Vertical Circular Motion Questions in English

(A) Let the mass of the body be $m$. At point $A$,the normal force $N$ provides the centripetal force required for circular motion. Since the weight $mg$ acts vertically downwards and the normal force acts horizontally towards the center $O$,the net force on the body at $A$ is the vector sum of $N$ and $mg$.
The magnitude of the net force is $F_A = \sqrt{N^2 + (mg)^2}$.
Given $F_A = \sqrt{2} mg$,we have $\sqrt{N^2 + (mg)^2} = \sqrt{2} mg$.
Squaring both sides,$N^2 + m^2g^2 = 2m^2g^2$,which gives $N^2 = m^2g^2$,so $N = mg$.
The centripetal force equation at $A$ is $N = \frac{mv_A^2}{R}$,so $mg = \frac{mv_A^2}{R}$,which implies $v_A^2 = Rg$.
By the law of conservation of energy,the total energy at the starting point $P$ (height $H$) equals the total energy at point $A$ (height $R$):
$mgH = \frac{1}{2}mv_A^2 + mgR$.
Substituting $v_A^2 = Rg$:
$mgH = \frac{1}{2}m(Rg) + mgR = \frac{3}{2}mgR$.
Therefore,$H = \frac{3}{2}R$.

(B) At the highest point of the vertical circular path,the forces acting on the passenger are the gravitational force $mg$ (acting downwards) and the normal reaction $N$ (acting downwards).
The net centripetal force required for circular motion is provided by the sum of these forces:
$N + mg = \frac{mv^2}{R}$
For the passenger not to fall out,the minimum condition is that the normal reaction $N$ becomes zero at the top point.
Setting $N = 0$,we get:
$mg = \frac{mv^2}{R}$
$v^2 = gR$
$v = \sqrt{gR}$
Given $g = 10 \ m \ s^{-2}$ and $R = 10 \ m$,we substitute these values:
$v = \sqrt{10 \times 10} = \sqrt{100} = 10 \ m \ s^{-1}$.
Thus,the minimum speed required is $10 \ m \ s^{-1}$.

(C) At the bottom of the incline (point $P$), the potential energy of the ball is converted into kinetic energy:
$\frac{1}{2} m v^2 = m g h$
$\Rightarrow v = \sqrt{2 g h} \quad \dots (i)$
To complete the vertical circular motion, the velocity of the ball at the bottom must be at least $\sqrt{5 g R}$.
$\Rightarrow v \geq \sqrt{5 g R}$
Substituting the value of $v$ from equation $(i)$:
$\sqrt{2 g h} \geq \sqrt{5 g R}$
Squaring both sides:
$2 g h \geq 5 g R$
$h \geq \frac{5}{2} R$
Given $R = 20 \,cm$, the minimum height $h_{\min}$ is:
$h_{\min} = \frac{5}{2} \times 20 \,cm = 50 \,cm$

(D) When a particle moves along a vertical circle,the minimum speed required at the highest point to complete the loop is given by the formula:
$v_{\min} = \sqrt{gR}$
From this expression,we can see that the minimum speed is directly proportional to the square root of the radius:
$v_{\min} \propto \sqrt{R}$
Let the initial radius be $R_1 = R$ and the initial minimum speed be $(v_{\min})_1 = v$.
Let the new radius be $R_2 = 2R$ and the new minimum speed be $(v_{\min})_2$.
Using the proportionality relation:
$\frac{(v_{\min})_1}{(v_{\min})_2} = \sqrt{\frac{R_1}{R_2}}$
Substituting the known values:
$\frac{v}{(v_{\min})_2} = \sqrt{\frac{R}{2R}} = \frac{1}{\sqrt{2}}$
Solving for the new minimum speed:
$(v_{\min})_2 = \sqrt{2}v$

(B) Given: Length of the rigid wire $l = 1 \,m$, mass of the ball $m = 500 \,g = 0.5 \,kg$, initial velocity at the lowest point $u = 6 \,m/s$. Let $v$ be the velocity of the ball when the wire makes an angle $\theta = 60^{\circ}$ with the upward vertical.
Using the law of conservation of energy between the lowest point (let it be $P$) and the point $C$ where the wire makes $60^{\circ}$ with the upward vertical:
Total energy at $P$ = Total energy at $C$
$\frac{1}{2}mu^2 + 0 = \frac{1}{2}mv^2 + mgh$
where $h$ is the height of point $C$ above the lowest point. From the geometry, $h = l + l \cos 60^{\circ} = l(1 + \cos 60^{\circ}) = 1(1 + 0.5) = 1.5 \,m$.
Substituting the values:
$\frac{1}{2} \times (6)^2 = \frac{1}{2}v^2 + g \times 1.5$
$18 = \frac{1}{2}v^2 + 10 \times 1.5$
$18 = \frac{1}{2}v^2 + 15$
$\frac{1}{2}v^2 = 3 \implies v^2 = 6 \,m^2/s^2$.
The radial component of acceleration (centripetal acceleration) is given by $a_r = \frac{v^2}{l}$.
$a_r = \frac{6}{1} = 6 \,m/s^2$.

(C) For water not to spill out of the bucket, the bucket must complete the vertical circular motion.
At the highest point, the minimum velocity required is $v_{top} = \sqrt{gR}$.
Using the principle of conservation of energy between the lowest point (bottom) and the highest point (top):
$\frac{1}{2}mv_{bottom}^2 = \frac{1}{2}mv_{top}^2 + mg(2R)$
$v_{bottom}^2 = v_{top}^2 + 4gR$
$v_{bottom}^2 = gR + 4gR = 5gR$
$v_{bottom} = \sqrt{5gR}$
Given $g = 10 \,m/s^2$ and $R = 0.5 \,m$:
$v_{bottom} = \sqrt{5 \times 10 \times 0.5} = \sqrt{25} = 5 \,m/s$.

(C) To complete a vertical circle,the minimum velocity at the bottom must be $v = \sqrt{5Rg}$,where $R$ is the radius of the circle.
Applying the law of conservation of mechanical energy between the top of the incline and the bottom:
$P_1 + K_1 = P_2 + K_2$
$mgh + 0 = 0 + \frac{1}{2} m v^2$
$mgh = \frac{1}{2} m (\sqrt{5Rg})^2$
$mgh = \frac{1}{2} m (5Rg)$
$h = \frac{5R}{2}$
Given $R = 2 \ m$,we have:
$h = \frac{5 \times 2}{2} = 5 \ m$

(A) For a bob to complete a vertical circle,the minimum velocity at the lowest point is $v_0 = \sqrt{5gL}$.
Applying the law of conservation of energy between the lowest point and the point where the string is horizontal (at height $L$):
$\frac{1}{2}mv_0^2 = \frac{1}{2}mv^2 + mgL$
$\frac{1}{2}m(5gL) = \frac{1}{2}mv^2 + mgL$
$\frac{5}{2}gL = \frac{1}{2}v^2 + gL \implies \frac{3}{2}gL = \frac{1}{2}v^2 \implies v^2 = 3gL$.
At the horizontal position,the forces acting on the bob are:
$1$. Centripetal force (horizontal): $F_x = \frac{mv^2}{L} = \frac{m(3gL)}{L} = 3mg$.
$2$. Gravitational force (vertical): $F_y = mg$.
The net force is $F_{\text{net}} = \sqrt{F_x^2 + F_y^2} = \sqrt{(3mg)^2 + (mg)^2} = \sqrt{9m^2g^2 + m^2g^2} = \sqrt{10}mg$.

(B) Let $v$ be the velocity at the highest point. The forces acting on the particle at the highest point are the gravitational force $Mg$ (downwards),the electrostatic repulsive force $F_e = \frac{1}{4 \pi \epsilon_0} \frac{Q^2}{R^2}$ (upwards),and the tension $T$ (downwards).
Given $Mg = \frac{Q^2}{4 \pi \epsilon_0 R^2}$,the net force towards the center is $F_{net} = Mg - F_e + T = \frac{M v^2}{R}$.
At the highest point,$T = 0$,so $Mg - \frac{Q^2}{4 \pi \epsilon_0 R^2} = \frac{M v^2}{R}$.
Since $Mg = \frac{Q^2}{4 \pi \epsilon_0 R^2}$,we have $0 = \frac{M v^2}{R}$,which implies $v = 0$.
Now,applying the law of conservation of energy between the lowest point (velocity $v_0$) and the highest point (velocity $v = 0$):
Total energy at lowest point = Total energy at highest point.
$\frac{1}{2} M v_0^2 + U_{lowest} = \frac{1}{2} M v^2 + U_{highest}$.
The potential energy includes both gravitational and electrostatic components.
$U_{grav} = Mgh$,$U_{elec} = \frac{1}{4 \pi \epsilon_0} \frac{Q^2}{r}$.
Change in energy: $\frac{1}{2} M v_0^2 = Mg(2R) + \Delta U_{elec}$.
Since the distance $R$ from the center is constant,the electrostatic potential energy remains constant throughout the circular path.
Thus,$\frac{1}{2} M v_0^2 = Mg(2R)$.
$v_0^2 = 4gR$.
$v_0 = 2 \sqrt{gR}$.

(B) Let the particle be at a point $P$ where the thread makes an angle of $30^{\circ}$ with the horizontal. The angle with the vertical is $\theta = 90^{\circ} - 30^{\circ} = 60^{\circ}$.
At this point, the forces acting along the radial direction are the tension $T$ and the component of gravity $mg \cos 60^{\circ}$.
The equation of motion is $T + mg \cos 60^{\circ} = \frac{mv^2}{r}$.
Given that the tension $T = 0$ at this point, we have $mg \cos 60^{\circ} = \frac{mv^2}{r}$.
Since $\cos 60^{\circ} = \frac{1}{2}$, we get $mg(\frac{1}{2}) = \frac{mv^2}{r}$, which simplifies to $v^2 = \frac{gr}{2}$.
Now, apply the law of conservation of mechanical energy between the bottom point $A$ and point $P$.
The height of point $P$ above point $A$ is $h = r + r \sin 30^{\circ} = r + r(\frac{1}{2}) = \frac{3r}{2}$.
Let $u$ be the velocity at point $A$. Then, $\frac{1}{2}mu^2 = \frac{1}{2}mv^2 + mgh$.
Substituting the values: $\frac{1}{2}mu^2 = \frac{1}{2}m(\frac{gr}{2}) + mg(\frac{3r}{2})$.
$u^2 = \frac{gr}{2} + 3gr = \frac{7gr}{2}$.
Therefore, the velocity at the bottom point is $u = \sqrt{\frac{7}{2}gr}$.

(B) At the highest point of the vertical circular loop,the forces acting on the body are the normal force $N$ (directed downwards) and the gravitational force $mg$ (directed downwards).
These forces provide the necessary centripetal force: $N + mg = \frac{mv^2}{R}$.
Given that the body exerts a force of three times its weight on the plane,the normal force $N = 3mg$.
Substituting this into the centripetal force equation: $3mg + mg = \frac{mv^2}{R} \Rightarrow 4mg = \frac{mv^2}{R} \Rightarrow v^2 = 4gR$.
Now,we apply the law of conservation of mechanical energy between the starting point at height $h$ and the highest point of the loop at height $2R$.
The total energy at the start is $mgh$ (potential energy).
The total energy at the highest point is $mg(2R) + \frac{1}{2}mv^2$ (potential energy + kinetic energy).
Equating the energies: $mgh = 2mgR + \frac{1}{2}m(4gR) = 2mgR + 2mgR = 4mgR$.
Thus,$h = 4R$.
Comparing this with $h = \alpha R$,we get $\alpha = 4$.