/s$.
Rounding to the nearest integer provided in the options,the correct velocity is $14 \ m/s$.

(C) For a car moving on a horizontal circular path,the limiting factor for speed is either skidding or overturning.
$1$. Condition for skidding: The centripetal force is provided by friction,so $mv^2/r \leq \mu mg$,which gives $v \leq \sqrt{\mu gr}$.
Substituting the values: $v = \sqrt{0.8 \times 9.8 \times 15} = \sqrt{117.6} \approx 10.84 \ m/s$.
$2$. Condition for overturning: The car will overturn if the torque due to centripetal force about the outer wheels exceeds the torque due to gravity. The condition is $v \leq \sqrt{gr(w/2h)}$,where $w = 1.1 \ m$ and $h = 0.62 \ m$.
$v \leq \sqrt{9.8 \times 15 \times (1.1 / (2 \times 0.62))} = \sqrt{147 \times 0.887} \approx \sqrt{130.4} \approx 11.42 \ m/s$.
Since the speed must satisfy both conditions,the maximum safe speed is the smaller of the two values,which is $10.84 \ m/s$.

(A) To prevent slipping,the required centripetal force must be provided by the static frictional force.
The condition for no slipping is: $F_c \leq f_s$
Where $F_c = m R \omega^2$ is the centripetal force and $f_s \leq \mu N$ is the maximum static friction.
Since the disc is horizontal,the normal force $N = mg$.
Therefore,$m R \omega^2 \leq \mu mg$.
Canceling mass $m$ from both sides,we get $R \omega^2 \leq \mu g$.
Solving for $R$,we get $R \leq \frac{\mu g}{\omega^2}$.
Given $\mu = 0.5$,$\omega = 5 \ rad/s$,and taking $g = 10 \ m/s^2$:
$R \leq \frac{0.5 \times 10}{5^2} = \frac{5}{25} = 0.2 \ m$.
Thus,the man must stand at a distance $R \leq 0.2 \ m$ from the center.

(D) For a car moving on a banked circular track with friction,the maximum speed $v_{max}$ is given by the formula:
$v_{max} = \sqrt{gR \left( \frac{\mu + \tan \theta}{1 - \mu \tan \theta} \right)}$
Given the banking angle $\theta = 30^o$,we know that $\tan 30^o = \frac{1}{\sqrt{3}}$.
Substituting this into the formula:
$v_{max} = \sqrt{gR \left( \frac{\mu + \frac{1}{\sqrt{3}}}{1 - \mu \frac{1}{\sqrt{3}}} \right)}$
Multiplying the numerator and denominator inside the square root by $\sqrt{3}$:
$v_{max} = \sqrt{gR \left( \frac{\mu \sqrt{3} + 1}{\sqrt{3} - \mu} \right)}$
Comparing this with the given options,none of the provided expressions match this result. Therefore,the correct choice is $D$.

(C) The position vector of the particle at point $P(a \cos \theta, a \sin \theta)$ is $\vec{r} = a \cos \theta \hat{i} + a \sin \theta \hat{j}$.
The unit radial vector is $\hat{r} = \cos \theta \hat{i} + \sin \theta \hat{j}$.
Since the particle is moving in a circle,the velocity vector $\vec{v}$ is tangent to the circle. The direction of motion is anticlockwise,so the velocity vector is in the direction of $\hat{\theta} = -\sin \theta \hat{i} + \cos \theta \hat{j}$.
Friction acts in the direction opposite to the velocity of the particle. Therefore,the direction of friction is $-\hat{\theta} = \sin \theta \hat{i} - \cos \theta \hat{j}$.

(B) Let $m$ be the mass of the particle,$r$ be the radius of the circular path,$\omega$ be the angular velocity,and $\theta = 45^\circ$ be the semi-vertical angle. The height $h = 1 \ m$. Since $\tan \theta = r/h$,we have $r = h \tan 45^\circ = 1 \ m$.
For maximum angular velocity,the friction force $f = \mu N$ acts downwards along the slant surface to prevent the particle from sliding up.
The forces acting on the particle are: Normal force $N$ perpendicular to the surface,gravitational force $mg$ downwards,and friction $f = \mu N$ downwards along the slope.
Resolving forces horizontally: $N \sin \theta + f \cos \theta = m \omega^2 r$.
Resolving forces vertically: $N \cos \theta - f \sin \theta = mg$.
Substituting $f = \mu N$: $N(\sin \theta + \mu \cos \theta) = m \omega^2 r$ and $N(\cos \theta - \mu \sin \theta) = mg$.
Dividing the two equations: $\frac{\sin \theta + \mu \cos \theta}{\cos \theta - \mu \sin \theta} = \frac{\omega^2 r}{g}$.
Given $\theta = 45^\circ$,$\sin 45^\circ = \cos 45^\circ = 1/\sqrt{2}$,and $\mu = 0.5$:
$\frac{1/\sqrt{2} + 0.5/\sqrt{2}}{1/\sqrt{2} - 0.5/\sqrt{2}} = \frac{\omega^2 (1)}{10} \implies \frac{1.5}{0.5} = \frac{\omega^2}{10} \implies 3 = \frac{\omega^2}{10}$.
Thus,$\omega^2 = 30$,so $\omega = \sqrt{30} \ \text{rad/s}$.

(D) For the mass $M$ to remain suspended on the inner wall of the rotating cylinder,the upward frictional force $f_s$ must balance the downward gravitational force $mg$.
$f_s \geqslant mg$
Since the frictional force $f_s = \mu N$,where $N$ is the normal force provided by the wall of the cylinder,we have:
$\mu N \geqslant mg$
The normal force $N$ provides the necessary centripetal force for the mass $M$ to move in a circle of radius $R$ with angular velocity $\omega$:
$N = M R \omega^2$
Substituting this into the inequality:
$\mu (M R \omega^2) \geqslant Mg$
$\mu \geqslant \frac{g}{R \omega^2}$
Therefore,the minimum coefficient of static friction is:
$\mu_{\min} = \frac{g}{R \omega^2}$

(D) The mass slides off when the centripetal force required exceeds the maximum static frictional force.
$\frac{mv^2}{R} = \mu mg$
$v = \sqrt{\mu Rg}$
Once the mass leaves the disc,it undergoes projectile motion with an initial horizontal velocity $v$ and zero initial vertical velocity.
The time taken to fall a height $h$ is given by $h = \frac{1}{2}gt^2$,so $t = \sqrt{\frac{2h}{g}}$.
The horizontal distance $d$ traveled is $d = v \times t$.
$d = \sqrt{\mu Rg} \times \sqrt{\frac{2h}{g}} = \sqrt{2\mu Rh}$.

(A) For a particle moving on a circular overbridge with constant speed,the forces acting along the tangent to the path are the component of gravity $mg \sin \theta$ and the frictional force $f$.
Since the speed is constant,the net tangential force must be zero.
Therefore,the frictional force must balance the tangential component of gravity:
$f = mg \sin \theta$
As $\theta$ increases from $0$ to $\pi/2$,the value of $\sin \theta$ increases from $0$ to $1$.
Thus,the magnitude of friction $f$ increases with $\theta$ following a sine curve.
Looking at the provided options,the graph that represents $f = mg \sin \theta$ is the one that starts from zero at $\theta = 0$ and increases.

(D) When a cyclist turns on a curved road,they must lean inwards to maintain balance against the centrifugal force.
Let $m$ be the mass of the cyclist,$v$ be the velocity,and $R$ be the radius of the curve.
The vertical component of the normal force $N$ balances the weight of the cyclist: $N \cos \theta = mg$.
The horizontal component of the normal force provides the necessary centripetal force: $N \sin \theta = \frac{mv^2}{R}$.
Dividing the second equation by the first,we get: $\frac{N \sin \theta}{N \cos \theta} = \frac{mv^2 / R}{mg}$.
Therefore,$\tan \theta = \frac{v^2}{Rg}$.

(C) The formula for the maximum safe speed $v$ on a banked road with friction is given by:
$v = \sqrt{gr \left( \frac{\mu + \tan \theta}{1 - \mu \tan \theta} \right)}$
Given:
Radius $r = 1000 \, m$
Banking angle $\theta = 45^o$
Coefficient of friction $\mu = 0.5$
Acceleration due to gravity $g = 9.8 \, m/s^2$
Substituting the values:
$v^2 = 9.8 \times 1000 \times \left( \frac{0.5 + \tan 45^o}{1 - 0.5 \times \tan 45^o} \right)$
Since $\tan 45^o = 1$:
$v^2 = 9800 \times \left( \frac{0.5 + 1}{1 - 0.5 \times 1} \right)$
$v^2 = 9800 \times \left( \frac{1.5}{0.5} \right)$
$v^2 = 9800 \times 3 = 29400$
$v = \sqrt{29400} \approx 171.46 \, m/s \approx 172 \, m/s$.

(A) For a rotating ring,the tension $T$ in the ring is related to its linear velocity $v$ and linear mass density $\mu$ by the formula $T = \mu v^2$.
Given,$T = 10\, N$ and $\mu = 0.1\, kg/m$.
Rearranging the formula to solve for $v$,we get $v = \sqrt{\frac{T}{\mu}}$.
Substituting the given values: $v = \sqrt{\frac{10}{0.1}} = \sqrt{100}$.
Therefore,the maximum velocity is $v = 10\, m/s$.

(B) For a motorcyclist to stay on the vertical wall of the "well of death" without sliding down,the frictional force must balance the gravitational force.
$f = mg$
Since $f = \mu N$ and the normal force $N$ provides the centripetal force $N = \frac{mv^2}{r}$,we have:
$\mu \left( \frac{mv^2}{r} \right) = mg$
$v^2 = \frac{rg}{\mu}$
$v = \sqrt{\frac{rg}{\mu}}$
Given diameter $d = 18\, m$,so radius $r = 9\, m$.
Given $\mu = 0.8$ and $g = 10\, m/s^2$.
$v = \sqrt{\frac{9 \times 10}{0.8}} = \sqrt{\frac{90}{0.8}} = \sqrt{112.5} \approx 10.6\, m/s$.
Rounding to the nearest provided option,the minimum speed is $10.5\, m/s$.

(A) For a vehicle moving on a banked curved road,the condition for banking is given by $\tan \theta = \frac{v^2}{Rg}$.
Here,$v$ is the velocity,$R$ is the radius of curvature,and $b$ is the width of the road.
Let $h$ be the elevation difference between the outer and inner edges.
For small angles of banking,$\tan \theta \approx \sin \theta = \frac{h}{b}$.
Equating the two expressions for $\tan \theta$,we get $\frac{h}{b} = \frac{v^2}{Rg}$.
Therefore,the required elevation difference is $h = \frac{v^2b}{Rg}$.

(D) The centripetal force required for the coin to rotate with the disc is provided by the static frictional force.
For the coin to remain at rest relative to the disc,the frictional force must be equal to or greater than the required centripetal force: $f = mr\omega^2$.
The maximum static friction is given by $f_{max} = \mu mg$.
Equating the two,we get $\mu mg = mr\omega^2$,which simplifies to $\mu = \frac{r\omega^2}{g}$.
Given:
Frequency $n = 3.5\,rev/s$.
Angular velocity $\omega = 2\pi n = 2 \times \pi \times 3.5 = 7\pi\,rad/s$.
Radius $r = 1.25\,cm = 1.25 \times 10^{-2}\,m$.
Acceleration due to gravity $g = 10\,m/s^2$.
Substituting the values:
$\mu = \frac{(1.25 \times 10^{-2}) \times (7\pi)^2}{10}$.
Using $\pi \approx \frac{22}{7}$,we have $(7\pi)^2 = (7 \times \frac{22}{7})^2 = 22^2 = 484$.
$\mu = \frac{1.25 \times 10^{-2} \times 484}{10} = \frac{1.25 \times 4.84}{10} = 0.605 \approx 0.6$.

(D) Consider a small elemental ring of radius $x$ and width $dx$ on the circular mop. The area of this ring is $dA = 2\pi x dx$. The total area of the mop is $A = \pi R^2$. Since the force $F$ is distributed uniformly,the normal force $dN$ on this elemental ring is $dN = (F/A) dA = (F/(\pi R^2)) \times 2\pi x dx = (2F/R^2) x dx$. The frictional force on this ring is $df = \mu dN = \mu (2F/R^2) x dx$. The torque $d\tau$ due to this frictional force about the axis of rotation is $d\tau = x df = x \times \mu (2F/R^2) x dx = (2\mu F/R^2) x^2 dx$. To find the total torque $\tau$,we integrate $d\tau$ from $x = 0$ to $x = R$: $\tau = \int_0^R (2\mu F/R^2) x^2 dx = (2\mu F/R^2) [x^3/3]_0^R = (2\mu F/R^2) (R^3/3) = \frac{2}{3}\mu FR$.

(C) The maximum speed $v$ of a car on a circular level road is given by the formula $v = \sqrt{\mu r g},$ where $\mu$ is the coefficient of friction,$r$ is the radius of curvature,and $g$ is the acceleration due to gravity.
Given: $\mu = 0.3,$ $r = 300\,m,$ $g = 10\,m/s^2.$
Substituting the values: $v = \sqrt{0.3 \times 300 \times 10} = \sqrt{900} = 30\,m/s.$
To convert the speed from $m/s$ to $km/hr,$ multiply by $\frac{18}{5}:$
$v = 30 \times \frac{18}{5} = 6 \times 18 = 108\,km/hr.$

(C) The maximum speed $v_{max}$ for a car on an unbanked curve is given by the formula $v_{max} = \sqrt{\mu rg}$.
Here,the coefficient of static friction $\mu = 0.75$,the radius $r = 60\,m$,and the acceleration due to gravity $g = 9.8\,m/s^2$.
Substituting these values into the formula:
$v_{max} = \sqrt{0.75 \times 60 \times 9.8}$
$v_{max} = \sqrt{45 \times 9.8}$
$v_{max} = \sqrt{441}$
$v_{max} = 21\,m/s$.

(A) The condition for the object to just slip is that the required centripetal force must be equal to the maximum static frictional force.
$F_{c} = f_{s,max}$
$m r \omega^{2} = \mu m g$
From this,we get the relation $r \omega^{2} = \text{constant}$,which implies $r \propto \frac{1}{\omega^{2}}$.
Therefore,$\frac{r_{2}}{r_{1}} = \left( \frac{\omega_{1}}{\omega_{2}} \right)^{2}$.
Given $r_{1} = 4\, cm$ and $\omega_{2} = 2\omega_{1}$.
Substituting the values: $\frac{r_{2}}{4} = \left( \frac{\omega_{1}}{2\omega_{1}} \right)^{2} = \left( \frac{1}{2} \right)^{2} = \frac{1}{4}$.
$r_{2} = 4 \times \frac{1}{4} = 1\, cm$.