Circular Motion (Kinematics, Dynamics)

$v \approx 30$ km/s.

$v \approx 316$ m/s; the point describes a helical line with a pitch $h \approx 1.35$ m.

The velocity $v$ at this moment is possessed by points lying on an arc of radius $R$, whose center is at the point of contact between the disk and the plane (instantaneous center of rotation).

Let the roller move translationally with velocity $v$ and rotate about its axis with angular velocity $\omega$. The linear velocity of points on the roller in the coordinate system associated with its axis is $\omega R$. The points of contact between the roller and the racks also have this velocity. Since there is no slipping, this is also the velocity of the racks. In the coordinate system associated with the Earth, the velocities of the racks are equal to the sum of their velocities relative to the cylinder axis and the velocity of the cylinder axis relative to the Earth. Therefore, $v_1 = v + \omega R,\qquad v_2 = v - \omega R.$ From this, $v = \frac{1}{2}(v_1 + v_2),\qquad \omega = \frac{1}{2R}(v_1 - v_2).$ For the case where the rack velocities are directed in opposite directions ($v_1 = v + \omega R$, $v_2 = \omega R - v$), we get: $v = \frac{1}{2}(v_1 - v_2);\qquad \omega = \frac{v_1 + v_2}{2R}.$

$v_A = 2v_c \cos \alpha$. The accelerations of all points on the rim are centripetal and equal to $a_{cp} = v_c^2/R$.

The angular velocities of rotation of all points on the rim relative to the axis passing through the point of contact of the hoop with the plane (instantaneous center of rotation) are identical and equal to $\omega = v_c/R$. The linear velocity of point $A$ is $v_A = \omega r = \omega (2R \cos \alpha) = \frac{v_c}{R} 2R \cos \alpha = 2v_c \cos \alpha$ and is perpendicular to OA (see figure). Since the translational motion of the hoop is uniform, its points only have centripetal acceleration. This becomes obvious when considering the motion of the hoop in an inertial frame of reference moving with velocity $v_c$.

$n \approx 8.84 \text{ s}^{-1}$;

$a_{cp} \approx 926 \text{ m/s}^2$ (this acceleration is 94.4 times greater than $g$).

$v = v_0/2$.

1st method. When slipping, a constant friction force acts on the cylinder from the plane, which accelerates its translational motion and decelerates its rotational motion. After time $t$, the translational velocity will be $v = at$, and the linear velocity of rotational motion relative to the cylinder axis will be $v_r = v_0 - at$ (since the cylinder is thin-walled, all elementary masses have the same acceleration $a$ relative to the cylinder axis, equal in magnitude to the acceleration of the translational motion).

The velocity of the point on the cylinder touching the plane is equal to the difference in velocities $v_r - v$. Slipping will stop when this velocity becomes zero:
$v_r - v = (v_0 - at) - at = 0,$
whence $at = v_0/2$. Thus, the translational velocity (velocity of the cylinder axis) at this moment will be $v_0/2$ and will not change further.

2nd method. While slipping, the cylinder performs work against the friction force $F$ applied to it: $A = F S_{sl},$
where $S_{sl}$ is the path traveled by the point (more precisely, the line) of application of the friction force on the cylinder surface during its sliding on the plane over time $t$: from the moment of contact, when the sliding velocity $v_{sl} = v_0$, until the moment slipping stops ($v_{\text{sl}} = 0$). Since the motion is uniformly accelerated, the average sliding velocity is $v_0/2$. Thus,
$A = F S_{\text{sl}} = F v_0 \frac{t}{2}.$
According to the work-energy theorem, the work $A$ equals the decrease in the kinetic energy of the cylinder:
$\frac{mv_0^2}{2} - (\frac{mv^2}{2} (\text{trans}) + \frac{mv^2}{2} (\text{rot})) = F \frac{v_0 t}{2}$
(cylinder is thin-walled, therefore $T_{trans} = T_{rot}$, see problem 50). The friction force $F$, which brakes the cylinder's rotation to velocity $v$ over time $t$, simultaneously imparts translational motion with the same velocity $v$. Writing Newton's second law for the translational motion in impulse form $Ft = mv$ and substituting $Ft$ with $mv$ in equation (1), we obtain an equation from which we find $v = v_0/2$.

No. This force, according to the problem statement, ensures the uniform rotation of the body, is a constant magnitude centripetal force, which at each moment in time is directed perpendicular to the displacement of the body and does no work.

$\omega = \sqrt{\frac{1}{3} \frac{k}{m}}$.

The centripetal acceleration $a_{\text{cp}} = \omega^2 (l_0 + \Delta l) = 1.5 l_0 \omega^2$ ($l_0$ is the initial length of the spring; $\Delta l$ is its extension) is imparted to the mass by the tension force of the spring, which, according to Hooke's law, is $F = k \Delta l = 0.5 l_0 k.$ Therefore, $1.5 m \omega^2 l_0 = 0.5 l_0 k.$ Hence $\omega = \sqrt{\frac{1}{3} \frac{k}{m}}.$

$T_1 = (m_1 l_1 + m_2 l_2) \omega^2$;

$T_2 = m_2 \omega^2 l_2$.

The force $T_2$ imparts centripetal acceleration to the outermost mass; the force $(T_1 - T_2)$ imparts it to the other mass (see the following figure).

$n = 6.75 \text{ min}^{-1}$.

A person cannot stay on the platform if the maximum possible static friction force with the platform is insufficient to provide the necessary centripetal acceleration, i.e., $F_{fr} = kmg < m\omega^2 R$ or $n = \omega/(2\pi) > \frac{1}{2\pi} \sqrt{kg/R}.$

See the following figure.

Before the body starts sliding, the friction force provides it with centripetal acceleration $a = \omega^2 r.$ Therefore, considering the smallness of the tangential acceleration according to the condition, $F_{fr} \approx m\omega^2 r \le kmg.$ The body starts sliding at $\omega = \omega_0$, such that $F_{\text{cp}} = kmg$, i.e., when $\omega_0^2 r = kg$.

$h = l(T - mg)/(2mg) \approx 2.0$ m.

The kinetic energy of the stone at the lowest point of the circle completely converts into its potential energy during the ascent: $mv^2/2 = mgh$, where $h$ is the height of ascent above the lowest point of the circle. Further, write the equation of dynamics of rotational motion for the moment of passing the lowest point $mv^2/l = T - mg$ and solve both equations together to find the answer.

$T = mg[\frac{l}{R} - (1 + \frac{\sqrt{2}}{2})] \approx 2205$ N.

The flight distance $l$ will be maximum if the hammer's velocity is directed at an angle $\alpha = 45^\circ$ to the horizontal. In this case (see problem 63 in Chapter: Kinematics) $l_0 = \sqrt{v_0^2/g}$, where $l_0$ is the flight distance when the points of takeoff and landing are at the same height above the Earth's surface. In our case (see figure) $l_0 = l - R$.

a) $F_1 = mg \approx 29400$ N;

b) $F_2 = mg (\cos \alpha - \frac{v^2}{gR}) \approx 24000$ N;

c) $F_3 = mg (\cos \alpha + \frac{v^2}{gR}) \approx 34000$ N.

At the moment the car passes the middle of the bridge, $F_2 = 24400$ N; $F_3 = 34400$ N.

Two forces act on the car: the force of gravity $F_t = mg$ and the normal reaction force of the bridge $N$, which, according to Newton's third law, is equal in magnitude to the force $F$ of the car's pressure on the bridge. Along the curved bridge, the car moves with centripetal acceleration, imparted to it by the resultant of the force $N$ and the radial component of the gravity force, equal to $mg \cos \alpha$ (see figure). In accordance with Newton's second law, one can write $mv^2/R = mg \cos \alpha - N_2$ for a convex bridge and $mv^2/R = N_3 - mg \cos \alpha$ for a concave bridge.

$\alpha = \arccos (\frac{F}{mg} + \frac{v^2}{gR}) \approx 8.2^\circ$.

$A = mgl (1 + \frac{1}{2} \tan \alpha \sin \alpha - \cos \alpha) \approx 1.23$ J.

The work goes into increasing the potential and kinetic energy. The speed of the ball is found from the condition $mv^2/(l \sin \alpha) = mg \tan \alpha$, and $h = l (1 - \cos \alpha)$.

$v = \sqrt{kgR} \approx 89$ km/h.

See problem 11.

1. $k = v^2/(gR) \approx 0.4$.

2. $v = \dfrac{\sqrt{2kgR\alpha}}{\sqrt{1 + 4\alpha^2}} \approx 14.5$ m/s.

• See problem 11.

The acceleration $a$ of the car is composed of the constant tangential acceleration $a_t$, which provides the car's acceleration, and the centripetal acceleration $a_{cp} = v^2/R$ ($v$ is the car's speed). Therefore
$a = \sqrt{a_t^2 + \frac{v^4}{R^2}}.$
The external force providing acceleration to the car is the friction force of the wheels on the road. Since the friction force cannot exceed $kmg$ ($m$ is the mass of the car), the maximum acceleration $a$ at the moment the car transitions to the horizontal section of the road cannot be greater than $kg$:
$\sqrt{a_t^2 + \frac{v^4}{R^2}} = kg.$
Since
$v = \sqrt{2a_t l}$
($l = R\alpha$ is the path traveled by the car; $\alpha = \pi/6$), then
$a_t = \frac{v^2}{2l} = \frac{v^2}{2R\alpha},$
and after substitution we get
$\frac{v^2}{R} \sqrt{\frac{1}{4\alpha^2} + 1} = kg.$
Hence $v = \frac{\sqrt{2kgR\alpha}}{\sqrt{1 + 4\alpha^2}}$.

$\Delta h \approx 7.65$ cm.

The resultant force $\vec{F}$ of the wagon's weight $m\vec{g}$ and the reaction force of the rails $\vec{N}$ imparts centripetal acceleration $a_{cp}$ to the wagon; $m a_{\text{cp}} = m\vec{g} + \vec{N};$ the resultant is directed horizontally (since its direction must coincide with the direction of the train's acceleration) and is equal to $mg \tan \alpha$ (see figure): $mv^2/R = mg \tan \alpha.$ Hence $\tan \alpha = \frac{v^2}{gR}.$ The upper rail exceeds the lower one by the amount $\Delta h = d \tan \alpha = d v^2/(gR).$

• In this section, the dimensions of the bodies are considered much smaller than the radius of rotation, i.e., the entire mass of the body is considered concentrated at one point - its center of mass.

$\alpha \approx 22^\circ$.

For a motorcyclist to move in a circle, he must lean so that the resultant of the three forces applied to him: gravity $m\vec{g}$, normal reaction force of the road $\vec{N}$, and friction force with the road $\vec{F}_{fr}$ (see figure) - provides him with centripetal acceleration. Since the center of mass of the motorcyclist does not move vertically, $N = mg$. The centripetal acceleration is ultimately provided by the friction force: $mv^2/R = F_{fr}$.

The resultant $\vec{R} = \vec{N} + \vec{F}{fr}$, obviously, must pass through the center of mass of the motorcyclist (the moment of the resultant relative to the center of mass must be zero). Therefore
$\cot \alpha = N/F{fr} = mg / (mv^2/R) = gR/v^2.$

1. $v \approx 18.8$ m/s.

2. $\phi \approx 21.8^\circ$.
3. $v_{\text{max}} \approx 33.5$ m/s.
4. $\alpha_0 = \arctan (1/k)$.

• Centripetal acceleration is provided by the friction force (see previous problem), therefore $mv^2/R = F_{fr}$. But the friction force cannot exceed the value
  $F_{\text{fr.max}} = kN = kmg,$
  meaning the motorcyclist's speed cannot be greater than some limiting value, determined by this maximum possible value of the friction force:
  $mv_{\text{max}}^2/R = kmg.$
  Hence $v_{\text{max}} = \sqrt{kgR}.$
• The angle of inclination of the motorcyclist at this speed must be equal to:
  $\phi = \arctan \frac{v_{\text{max}}^2}{gR} = \arctan k.$
• In creating the centripetal acceleration, the horizontal components of the friction force and the track reaction force will participate (see figure):
  $mv_{\text{max}}^2/R = F_{fr.max} \cos \alpha + N \sin \alpha. \tag{1}$
  The sum of the vertical components of the forces acting on the motorcycle must be zero (there is no acceleration in the vertical direction):
  $F_{fr.max} \sin \alpha + mg - N \cos \alpha = 0.\tag{2}$
  Maximum friction force
  $F_{fr.max} = kN.\tag{3}$
  Solving equations (1), (2), and (3) together, we get
  $v_{\text{max}} = \sqrt{gR \frac{k + \tan \alpha}{1 - k \tan \alpha}}.$

• Hint. From part 3, it follows that
  $v_{\text{max}} \to \infty$ when $1 - k \tan \alpha \to 0$, whence
  $\alpha_0 = \arctan (1/k).$

  

$R \approx 5780$ m.

The aircraft changes its direction of flight by transitioning to motion along an arc of a circle by banking its fuselage around the direction of flight. In this case, the centripetal acceleration is imparted to the aircraft by the resultant of the gravity force and the lift force, perpendicular to the plane of the wings. This resultant is equal to $mg \tan \alpha$ (see figure). According to Newton's second law $mg \tan \alpha = mv^2/R,$ whence $R = v^2/(g \tan \alpha).$

$v \approx 26.1$ m/s.

When the car turns, the pressure on its wheels, and hence the forces acting on the car wheels from the road, are redistributed. At the maximum permissible speed of the car, the forces acting on it from the road will be applied to the outer wheels and are equal to: a) the reaction force $N = mg$ (numerically equal to the gravity force acting on the car, as there is no acceleration in the vertical direction) and b) the friction force $F_{fr} = mv^2/R$ (as it provides the car with centripetal acceleration). The car will overturn if the direction of the resultant of these forces passes below the center of mass of the car - in this case, there is an overturning moment of force relative to the center of mass. At the critical speed, this resultant passes through the center of mass of the car and therefore (see figure) $N/F_{fr} = \tan \alpha = (a/2)/h = a/(2h),$ or $mg / (mv^2/R) = a/(2h),$ whence $v = \sqrt{gRa/(2h)}.$

Brake.

During braking, the friction force $F_{fr}$ must perform work $A = F_{fr} x$ ($x$ is the braking distance), equal to the kinetic energy the car had before braking: $F_{fr} x = mv^2/2$. Hence the braking distance $x = mv^2/(2 F_{fr})$. If the driver, without braking, turns aside, the car will move along an arc of a circle, and the centripetal acceleration $a_{cp} = v^2/R$ ($R$ is the radius of curvature) will be provided to the car by the same friction force. From the equation of motion $F_{fr} = mv^2/R$, we find that the turning radius will be $R = mv^2/F_{fr}$, i.e., it is twice the 'braking distance'. This means it is more advantageous to brake than to turn.

$T \approx 51$ N.

Centripetal acceleration is imparted to the suspended mass by the resultant of the gravity force and the tension force $T$ of the spring. This resultant is equal to $mg \tan \alpha$ (see figure). According to Newton's second law $mg \tan \alpha = mv^2/R$, hence $\tan \alpha = v^2/(gR)$. The tension force of the spring $T = mg/\cos \alpha = mg \sqrt{1 + \tan^2 \alpha} = mg \sqrt{1 + \frac{v^4}{g^2 R^2}}.$

a) $F_{sep} = (\rho_m - \rho_c) g \approx 980 \text{ N/m}^3$;

b) $F_{sep} = (\rho_m - \rho_c) \omega^2 r \approx 3.94 \times 10^6 \text{ N/m}^3$.

a) In a stationary vessel, the separating force is the Archimedean (buoyant) force.

b) If the volume $V$, occupied by a particle of cream, were occupied by milk, it would be in equilibrium. This means that a force $F_1 = \rho_m \omega^2 r V,$ directed towards the axis of rotation, acts on the cream particle from the surrounding milk. This force turns out to be greater than the force required for the circular motion of the cream particle, which is $F_2 = \rho_c \omega^2 r V.$
The difference between these forces is the separating force.

$F_{\text{top}} = \frac{Mv^2}{R} - Mg \approx 4030$ N, $F_{\text{bottom}} = \frac{Mv^2}{R} + Mg \approx 5630$ N.

See problem 15. The centripetal acceleration at the top and bottom points of the 'dead loop' is imparted to the pilot by the resultant of the gravity force and the reaction force of the seat.

$T = Mg/\cos \alpha \approx 990$ N; $\omega = \sqrt{g/(l\cos\alpha)} \approx 1.68$ rad/s.

Centripetal acceleration is imparted to the person by the resultant of the gravity force and the tension force of the rope; this resultant is horizontal.

$T = 2\pi \sqrt{l\cos\alpha/g}$. For small angles $T \approx 2\pi \sqrt{l/g}$.

Centripetal acceleration is imparted to the pendulum by the resultant of the gravity force $m\vec{g}$ and the tension force of the string $\vec{T}$ (see figure). This resultant lies in the plane of motion of the pendulum and is equal to $mg \tan \alpha$. According to Newton's second law, $m\omega^2 R = mg \tan \alpha$ or $\frac{4\pi^2}{T^2} l \sin \alpha = g \tan \alpha$ (since $R = l \sin \alpha$ and $\omega = 2\pi/T$). Hence $T = 2\pi \sqrt{l \cos \alpha / g}.$

$n = \frac{1}{2\pi} \sqrt{g/h}$. (The answer does not depend on the length of the suspension.)

$h \le (T - mg) l / (2mg) \approx 2.5$ m.

Two forces act on the chandelier at any point of the trajectory (see figure): the force of gravity $mg$ and the tension force of the chain $T$. When the chandelier passes the equilibrium position (point A), the forces are directed along the same line. Let's write the equation of Newton's second law for the chandelier at the moment it passes point A: $ma_{cp} = T - mg,$ where $a_{cp} = v^2/l$ is the centripetal acceleration. Then $mv^2/l = T - mg.\tag{1}$ From the law of conservation of energy, we find the speed of the chandelier at point A: $mgh = mv^2/2,$ or $v^2 = 2gh.\tag{2}$ Substituting expression (2) into equation (1), we have $m(2gh)/l = T - mg,$ whence $h = (T - mg) l / (2mg).$ According to the condition, $T < 2mg$, therefore, $h < 2.5 \text{ m}$.

$\alpha_{\text{min}} = \arccos 0.75 \approx 41.4^\circ$.

See the previous problem.

$\alpha = \arccos (1/3)$.

$6mg$.

See the hint for problem 28. The relationship between the speeds of the mass at the top $v_t$ and bottom $v_b$ points is determined by the law of conservation of energy: $\frac{mv_b^2}{2} - \frac{mv_t^2}{2} = mg \cdot 2R.$

$5mg$.

See the previous problem and problem 28.

$v_{\text{min}} = \sqrt{5gl}$ for the string; $v_{\text{min}} = 2\sqrt{gl}$ for the rod.

In the case of a massless rod, the minimum speed $v_{\text{min}}$ of the mass at the lowest point is determined by the condition: the change in its potential energy upon rising from the lowest to the highest point equals the initial kinetic energy of the mass: $mg \cdot 2l = mv^2/2.$ The speed of the mass at the highest point is zero. In the case where the mass rotates on an inextensible string, this condition is insufficient; another condition is needed: the string must remain taut up to the highest point. The speed of the mass at the highest point cannot be zero, but must be such that the centripetal acceleration is provided to the mass only by the force of gravity: $mv_{top}^2/l = mg$. At this moment $T = 0$, which is the condition for the minimum speed.

$T = 3Mg \sin \alpha$; $T = 3Mg$.

See problem 32.

$\displaystyle T = Mg(3 \cos \phi - 2 \cos \phi_0 + \frac{v_0^2}{gl})$ until it comes to rest.

See problem 32. Consider the initial kinetic energy of the pendulum.

$\cos \alpha = \sqrt{3}/3$; $\alpha \approx 54.7^\circ$.

The vertical component of the bob's velocity will increase until the vertical component of the resultant of the applied forces, gravity $m\vec{g}$ and string tension $\vec{T}$ (see figure), becomes zero: $T \cos \alpha - mg = 0.\tag{1}$ After this, the vertical component of the bob's velocity will decrease. Since the bob moves in a circle, according to Newton's second law $mv^2/R = T - mg \cos \alpha.$ Substituting the bob's velocity found from the law of conservation of energy: $mv^2/2 = mgR \cos \alpha$ [assuming release from horizontal position], into this equation, and solving the resulting equation together with equation (1), we find: $\cos \alpha = 1/\sqrt{3}.$

a) $F = 2mg \cos^2 \alpha$;

b) $F = 2mg (3 - 2 \cos \alpha)$;

c) $F = 2mg$.

a) The resultant of the string tension force and the ball's weight force is directed perpendicular to the string and imparts an initial acceleration $a = g \sin \alpha$ to the ball.

b) The speed of each ball at the initial moment of impact can be found using the law of conservation of mechanical energy. Then, from the laws of conservation of energy and momentum, it follows (see problems 28 and 39 in Chapter: Conservation of Energy) that the speeds of the balls at the final moment of impact will be equal in magnitude to their speeds at the initial moment of impact and directed in opposite directions. Knowing the speeds of the balls, it is easy to find the tension forces of the strings (see, for example, problem 32).

c) At the moment of maximum deformation of the balls, the speed of each ball is zero.

$h = \sqrt{3}l$; parabolic trajectory (trajectory of a body thrown at an angle to the horizon); $H = 40l/27$.

See problems 37 and 32. At the point of maximum ascent of the pendulum, the tension force of the string becomes zero, and the centripetal acceleration is provided to the pendulum by the radial component of the gravity force.

Below the suspension point by $l/6$; parabolic; $2l/27$ below the suspension point.

See the previous problem.

$\omega > \sqrt{2g\tan\alpha/D} \approx 13$ rad/s.

Two forces act on the ball located on the wall of the rotating vessel: the force of gravity $mg$ and the reaction force $N$ of the vessel wall (see figure). At equilibrium of the ball relative to the vessel, the resultant of these forces provides the ball with centripetal acceleration, i.e., $m\omega^2 R = mg \tan \alpha.$ From this, it follows that the ball can be in equilibrium at a distance $R$ from the axis of rotation of the vessel if the vessel rotates with angular velocity $\omega = \sqrt{g\tan\alpha/R}.$ At the bottom of the vessel, the ball is still in equilibrium, but no longer presses on the bottom at $\omega_{cr} = \sqrt{2g\tan\alpha/D};$ if $\omega$ is increased, then from the equation given above, it follows that for no $R \ge D/2$ will it be in equilibrium and will be ejected.

$h \approx 1$ m; $N \approx 0.4$ N.

At equilibrium of the ball relative to the sphere, the resultant of the reaction force $N$ of the sphere and the gravity force $mg$ of the ball is directed towards the axis of rotation and provides the ball with centripetal acceleration. The magnitude of this resultant $Q = mg \tan \alpha$ (see figure).

According to Newton's second law $m\omega^2 l = mg \tan \alpha,$ where $l = R \sin \alpha$ is the distance of the ball from the axis of rotation;
$\omega^2 R \sin \alpha = g \tan \alpha.$
This equation has two roots for $\alpha$:

1. $\sin \alpha = 0$, $\alpha = 0$;
2. $\cos \alpha = g/(\omega^2 R)$.

The position of the ball with $\alpha = \arccos\left(\frac{g}{\omega^2 R}\right)$ is possible only when
$\omega \ge \sqrt{g/R}.$
The position $\alpha = 0$ is possible at any angular velocity of the sphere, but for $\omega < \sqrt{g/R}$ the equilibrium is stable, and for $\omega \ge \sqrt{g/R}$ it is unstable. In our case $\omega > \sqrt{g/R} \approx 2.2 \text{ s}^{-1}$. Height $h = R - R \cos \alpha = R\left(1 - \frac{g}{\omega^2 R}\right)$, and the reaction force of the sphere
$N = mg/\cos \alpha = m\omega^2 R.$

$T = 2\pi \sqrt{R\tan\alpha/(a + g)}$.

The forces acting on the ball must provide it with both horizontal and vertical acceleration. Let's resolve the force $F$, acting on the ball from the side of the conical surface, into vertical $F' = F \sin \alpha$ and horizontal $F'' = F \cos \alpha$ components (see figure). The resultant of the vertical component of the force $F'$ and the gravity force provides the ball with acceleration $a$ upwards. The horizontal component of the force $F''$ provides the ball with centripetal acceleration $a_{cp} = \omega^2 R$ ($\omega$ is the angular velocity of rotation). Therefore, according to Newton's second law $ma = F \sin \alpha - mg$ and $m\omega^2 R = F \cos \alpha$. Eliminating $F$ from these equations, we find: $\omega = \sqrt{(a + g)/(R\tan\alpha)},$ $T = 2\pi/\omega = 2\pi \sqrt{R\tan\alpha/(a + g)}.$

a) $h = 2.5R$;

b) $F = 3mg (1 - \cos \alpha)$.

a) The body will not detach from the loop if at the highest point it has a speed $v \ge \sqrt{gR}$ (see problem 37). Writing the law of conservation of energy $\frac{1}{2}mv^2+ mg \cdot 2R = mgh,$ we find that the body must slide down from a height $h = 2.5 R$.

b) At the point whose radius vector makes an angle $\alpha$ with the vertical (see figure), the centripetal acceleration will be provided to the body by the resultant of the reaction force $N$ of the platform and the radial component of the gravity force $(mg)_r = mg \cos \alpha$.

According to Newton's second law
$mv^2/R = N + mg \cos \alpha.$
The speed of the body at this point can be easily determined using the law of conservation of energy:
$\frac{5}{2} mgR = \frac{mv^2}{2} + mg (R + R \cos \alpha),$
$v = \sqrt{gR (3 - 2 \cos \alpha)}.$
Substituting this expression for $v$ into Newton’s second law, we get:
$
N = \frac{m}{R} \left[ gR(3 - 2\cos\alpha) \right] - mg \cos\alpha = 3mg(1 - \cos\alpha)
$
The normal force $F$ exerted by the body on the track numerically equals the normal reaction $N$.
(Determine the value of $N$ for $\alpha = 0$ and $\alpha = \pi$.)

$v_{\text{min}} = \sqrt{gR \cos \alpha}$.

Separation of the ore from the belt means that the ore has lost contact with the belt and only the force of gravity acts on the ore. Its radial component provides the ore particles with centripetal acceleration at the moment the belt runs onto the drum (see figure) $mv^2/R = mg \cos \alpha.$

$h = R/3$.

Before detaching from the surface of the sphere, the body is acted upon by its weight $mg$ and the reaction force of the support $N$. The resultant of the radial component of the weight force and the support reaction force provides the body with centripetal acceleration $mv^2/R = mg \cos \alpha - N.$ As the body slides down, the component of the weight force $mg \cos \alpha$ (where $\alpha$ is the angle between the vertical and the direction from the center of the sphere to the body) decreases, and its speed increases, which leads to an even faster decrease in the support reaction force (the support reaction force depends on the nature of the body's motion). At the moment of detachment, $N$ becomes zero and then $mv^2/R = mg \cos \alpha.\tag{1}$ Based on the law of conservation of energy $mv^2/2 = mgh.\tag{2}$ Also, it is obvious that $\cos \alpha = (R - h)/R$. (3) Solving equations (1), (2), (3) together, we get $h = R/3.$

$mv^2$.

Each point of the hoop participates in two motions: translational motion with velocity $v$ and rotational motion with angular velocity $\omega$ about the center of the hoop. The linear velocity of rotational motion is $v_{rot} = \omega R$. In the absence of slipping $\omega = v/R$ and therefore $v_{rot} = v$. Consider two diametrically opposite identical small elements of the hoop, having mass $\Delta m$ (see figure). Their velocities are equal to the vector sum of the velocities of translational and rotational motions, and therefore: $v_A^2 = v^2 + v_{rot}^2 - 2vv_{rot} \cos \alpha = 2v^2 (1 - \cos \alpha);$ $v_B^2 = v^2 + v_{rot}^2 + 2vv_{rot} \cos \alpha = 2v^2 (1 + \cos \alpha).$ The total kinetic energy of these elements of the hoop is $\frac{\Delta m v_A^2}{2} + \frac{\Delta m v_B^2}{2} = 2 \Delta m v^2.$ This relationship does not depend on $\alpha$, i.e., on the choice of the diameter at the ends of which the chosen elements of the hoop lie. Therefore, the kinetic energy of the entire hoop will be equal to $mv^2$.

$h = (R - r)/2$.

Sliding down, the hoop moves along a circle. Centripetal acceleration is imparted to it by the sum of the radial components of the forces applied to the hoop (see figure), equal to $N - mg \cos \alpha$ ($N$ is the reaction force of the hemisphere, $m$ is the mass of the hoop, and $\alpha$ is the angle formed by the radius vector of the hoop with the vertical). According to Newton's second law $mv^2/(R - r) = N - mg \cos \alpha,\tag{1}$ where $v$ is the velocity of the center of the hoop. From the law of conservation of energy, it follows that $mg(R-r)(1-\cos\alpha) = \frac{mv^2}{2} + \frac{I\omega^2}{2}.$ For a hoop $I=mr^2$. With no slip $v=\omega r$. So $mg(R-r)(1-\cos\alpha) = \frac{mv^2}{2} + \frac{m(v/r)^2 r^2}{2} = mv^2.\tag{2}$ Here $mv^2/2$ corresponds to the kinetic energy of the translational motion of the center of mass of the hoop; $\frac{m(\omega r)^2}{2}$ corresponds to the kinetic energy of rotation of the hoop relative to the center of mass (see previous problem); $mg(R-r)(1-\cos\alpha)$ is the change in potential energy. From equation (2) we find $v^2 = g (R - r) (1-\cos \alpha)$. Substituting this expression for $v^2$ into equation (1) and setting $N = 0$ (condition for detachment), we finally get $\cos \alpha = 1/2$ and $h = (R - r) \cos \alpha = (R - r)/2.$

a) $mgR$;

b) half;

c) $2mg$.

$p = \frac{4M^2}{\pi \rho a^3 R t^2} \approx 1.2 \times 10^5$ Pa.

The centripetal acceleration of the mass of the selected volume of water (see figure) is provided by the force of lateral pressure $F$ from the pipe walls: $mv^2/R = F.$ The force $F$ is equal to the product of the pressure $p$ and the area of the diametral section ABCD, perpendicular to the radius of the ring, i.e., $F = pdl.$ The mass of the selected volume $m = \rho \pi d^2 l / 4.$ The flow velocity of water is determined from the formula $M = \rho \pi d^2 v t / 4\quad (\tex{time } t = 1 \text{ h}),$ whence $v = \frac{4M}{\rho \pi d^2 t}.$ Substituting the values of $m$ and $v$ into the formula for $F$, we get the answer.

When moving along a convex arc.

Indeed, in this case, the pressure of the body on the surface at each point is less than the pressure on the surface at a point located at the same distance from point A, in the case of the body moving along a concave arc (see problem 15). This means that the friction force acting on the body at each point, and consequently, the work done against friction over the entire path, is less than when moving along the concave arc, and the body at point B has a greater kinetic energy in this case.

$\omega = 2 \sqrt{\dfrac{(m_1 - m_2)g}{(m_1 + m_2)l}}$;

$F = g \dfrac{3(m_1^2 + m_2^2) - 2m_1 m_2}{m_1 + m_2}$.

The two masses move with the same angular velocity. Let us choose the potential energy reference level at the height where the mass $m_1$ is located at the moment it passes through the equilibrium position. Then, by the law of conservation of energy, we can write: $ (m_1 + m_2)g\frac{l}{2} = m_2 g l + \frac{1}{2} m_1 v_1^2 + \frac{1}{2} m_2 v_2^2 $

Substituting $v_1 = v_2 = \omega \frac{l}{2}$, we find:
$
\omega = 2 \sqrt{ \frac{(m_1 - m_2)g}{(m_1 + m_2)l} }
$

The force $F$ exerted on the axis equals the sum of the tension forces $T_1$ and $T_2$ in each half of the rod. We assume both rods are under tension. Then, according to Newton’s second law:
$
T_1 - m_1 g = m_1 \omega^2 \frac{l}{2}, \quad m_2 g + T_2 = m_2 \omega^2 \frac{l}{2}
$

Substituting the expression for $\omega$ at the moment both masses pass through the equilibrium position and solving these equations for $T_1$ and $T_2$, we obtain:
$
T_1 = \frac{m_1 g}{m_1 + m_2}(3m_1 - m_2), \quad T_2 = \frac{m_2 g}{m_1 + m_2}(m_1 - 3m_2)
$
(When $m_1 < 3m_2$, we get $T_2 < 0$. This means that the half of the rod to which the mass $m_2$ is attached is under compression rather than tension, contrary to our assumption.)

The total force on the axis (assuming both rods are in tension):
$
F = T_1 - T_2 = \frac{3\left(m_1^2 + m_2^2\right) - 2m_1 m_2}{m_1 + m_2} g
$

$F \approx mg \sqrt{1 + \dfrac{4h^2 n^2}{R^2}}$.

After making $n$ revolutions, the ring descends by a height $nh$, and from the law of conservation of energy $mghn = \frac{1}{2}mv^2$, it follows that at this point the ring will have a speed: $ v = \sqrt{2ghn} $

This speed is directed tangent to the spiral and can be considered as the sum of a vertical component $v_0$ and a horizontal (rotational) component $v_1$.

In particular:
$
v_1 = v \cos\alpha,
$
where $\alpha$ is the inclination angle of the spiral;
$\alpha \approx \sin\alpha = \frac{h}{2\pi R}$.

Since $h \ll R$, we can take $\cos\alpha \approx 1$ and $v_1 \approx v$.

The horizontal component of the spiral’s reaction force must provide the ring with centripetal acceleration:
$
a_{\text{cp}} = \frac{v_1^2}{R} \approx \frac{2ghn}{R}
$
Therefore, the horizontal reaction force is:
$
F_{\text{hor}} = ma_{\text{cp}} = \frac{2mghn}{R}
$

The ring presses on the spiral with the same force in the horizontal direction.
Thus, the total pressure force $F$ of the ring on the spiral consists of the horizontal component $F_{\text{hor}}$ and the vertical component of gravity, $mg \cos\alpha \approx mg$:
$
F \approx \sqrt{(mg)^2 + \left(\frac{2mghn}{R}\right)^2}
= mg \sqrt{1 + \frac{4h^2 n^2}{R^2}}
$

$T \approx m n^2 l \approx 12$ N.

Consider a small element of the chain of length $\Delta l = R \Delta \alpha$. The mass of this element $\Delta m = m \Delta l / l = m R \Delta \alpha / l.$ During the rotation of the chain, the centripetal acceleration is imparted to the selected element by the tension forces acting on it from the rest of the chain. The resultant of these forces is $2T \sin (\Delta \alpha / 2)$ (see figure).

According to Newton's second law
$2T \sin (\Delta \alpha / 2) = (\Delta m) \omega^2 R.$
Substituting $\omega = 2\pi n$ here and taking into account that for small angles $\sin (\Delta \alpha / 2) \approx \Delta \alpha / 2$, we get
$2T \Delta \alpha / 2 \approx m R^2 \Delta \alpha (2\pi n)^2 / l,$
whence $T \approx m n^2 (2\pi R)^2 / l = m n^2 l.$

$\Delta F_{lift} \approx 1.74 \times 10^3$ N.

The resultant of the gravity force of the aircraft and the lift force $F_{lift}$ provides it with centripetal acceleration. When moving from west to east, the velocities of the aircraft and the Earth add up. Therefore, we can write $m (\omega R + v)^2 / R = mg - F_{1lift}\tag{1}$ where $\omega$ is the angular velocity of the Earth; $R$ is its radius. If the aircraft flies in the opposite direction, then $m (\omega R - v)^2 / R = mg - F_{2lift}. \tag{2}$ From equations (2) and (1) we find the change in lift force: $\Delta F_{lift} = F_{2lift} - F_{1lift} = 4mv\omega = 4m (2\pi v / T),$ where $T$ is the period of rotation of the Earth, i.e., one day.