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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 ...

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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. ...

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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 ...

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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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You say: the motor's velocity is always $>=0$ but this is not true. The magntude of the velocity is indeed always greater than or equal to zero, but velocity is a vector so the direction matters as well. The motor will oscillate alternately up and down as the masses rotate around it. The combined centre of mass of the motor and the two masses will ...

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As the particle moves on circle, $v_2=r\dot{\theta}$ ($\dot{\theta}=\frac d{dt}\theta$) $v_2=\large{\frac{v_1}{\cos \theta}}=r\dot{\theta}\;\Longrightarrow\;\dot{\theta}=\large{\frac{v_1}{r\cos\theta}}$ $y=r\sin\theta=v_1t$ (because $v_1$ is constant) $x=r\cos\theta=\sqrt{r^2-v_1^2t^2}$ , $0\le \theta\le \pi/2$ After $\theta=\pi/2$, the particle cannot ...

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If you are thinking about terminology of the case, then I'd suggest to call it '1D motion' to avoid confusion with true straight. With this terminology it is no matter if trajectory is planar or goes arbitrary in 3D space, or even makes loops.

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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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You can do this derivation by breaking the position of the orbiting particle down into components. It isn't short, but I think it's useful because it supplements the algebra with concrete physical analogies. I'll organize it into four parts: decomposition, oscillation, energy, and symmetry. Decomposition The position of a particle moving along a circular ...

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If what you mean is Center of Mass, then, no, it will not change the angular velocity (if there are no other external torque acting on the object, and the object is not attached in any way to other objects, i.e, glued, or attached to a fixed rotating rod or another object, who's rotating axis does not intersect the Center of Mass of the object attached). In ...

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Point C is sure instant centre of rotation for both gears, otherwise they would get teeth broken if any relative slide to each other. Analogy is a wheel on a road having instant centre of rotation at the bottom point thus velocity of top point is twice more than of the car. As far as I understand, confusion point is that first you calculate VO in respect ...

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If the center of mass is the same as the center of gravity then it does not change the angular velocity. As far as we know the inertial mass is the same as gravitational mass and hence the center of mass is always also a center of gravity (defined as a center of gravitational mass). Some people define a center of gravity as point where the average ...

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