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this question is really severely damaged: the title (top/bottom quarks) does not match the question being asked (up/down quarks plus tau electrons), and the question literally being asked has a meaningful typo (tau selectrons) which invokes ideas from the still-speculative physics of supersymmetry, which is even crazier. To answer the question you literally ...


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The pulley (and the attachment to the ceiling) are part of the system here. Because of this, you cannot simply use conservation of momentum on the three given masses. If the final velocity were $v$, then the total energy of the system would have increased since both the pan and counterweight would be moving and the other mass would not have slowed. You ...


2

According to me, the momentum of the mass on the other side shouldn't be negative because the system considered is a connected system, i.e, the momentum transfer to the mass hanging on the other side of the pulley is momentum transferred through the pulley's string. I agree with @BowlOFRed about the contribution of the momentum transferred to the ceiling, ...


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If this is a correct description of what happens, can we conclude that g does same work on P and on P'? Yes. This is correct. If g acts perpendicularly to the velocity, it performs work of magnitude zero. This is also correct. The reason the two statements above are not contradictory is that the work done by the gravity changes the direction of ...


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Consider this previous exchange on the Cosmic Microwave Background The crucial bit is this: However, the crucial assumption of Einstein's theory is not that there are no special frames, but that there are no special frames where the laws of physics are different. There clearly is a frame where the CMB is at rest, and so this is, in some sense, the ...


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On an intuitive level if $a\ll R$ then nearly all of the deformation will occur close to the surface. Imagine for a bit that R is radius of the earth and you're pushing on some dirt with base ball so there's a circular contact patch with a radius of about 1/2 an inch. Now if earth were half as big would the forces/stresses/strain/contact area be any ...


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While not necessary to solve this problem, I want you to know that... Concept # 0: the angular velocity of circular motion is directly proportional to the linear velocity of motion, $$ v = \omega r $$ where $v$ is the linear velocity, $\omega$ is the angular frequency, and $r$ is the radius of circular motion. Concept # 1: whenever an object exhibits ...


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Faced by the same question and a background that includes courses in vector calculus, I have sought a simpler answer. My answer is much that same as to why one can easily balance on a typical bicycle. Bicycles are constructed so that the point where the extension of the front fork pivot would hit the ground is in front of the point where the front tire ...


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1."The particle will complete the circle when at the highest point if the string doesn't slack at the highest point when θ=π." - Is it because if the rope slacks, then the mg component will pull it inwards and so the particle will not move as a circle? Yes. Gravity (the weight $W=mg$) is then strong enough to pull it back from the "swing". The rope is ...


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Let me first do this the way that I know is correct: with Lagrangian mechanics. This says that all of the physics you need is contained in the Lagrangian, which is the kinetic energy minus the potential energy. Your three masses Left, Right, and Bottom make the kinetic energy $K = \frac 12 m (v_L^2 + v_R^2 + v_B^2),$ where $m = \text{1 kg}.$ Defining ...


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Your moment of inertia is incorrect. You must calculate is based on the individual masses and their distances from the pivot: $$\mathcal{I}=\Large\Sigma \large\left( m_ir_i^2\right).$$ If you do this you should get an answer that agrees with what @ChrisDrost did, 2.47 s And you shouldn't assume that the center of mass is 1/3 of the way below the rotation ...



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