Circular Motion (Kinematics, Dynamics)

km/s.

m/s; the point describes a helical line with a pitch m.

The velocity at this moment is possessed by points lying on an arc of radius , 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 and rotate about its axis with angular velocity . The linear velocity of points on the roller in the coordinate system associated with its axis is . 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, From this, For the case where the rack velocities are directed in opposite directions (, ), we get:

. The accelerations of all points on the rim are centripetal and equal to .

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 . The linear velocity of point is 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 .

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(this acceleration is 94.4 times greater than ).

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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 , the translational velocity will be , and the linear velocity of rotational motion relative to the cylinder axis will be (since the cylinder is thin-walled, all elementary masses have the same acceleration 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 . Slipping will stop when this velocity becomes zero:

whence . Thus, the translational velocity (velocity of the cylinder axis) at this moment will be and will not change further.

2nd method. While slipping, the cylinder performs work against the friction force applied to it:
where 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 : from the moment of contact, when the sliding velocity , until the moment slipping stops (). Since the motion is uniformly accelerated, the average sliding velocity is . Thus,

According to the work-energy theorem, the work equals the decrease in the kinetic energy of the cylinder:

(cylinder is thin-walled, therefore , see problem 50). The friction force , which brakes the cylinder's rotation to velocity over time , simultaneously imparts translational motion with the same velocity . Writing Newton's second law for the translational motion in impulse form and substituting with in equation (1), we obtain an equation from which we find .

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.

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The centripetal acceleration ( is the initial length of the spring; is its extension) is imparted to the mass by the tension force of the spring, which, according to Hooke's law, is Therefore, Hence

;

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

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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., or

See the following figure.

Before the body starts sliding, the friction force provides it with centripetal acceleration Therefore, considering the smallness of the tangential acceleration according to the condition, The body starts sliding at , such that , i.e., when .

m.

The kinetic energy of the stone at the lowest point of the circle completely converts into its potential energy during the ascent: , where 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 and solve both equations together to find the answer.

N.

The flight distance will be maximum if the hammer's velocity is directed at an angle to the horizontal. In this case (see problem 63 in Chapter: Kinematics) , where 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) .

a) N;

b) N;

c) N.

At the moment the car passes the middle of the bridge, N; N.

Two forces act on the car: the force of gravity and the normal reaction force of the bridge , which, according to Newton's third law, is equal in magnitude to the force 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 and the radial component of the gravity force, equal to (see figure). In accordance with Newton's second law, one can write for a convex bridge and for a concave bridge.

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J.

The work goes into increasing the potential and kinetic energy. The speed of the ball is found from the condition , and .

km/h.

See problem 11.

1. .

2. m/s.

• See problem 11.

The acceleration of the car is composed of the constant tangential acceleration , which provides the car's acceleration, and the centripetal acceleration ( is the car's speed). Therefore

The external force providing acceleration to the car is the friction force of the wheels on the road. Since the friction force cannot exceed ( is the mass of the car), the maximum acceleration at the moment the car transitions to the horizontal section of the road cannot be greater than :

Since

( is the path traveled by the car; ), then

and after substitution we get

Hence m a_{\text{cp}} = m\vec{g} + \vec{N};mv^2/R = mg \tan \alpha.\tan \alpha = \frac{v^2}{gR}.\Delta h = d \tan \alpha = d v^2/(gR).\cot \alpha = N/F{fr} = mg / (mv^2/R) = gR/v^2.F_{\text{fr.max}} = kN = kmg,mv_{\text{max}}^2/R = kmg.v_{\text{max}} = \sqrt{kgR}.\phi = \arctan \frac{v_{\text{max}}^2}{gR} = \arctan k.mv_{\text{max}}^2/R = F_{fr.max} \cos \alpha + N \sin \alpha. \tag{1}F_{fr.max} \sin \alpha + mg - N \cos \alpha = 0.\tag{2}F_{fr.max} = kN.\tag{3}v_{\text{max}} = \sqrt{gR \frac{k + \tan \alpha}{1 - k \tan \alpha}}.v_{\text{max}} \to \infty\alpha_0 = \arctan (1/k).mg \tan \alpha = mv^2/R,R = v^2/(g \tan \alpha).N/F_{fr} = \tan \alpha = (a/2)/h = a/(2h),mg / (mv^2/R) = a/(2h),v = \sqrt{gRa/(2h)}.T = mg/\cos \alpha = mg \sqrt{1 + \tan^2 \alpha} = mg \sqrt{1 + \frac{v^4}{g^2 R^2}}.F_1 = \rho_m \omega^2 r V,F_2 = \rho_c \omega^2 r V.m\omega^2 R = mg \tan \alpha\frac{4\pi^2}{T^2} l \sin \alpha = g \tan \alphaT = 2\pi \sqrt{l \cos \alpha / g}.ma_{cp} = T - mg,mv^2/l = T - mg.\tag{1}mgh = mv^2/2,v^2 = 2gh.\tag{2}m(2gh)/l = T - mg,h = (T - mg) l / (2mg).h < 2.5 \text{ m}\frac{mv_b^2}{2} - \frac{mv_t^2}{2} = mg \cdot 2R.mg \cdot 2l = mv^2/2.T \cos \alpha - mg = 0.\tag{1}mv^2/R = T - mg \cos \alpha.mv^2/2 = mgR \cos \alpha\cos \alpha = 1/\sqrt{3}.m\omega^2 R = mg \tan \alpha.\omega = \sqrt{g\tan\alpha/R}.\omega_{cr} = \sqrt{2g\tan\alpha/D};Q = mg \tan \alpham\omega^2 l = mg \tan \alpha,\omega^2 R \sin \alpha = g \tan \alpha.\omega \ge \sqrt{g/R}.N = mg/\cos \alpha = m\omega^2 R.\omega = \sqrt{(a + g)/(R\tan\alpha)},T = 2\pi/\omega = 2\pi \sqrt{R\tan\alpha/(a + g)}.\frac{1}{2}mv^2+ mg \cdot 2R = mgh,mv^2/R = N + mg \cos \alpha.\frac{5}{2} mgR = \frac{mv^2}{2} + mg (R + R \cos \alpha),v = \sqrt{gR (3 - 2 \cos \alpha)}.
N = \frac{m}{R} \left[ gR(3 - 2\cos\alpha) \right] - mg \cos\alpha = 3mg(1 - \cos\alpha)
mv^2/R = mg \cos \alpha.mv^2/R = mg \cos \alpha - N.mv^2/R = mg \cos \alpha.\tag{1}mv^2/2 = mgh.\tag{2}h = R/3.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).\frac{\Delta m v_A^2}{2} + \frac{\Delta m v_B^2}{2} = 2 \Delta m v^2.mv^2/(R - r) = N - mg \cos \alpha,\tag{1}mg(R-r)(1-\cos\alpha) = \frac{mv^2}{2} + \frac{I\omega^2}{2}.mg(R-r)(1-\cos\alpha) = \frac{mv^2}{2} + \frac{m(v/r)^2 r^2}{2} = mv^2.\tag{2}\cos \alpha = 1/2h = (R - r) \cos \alpha = (R - r)/2.mv^2/R = F.F = pdl.m = \rho \pi d^2 l / 4.M = \rho \pi d^2 v t / 4\quad (\tex{time } t = 1 \text{ h}),v = \frac{4M}{\rho \pi d^2 t}. (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
\omega = 2 \sqrt{ \frac{(m_1 - m_2)g}{(m_1 + m_2)l} }

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}

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)

F = T_1 - T_2 = \frac{3\left(m_1^2 + m_2^2\right) - 2m_1 m_2}{m_1 + m_2} g
v = \sqrt{2ghn}
v_1 = v \cos\alpha,

a_{\text{cp}} = \frac{v_1^2}{R} \approx \frac{2ghn}{R}

F_{\text{hor}} = ma_{\text{cp}} = \frac{2mghn}{R}

F \approx \sqrt{(mg)^2 + \left(\frac{2mghn}{R}\right)^2}
= mg \sqrt{1 + \frac{4h^2 n^2}{R^2}}
\Delta m = m \Delta l / l = m R \Delta \alpha / l.2T \sin (\Delta \alpha / 2)2T \sin (\Delta \alpha / 2) = (\Delta m) \omega^2 R.2T \Delta \alpha / 2 \approx m R^2 \Delta \alpha (2\pi n)^2 / l,T \approx m n^2 (2\pi R)^2 / l = m n^2 l.m (\omega R + v)^2 / R = mg - F_{1lift}\tag{1}m (\omega R - v)^2 / R = mg - F_{2lift}. \tag{2}\Delta F_{lift} = F_{2lift} - F_{1lift} = 4mv\omega = 4m (2\pi v / T),$ where is the period of rotation of the Earth, i.e., one day.