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(II) A pendulum consists of a mass $M$ hanging at the bottom end of a massless rod of length $\ell,$ which has a frictionless pivot at its top end. A mass $m,$ moving as shown in Fig. 42 with velocity $v$impacts $M$ and becomes embedded. What is the smallest value of $v$ sufficient to cause the pendulum (with embedded mass $m$ ) to swing clear over the top of its arc?
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Physics 101 Mechanics
Motion Along a Straight Line
Moment, Impulse, and Collisions
University of Michigan - Ann Arbor
Simon Fraser University
In physics, a conservative force is a force that is path-independent, meaning that the total work done along any path in the field is the same. In other words, the work is independent of the path taken. The only force considered in classical physics to be conservative is gravitation.
In physics, the kinetic energy of an object is the energy which it possesses due to its motion. It is defined as the work needed to accelerate a body of a given mass from rest to its stated velocity. Having gained this energy during its acceleration, the body maintains this kinetic energy unless its speed changes. The same amount of work is done by the body in decelerating from its current speed to a state of rest. The kinetic energy of a rotating object is the sum of the kinetic energies of the object's parts.
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to this answer. First part is energy conservation. And so we're assuming that there is no friction, no energy being lost did fiction in the process and that all of the potential energy Ah, all of the kinetic energy from the, uh from the system is converted to potential energy of the top. So, uh, Kay beauty so kinetic energy of the bottom is equal to you t potential. And you're the top, right? So the bottom you have the two masses sticking together. So 1/2 M plus big M times V d squared. And this is the velocity of the bottom eyes equal to the, uh, some of the masses times Judy times height to which it rises. So that's two times, Uh, that's that's that's two times out. Or I should say, two times little L on dso the We noticed that the sums of the massive some of the masses cancel out from both sides. So, Vehbi, what we get is ah, squared of four g times out our small out. That's the velocity of the bottom. Now, for part two, we use momentum conservation. We're assuming first of all, that block does not move during the collision up on. And then we're saying that initial momentum of the initial momentum up the the, uh, bullet is transferred of what? Whatever the small masses. Sorry. Eyes transferred. Thio converted to the final momentum off the block. Our guest pendulum Um, been small mass sister. Right? So initially, you have m time small V. And finally, you have big M plus small m times, velocity of the bottom. And so ve therefore is, uh, small M plus big m times velocity at the bottom over small him, and that's it.
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