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Using $$d\vec{s} = d\vec{s_{b/c}} + d\vec{s_{c/g}}$$ where, $d\vec{s}$ is the small displacement of the body wrt ground. $d\vec{s_{b/c}}$ is small displacement of the mass wrt vehicle. $d\vec{s_{c/g}}$ is small the displacement of vehicle wrt to the ground. we can write the work done by such forces as $$W_{F} = \int_{A}^{B}\vec{F}.d\vec{s} = \int_{A}^{B}\... 1 In general if the situations are more complex then we have conservation laws and work energy theorem in our pocket. Both of the situations are not much complex,in 1^{st} case: Working from ground frame we can use work energy theorem:$$\Delta K=W_{tension}+ W_{gravity}$$Here the problem shortens easily,because finding the work done by gravity is very easy ... 1 Work done by tension in car's frame of reference would be zero. Then we can apply conservation of energy where initial and final kinetic would be zero(since at start it didn't have any energy) then use pseudo force's work equal to work done by gravity. We get answer$$\theta = 2\arctan(\frac{a}{g})$$(where \arctan is \tan inverse function). 2 The short answer is that when you write$$T = \frac{1}{2}M v_{CM}^2 + \frac{1}{2}I\omega^2$$you are assuming that the moment of inertia is being calculated about the center of mass. It's true that the moment of inertia can be calculated about whatever axis you'd like; however, if you don't choose the center of mass, then the expression for the total ... 1 General relativity tells us that gravity is equivalent to acceleration. In this case the pendulum suddenly finds itself in a system where gravity has shifted backwards by an angle with a tangent of a/g. Being a pendulum it will swing toward this new equilibrium, and momentum will carry it an equal angle beyond. 2 For an extended body in a uniform gravitational field, its potential energy is the same as that of a point mass located at the bodies' center of mass. This can be seen by writing out the total gravitational potential energy in terms of the gravitational potential energies of its constituent masses:$$ U = \sum_{i} m_i g z_i = g \sum_i m_i z_i $$where z ... 1 In theory would be possible, but to do it you'll need to know the precise position of the centre of mass of the tower, which if it was a homogeneous cylinder would be pretty easy, and the angular velocity at which the tower leans. So, suppose that we know the position of the centre of mass, with which we can evaluate the maximum angle of lean before the ... 4 The tower of Pisa started leaning because of uneven settling of the soil underneath it. (It could be that it wasn't just a process of compression, but that there was also reflow. I don't know the details.) Decades ago it was discovered that this uneven settling was accelerating. That is, the lean wasn't increasing linear with time, it was speeding up. Study ... 0 I think that Callahan is probably just wrong. The strongest argument that he isn't seems to be that a monograph on special relativity published by a respected publishing house couldn't possibly be wrong about something so basic. But I have a very low opinion of SR pedagogy in general so I'm not much swayed by that argument. I haven't read the whole chapter, ... 0 this also true for two point mass system (each of mass m) connected by a string in which one point mass performs circular motion around the other. Then in this case is it possible to have different velocity of the point mass (which is performing circular motion around the other) in inertial and Center of mass frame of reference ? Yes, velocity will be ... 0 Yes velocity of a particle is different when you shift to com frame or any other moving frame whose velocity is different from the frame initially from where you were observing the particle. 0 Yes , it is possible that you get different values for inertial frame and frame of centre of mass When you solve the velocity of a point on a ball in frame of centre of mass we assume the ball is pursuing a rotational motion in the frame if reference of centre of mass because in this frame the translational motion of centre of mass and that point nullify In ... 1 Let the pseudoforce be f. Then the equation of motion in the non inertial frame is$$m \ddot{x} = -kx + f = -k\left(x - \frac{f}{k}\right)$$Change variables to X =x - \frac{f}{k}, then the equation of motion is$$m\ddot{X} = -kX$$which is the equation for shm with solution$$X = A \cos (\omega t + \phi)$$where \omega = \sqrt{k/m} and A is the ... 0 I suppose that you are in euclidean space and so placing yourself in a non-inertial frame corresponds to apply a global transformation, in the sense that every point of space has associated this acceleration vector that corresponds to the opposite of your acceleration seen from an inertial frame of reference. Consider first the case in which you have an ... 0 Other than the specified reasons in previous answers, the major reasons I choose to use the centre of mass frame of reference is because It is easier to visualize the motion of the object, as it performs pure rotation about the centre of mass. You do not need to bother about pseudo forces. You can see for yourself that torques due to all the pseudo forces ... 4 Why a body does not rotate if force is applied on the centre of mass? The question is: Why does it rotate if it is not applied to the center of a mass? First we look at two rules that students of mechanical engineering (I don't know about physics) learn at university: Multiple forces applied to the same point of a solid body have the same effect as a ... 0 Because the relative velocity can usually assumed to be approximately constant, which makes things like the Tsiolkovsky rocket equation a lot easier to derive (indeed the rocket equation as usually described assumes a constant relative velocity of the exhaust). And, of course, the relative velocity of the exhaust is the absolute velocity measured with ... -1 This can be explained with a simple case - torque with zero net force. Consider a rod in free space being acted upon by two equal and opposite forces at the ends. The subsequent motion of the rod is dictated by - No net force on the COM The rod is a rigid structure What it means is individual particles of rod must move without changing their relative ... 2 The rotation of a body depends upon the point at which it is hinged. So if a force applied at the center of mass (COM) then it may cause rotation about the axis (if it's not the one through COM) at which it is hinged by causing a torque to act. But what about the case when it isn't hinged at any point with the force still acting on the COM? A torque still ... -1 Center of mass is a point in space, that is it doesn't have a length. Also note that a point is not a circle, it doesn't have a circumference nor area. If you replace a collection of particles with centre of mass you will end up having a point with same mass. Now just try to imagine applying a force to the point . The moment you apply a force to it , if ... 6 Okay so let me give you the intuition. Now you would've heard of cases of a pendulum made of a rod instead of a string. (You can checkout this: http://hyperphysics.phy-astr.gsu.edu/hbase/penrod.html) Okay, so in this case you can clearly see that the torque is taken about the hinge. And you apply gravitational force at the centre of mass and take its torque ... 1 Because force is the time derivative of momentum, and momentum is linked to the motion of the center of mass. If you consider a rigid body as a collection of particles glued together and their position split into the position of the center of mass \boldsymbol{r}_{\rm COM} plus some other relative position \boldsymbol{d}_i, then$$ \boldsymbol{r}_i = \...

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Let $\vec r_1$ denote the position of the centre of mass of an object of mass $m$, given by the formula below. $$\vec r_1 = \frac{1}{m}\int \rho \vec r^\prime \mathrm{d^3} \vec r^\prime$$ If an object is not rotating, then all of its points must have the same acceleration. Therefore, if a single force applied to the object does not cause rotation, it must be ...

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I'm talking about the weight while the elevator is being pulled up steadily. If the elevator is travelling at constant speed in a constant gravitational field, then its weight will stay the same while it's travelling. However, as it moves upward, it is getting further from the centre of the earth. Hence the earth's gravitational attraction is reducing as it ...

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While the elevator is accelerating upward, which means that its speed upward is increasing, the shaft, elevator, motors, pulleys, etc., will weigh more. If the elevator falls with no resistance from the mechanism, the whole assembly will weigh less; but if the elevator's fall is slowed by a braking system the whole assembly will weigh more.

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I don't see how the collision could be elastic and symmetrical without the two particles having the same mass. You're right, but this isn't a mistake in the argument. The authors are assuming a specific situation and using that to derive general constraints. If you change the assumptions, you would get a different, more complicated setup, which would not be ...

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It's a multiple choice, with three obviously wrong answers. You can't show the exact position of the centre of mass without equations for the curves, but you can think about where the intuitive middle is, allowing that matter further from the centre has greater moment.

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Symmetries: Can't do without more information. If have an analytical formula for the shape (parabola?), you can use the integral version of the center of mass equation to obtain an exact center of mass. Experimental method: Print it and use a pair of scissors to cut it out. Suspend it at any point on a string. Draw a line with a pencil along the string and ...

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It can be shown by exclusion: A and D are not possible, because they are to far of. You can draw a ellipsoid around B, which contains the area left of B and is symmetric. Still, you are left with the two tails on the right. only C is left, therefore its C. Maybe one can use the fact, that this figure is "stackable" to archive a more rigorous proof ...

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Why didn't he consider the centripetal force as a force acting on that object. Because Centripetal force is not a force acting on the body. The Centripetal force is not necessarily a unique force in its own right. As an answer pointed out, The "centripetal force" is simply defined as the net force acting towards the centre. If there were some more ...

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First of all let me clear you that whenever a body moves in a circular path there is always an acceleration radially inwards which we refer to as Centripetal acceleration. And then there must be a force associated with this acceleration again radially inward. We call this force as Centripetal force. Here is the proof of the presence of a centripetal ...

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I figured what was wrong with my argument a while back. First of all J. Thomas was right, I do implicitly put the center of mass at the origin. I guess the way the one body problem equation was derived in my book confused me. If I put my origin at the center of mass, then the position vectors of $m_1$ and $m_2$ relative to the center of mass are $-\frac{m_2}{... 0 I have read that the Milky Way and Andromeda are falling toward each other and that the two of them are falling together toward a region of the sky which contains a large cluster of galaxies. It occurs to me that if you assume that the galaxies are uniformly distributed throughout the universe, then the net force on any one of them would be zero. That means ... 3 For objects moving under gravity, the centre of mass is in free fall so one cannot directly use an accelerometer to detect any acceleration of the centre of mass. If, however, the gravitational field is not unform (as is the case of the earth orbiting the sun) and the body is stuck together so as to move as a whole (as is the earth), then some parts of ... 0 When Einstein was working on the field equations of GR, the initial value that he had obtained for the bending of light was the same as the Newtonian value. He even hired a team of astronomers to measure it, but thy were captured during the World War (which turned out to be good for Einstein, as he had the wrong predictions.) It was only later, when refining ... 2 An rotating object will deform if it isn't resilient enough to provide the required centripetal force. Take for example the manual skill of spreading out pizza dough by tossing it in the air with a lot of rotation. A wooden disk would keep its shape. The dough does have elasticity (making it tedious to try and spread it out with a roller) but the dough does ... 0 It does not make sense to "stop time in an instant across the universe" because time does not stop. You can choose an instant across the universe on a spacelike hypersurface, but this is assuming the conclusion, since your spacelike hypersurface (your now moment) must be chosen in order to choose an instant. Moreover, there are an infinite number ... 0 should'nt the light beam or photon stay where it is Light, electromagnetic wave, is a classical manifestation of the superposition of zillions of photons. Photons are elementary particles. Both move with velocity c , and cannot "stay" anyplace. c is an enormous velocity with respect to motions of moons and spaceships. At relativistic speeds the ... 1 You are not using the correct relativistic formula for your speed and Bolt's speed in the reference frame of the observer. $$v_{\mathrm{net}} = \frac {( u + v )} {1 +uv}$$ where$v_{\mathrm{net}}$,$u$and$v$are expressed as fractions of the speed of light. If you use the correct formula, you will find that the train, yourself and Bolt, are moving ... 0 You will not see anything strange when you look at Bolt. The relative speed between the two of you is small, so time dilation will be undetectable. Your confusion arises because you assume that you perceive in Bolt the same time dilation that an observer at rest with earth sees. The guys on Earth see Bolt's clock running slower than yours, but they see Bolt ... 0 You can totally use$\vec{\omega} \times \vec{r}$to derive the kinetic energy of a pendulum. Note that$\vec{\omega} \perp \vec{r}$, so$||v||=||\vec{\omega} \times \vec{r}|| = ||\vec{\omega}|| \cdot ||\vec{r}||$, and since$||\vec{\omega}||=\dot{\theta}$and$||\vec{r}|| = \ell$, we have that the kinetic energy$T = \frac{1}{2} m v^2 = \frac{1}{2} m \dot{\...

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Assuming the string remains taut, I think you can still treat the simple pendulum as a rigid body, consisting of one point mass. The centre of mass of such a body is located at the point mass (e.g. the bob in this case) itself so $\mathbf{r} = \mathbf{0}$ since this must be measured from the centre of mass of the rigid body. Therefore, we already see that $\... 2 When acceleration gets confusing, I go back to position and work my way up. Let the position of the top of the plane, relative to the inertial frame, be$\overrightarrow r_p(t)$. And let the block's position relative to the top of the plane be$\overrightarrow r_{b|p}(t)$. Then, the inertial position$\overrightarrow r_b(t)$of the block is$\...

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In my understanding a frame of reference is a structure that quantifies the term "at rest". To go into detail we first have to clearify the framework we are in, i.e. the spacetime structure. For Newtonian mechanics this would be the Galileian spacetime. In beginner courses this is often introduced as $M=\mathbb{R}\times\mathbb{R}^d$, a Cartesian ...

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He is doing right . Centripetal force is not any particular force ,any force acting in the radial direction behaves as a centripetal force,so your thinking that we should add mv²/R as centripetal force in LHS always is wrong. The net of all the forces acting in the radial direction is the centripetal force.The role of centripetal force can be played by any ...

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Why is the centripetal force non conservative while the centrifugal isn't? As regards to the title of your post, the centripetal or centrifugal force may be conservative or non conservative. For example, if the centripetal force is gravity then it is conservative. The equal but opposite centrifugal force (pseudo force) that exists only in the rotating (non ...

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This is not really an answer. It’s more of a continuation of your thought experiment. Maybe it will help to think of the question in different ways. Imagine that every particle in the universe had its own personal timer that said “exactly 20 billion years after the big bang, I need to record what I’m doing at that instant. Then when someone asks me what I ...

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There are actually several ways one can derive the expression for the coordinates moving with constant acceleration, if you consider a bit more directly the geometry of curves in Minkowski space, but that's a deviation from your question. The formula you are interested in can be derived directly from the chain rule in calculus. It is just that the derivation ...

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I completely agree with the wonderful answer of @Bob D sir. However I would like to add a few things to help your understanding even better. First of all,what a lot of people seem to believe is that friction always opposes motion.This is not true.Friction may also help in motion. Eg We can walk only because friction pushes us forward when we push our feet ...

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The car takes the corner because its wheels are turned (by the steering mechanism) at an angle to its direction of motion. There is still rolling resistance that is trying to slow the wheels down, but this is overcome by the car's engine which exerts a rotational torque on the wheels via the gearbox and transmission (just as it does when the car is going in ...

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