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what does this value mean? This value means for maintaining of moon on the circular path, we need a force by amount of $\textrm{mass of the moon}\times\textrm{that value}$ How can this value be interpreted intuitively? According to first law of Newton, a moving body will move on a straight line uniformly unless a force is exerted on it. As you can ...

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Acceleration is defined as the time rate of change of velocity - be it in its direction or magnitude or both. Velocity $\bf v$ in polar coordinates is given by: $$\mathbf v= \dot {\mathbf r} = \dot r\mathbf{e_r} + r~\dot \theta\mathbf{e_{\theta}}\;.\tag 1$$ And then acceleration $\bf a$: $$\mathbf a = \dot{ \mathbf v} = \left[\ddot r - r\dot \theta ^2\... 2 It is not true that the same force has to create the same change in kinetic energy. For instance, if two equal forces of opposite directions are applied on a body, the body does not change its energy. Thus each force makes zero work, or zero change in kinetic energy. You could tell that both forces create kinetic energies in different directions and that ... 2 Energy is force times distance. You haven't specified a distance, so you can't say how much energy that force delivers. If you allow the force to rotate the sphere, the application point will accelerate faster than it would if the sphere does not rotate. So a constant force will deliver energy to the sphere faster than it would in the non-rotating case. ... 2 All vectors, except \:\mathbf{r}\:, are infinitesimals. I wonder if the author (Irodov) makes use of this result anywhere in his textbook. EDIT The infinitesimal rotation of a vector \:\mathbf{r}\: around the direction of a unit vector \:\mathbf{n}=\left(n_{1},n_{2},n_{3}\right)\: by an infinitesimal angle \:\mathrm{d}\theta\: may be represented ... 1 Take a rope with a small weight on one end and spin around on a point such that the rope is taut. You will feel force on your arm 'pulling' on it towards the weight. That force exists because that's what you are exerting on the object to prevent constantly change it's direction. The value you refer to is the acceleration portion of that force from F=ma. ... 1 At the top of the loop if the normal reaction on the ball due to the track is F_n down and the weight of the ball is F_g down then using Newton's second law$$F_n + F_g = m\frac {v^2}{R} where $v$ is the speed of the ball, $m$ is the mass of the ball and $R$ is the radius of the loop. This equation tells you that the faster the ball is moving the ...

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Whilst the motion you intend is not altogether clear, the motor will indeed move as you say. But the attached rigid bodies also move, such that the center of mass of the whole system is stationary (or, more precisely, its state of motion unperturbed by the system's internal motions). Work out the path of the center of mass in your system to check this ...

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From the point of the particle's reference frame she doesn't follow a curve, she is at rest. From a frame at rest on the lab, the particle's motion is curvilinear (it is unidimensional, but curvilinear though). If you know the curve you can parametrize it with only one parameter, because it is unidimensional (for instance for some circular motion your can ...

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The first step is to bring all terms with dv/dt to the left side of the equation. Do you then see the solution?

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