# Tag Info

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I am only going to leave a brief answer, seeing that the comments are very accurate. The paradox can simply be resolved by considering the elastic nature of all the objects. How so ever instantaneous might the $dt$ or the time of collision seem to the human eye, actually it occurs over a small duration, based on the elasticity of both the objects involved in ...

2

The upright pole is in a position of unstable equilibrium. If the pole deviates from the vertical by an angle $\theta$ then the torque rotating the pole away from the vertical is: $$T = mg \frac{\ell}{2} \sin\theta$$ The moment of inertia of a pole about one end is $m\ell^2/3$, so the angular acceleration will be: $$\frac{d^2\theta}{dt^2} = \frac{3 ... 2 If \Delta S = r cos\theta then dl=ds=rd\theta and F_g=mg$$W_{g}=\int_0^\pi mgrcos\theta d\theta $$If you're taking the angle from the center of the circle (which you are, since you said that \Delta S = r cos\theta, then the initial position of the ball is -R, since displacement is a vector quantity (and the final ... 2 This is only from intuitive meaning. Let us find the mathematical meaning of k/m . Here, the equation is \frac{d^2x}{dt^2}=-\frac{k}{m}x suppose \frac{k}{m}=K Then, the solution of this equation is x=A\sin{\sqrt{K}t} quantity inside the sine function is the angle. Thus, \sqrt{K} must be angular velocity. Thus,$$\sqrt{K}=\omega\\ ...

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Force is not divided, it applied to the first bag, and then the first bag will make a force on the second one, and the second on the third. The first bag feels two forces, the one you apply and the reaction from the second bag, the second bag in turns feels two forces, one from the front bag and one from the rear bag. If the bags are attached trough ropes, ...

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In the ideal case where there is no friction and no perturbations and the top starts to spin in a perfectly vertical alignment, the two configurations (inverted or not) of the top are completely identical. However, once you have the top start rotating with a tilt from the vertical axis, or consider perturbations that will tilt it even if it wasn't, then the ...

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As the ball swings downward, its gravitational potential energy is converted to kinetic energy. At the bottom of the swing, its velocity will all be in the horizontal direction. From this, you can calculate the velocity with which the ball strikes the block. In a perfectly elastic collision, both kinetic energy and momentum are conserved. From this you can ...

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Acceleration due to gravity remains roughly constant near the surface of the earth. Yes, $a = F/M$, but as mass increases, the force exerted by gravity increases too($F\ \alpha \ m1m2\over r^2$), keeping $F/M$ or $a$ roughly constant around the surface of the earth

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Here's how to intuitively understand that $a=g$. Take a metal ball having mass 1kg and drop it. Its downward acceleration is $9.8m/s^2$, right? Now take a second ball and drop it. Same thing, right? Now drop both at the same time. Same? Now connect them together (with a tiny drop of weld metal) into a single 2kg mass, and drop them. Do they suddenly slow ...

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Aren't circles special cases of ellipses? In general, orbits can be either, but are usually elliptical (at least ideally).

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Maybe this is a bit beyond what you wanted, but you are implicitly evoking Einstein's equivalence principle, which tells us that gravitational accelerations and inertial accelerations are equivalent. It's only because this principle applies that we can add a gravitational and an inertial acceleration. Gravtitational accelerations are measured relative to a ...

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It is correct except for one sign, note that the work done by friction is negative (since you move the block in the opposite direction w.r.t. the friction force) and thus it is equal to $$-\mu mg d \cos \theta$$ with this solving for $v$ gives you 11.49 m/s.

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Although this has been answered many times already, anywhere on this site, the following holds: First law (existence of inertial reference frames) There exist in the universe some very particular reference frames where a point particle not subject to external forces moves in a straight line, i. e. $\dot{\textbf{p}}=0$. Second law (equation of motion in ...

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If all the mass of a rigid body was squeezed down into a single point (at the center of mass) then the mass moment of inertia would be zero. But since the mass is distributed in space it would take a finite angular momentum to spin the body up about the center of mass. The importance of the center of mass is that it is the location when the mass moment of ...

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No the time taken does not depend of the velocity attained by the first ball(if they are ideally rigid) it rather depends on the elasticity or rigidity of the balls. So for ideally rigid bodies, the time taken to transfer approaches 0. Nothing would happen with an increase in distance between the two balls. See: Is the reaction force for a stone hitting a ...

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If all of them feel the same force, they would have an acceleration that would give them a speed and hence a kinetic enegy greater than the work done by the applied force. It would violate the conservation of energy and conservation of linear momentum principles. The force on each bag will depend on their individual masses, you can compute the acceleration ...

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Let me discuss a simpler version of your rocket-question: one where there is no gravity, so that we don't have to worry about gravitational potential energy. Consider a rocket in free space (vacuum), and consider that the rocket is at rest. Now the rocket fires it's engine for a short time. The engine accelerates the rocket. The rocket now has kinetic ...

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Kinetics: In physics and engineering, kinetics is a term for the branch of classical mechanics that is concerned with the relationship between the motion of bodies and its causes, namely forces and torques. Kinematics: Kinematics is the branch of classical mechanics which describes the motion of points, bodies (objects), and systems of bodies ...

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You can decouple the horizontal and vertical motion of your rocket. In the vertical direction you have vertical thrust and gravity and horizontally you only have thrust (I ignore air resistance here). As you are interested in the altitude only, we only look at the vertical problem. All kinetic energy in the vertical direction is converted to potential energy ...

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