# Tag Info

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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 ...

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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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The ceiling is applying a total upwards force on the pulley of magnitude $2T$, in order to counteract the force of equal magnitude applied by the masses. So the net force on the system (consisting of three masses and the massless pulley and string) is not zero. Since $\int {Tdt} = mV = mv/3$, $\int 2Tdt = {2 \over 3} mv$ which is the amount of momentum lost. ...

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I don't have a good knowledge of physics but the basic answer is yes, g-force is pretty much an acceleration force. For example 1g (Earth gravity) is basically an acceleration of 9.8m/s2 towards the Earth, you don't accelerate because the ground resists this force. In terms of whether some one could pass out then yes you could. In space the weightlessness ...

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You've almost got it! The constant thrust comes from a mass rate $\mu$ of fuel being expelled at a velocity $v_f$ as opposed to the speed of the rocket itself $v$. Therefore the equation is instead: $$(m_0 - \mu t) \frac{dv}{dt} = \mu v_f - \alpha \frac {(m_0 - \mu t)}{r^2},$$ where $\alpha = G M.$ Hence the gravitational term you wrote as $G m_e (W_0 + ... 1 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 ... 1 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 ... 1 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 ... 1 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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