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If you have found that the initial angular acceleration of the disc is $\alpha = \frac {2g}{3R}$ that must mean that the initial centre of mass acceleration $a = \frac {2g}{3}$ because $a = R \alpha$. If the force acting on the pivot is $X$ upwards then applying Newton's second law for the centre of mass motion gives $mg-X = ma$, so $X = \frac {mg}{3}$. ...

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I would think that the difficulty is related to gravity. Without gravity, water will go everywhere in all directions.

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There's a physical reason why the block slides down along the slope of the inclined plane. If you perform this experiment, and if frictional force is small enough, you will observe that the block slides down along the slope of the plane. It will never move in a direction perpendicular to the slope. For it to never move in the perpendicular direction, we ...

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The normal force is not playing a role in this case because the force is perpendicular to the moving direction of the box. By "not playing a role" I mean, during this motion, $\ F_N$ is only be used to balance out $\ mgcos\theta$ so that the object won't be able to move "into the inclined plane". To find the force causing the block to accelerate down, we ...

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Since acceleration is a vector you can decompose it in the coordinate system you find convenient. If you define a cartesian coordinate system whose axis are along the normal to the plane and the plane itself you see there is a component of the acceleration $g\sin\theta$ along the plane. This is why the block accelerate in this direction. Notice that along ...

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Astronauts do not feel weightless on the Moon where the gravitational field strength is one sixth of that on the Earth. They feel "lighter" in spite of the heavy suits that they have to wear. A spring balance would give a reading for the weight of an astronaut on the Moon which was one sixth of that on the Earth. An astronaut in orbit feels weightless and ...

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Gravitational acceleration is a vector and thus has a direction. But, at the center of a spherically symmetric mass distribution, there is no preferred direction while, if there were a non-zero gravitational acceleration vector, it would point in a direction. Now, if one rotated the mass distribution through some angle, the acceleration vector should ...

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Intuitively, of course, it makes sense that the force in all directions at the center of a massive body would be equal. There are a few ways of formalizing this. The most rigorous would be an integral: $$G\rho\int_{0}^{r}\int_{0}^{2\pi}\int_{0}^{\pi}\frac{\vec{r}}{|r|^3}|r|^2sin(\phi)d\theta d\phi dr$$ where $\rho$ is the mass density. You'll need to ...

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If you measure a force (weight) for a given acceleration (gravity) in order to determine the mass of an object and you haven't started measuring then the mass is undefined. As soon as you apply an acceleration $a>0$ and you measure corresponding force $F>0$ you can determine the mass. Equations are useful only when they can be used to measure things, ...

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In addition to the already given answears this also might be of interest: When hammer and feather are dropped simultaneously they arrive at the same time, when dropped independently the hammer attracts the planet more than the feather, so you are right, the total time until impact is then smaller for the hammer. If you pick up the hammer and let it fall to ...

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The external force acts only for the small time when the cue has been struck. Once it moves, there is no force. This means that the ball is moving with zero external force, which means according to Newton's second law, the velocity of the ball is the same. here the act of friction is of less importance as it requires in a billiard play. So the center of mass ...

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I think you should take the sample velocities and divide them by the respective times after minusing the previous velocities to obtain the accelerations. If the time interval between calculating the discrete sample velocities are too small, then the above-got values may be taken as instantaneous acceleration. Now plot these over graph wrt time. Now here we ...

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A fairly simple way of treating the data is to present them as a histogram: Each data point (here 5 data points) is the quotient of the distance moved in that interval, say $\Delta y$, by the time interval $\Delta t$ and is the average velocity during that time interval: $$v_i=\frac{\Delta y_i}{\Delta t_i}$$ Where $i$ indicates interval number $i$. For ...

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Without loss of generality, we assume that a particle is moving in a circle of radius $R$ centered at the origin and lying in the $x$-$y$ plane. Using cylindrical coordinates $(\rho, \phi, z)$, the angular velocity of the motion is $\boldsymbol\omega = \omega \hat{\mathbf z}$. The velocity is tangent to the circle and given by $\mathbf v = ... 1 Let two orthonormal systems$Oxyz$,$O'x'y'z'$with a general motion (translational plus rotational) between each other and a point particle$\rm P$, see Figure. Symbol Conventions : 1.The vectors for position$\mathbf{R}$, velocity$\mathbf{U}$and acceleration$\mathbf{A}$of a particle with respect to$Oxyz$expressed by coordinates of this same ... 1 The net force on the string is not F. You are pulling the string forward with force F but I think you are forgetting that the block is pulling the string backwards with a force that is almost equal to F. If the masses of the string and block are m and M then for the whole system (string plus block) F=(M+m)a. The net force on the string is F '=ma=mF/(M+m). ... 0 You wrote it yourself,$F=ma$, so with zero mass there is zero force needed for a finite acceleration. A finite force gives an infinite acceleration. ... Lol, I just reviewed your answer about the electrostatic field being normal on the surface of a conductor. That's exactly the same, here! With a non-zero force the acceleration would be infinite ($F/m$... 1 The only thing I've figured out, is that the object will rotate but not moving, if the center of mass is on the line defined by the$A$point, and the normal vector of$\vec{F⃗}$. This is not necessarily true. Look at the free body diagram. Decompose$F$into$x$and$y$components. Newton now tells us that: $$ma_x=\Sigma F_x$$ $$ma_y=\Sigma F_y$$ ... 0 The laws describing the movement of the body are: $$\sum \vec F=m\vec a_{cm},$$ and $$\sum \vec \tau= \frac{d\vec L}{dt}$$ wehre$\vec F$is the external force,$\vec \tau$the external torque,$\vec L$the angular momentum and$\vec a_{cm}$is the acceleration of the the center of mass. As you can see from the first formula, the center of mass will always ... -1 No, we don't know that for sure about gravity. One object cannot have 2 different masses: The force that makes things roll, and fall, is the same force; it is gravity. This is what my question is about. We don't know for sure that Newton was correct: The fact that something heavier rolls at a higher speed shows us that Newton's theory of gravity is wrong. ... 3 This is actually a famous theorem known as the Einstein Equivalence Postulate (sometimes Equivalence Principle). It's true that since Earth is spinning, acceleration in a spacecraft isn't quite the same situation we experience daily, but in general, yes, gravity is indistinguishable from uniform acceleration. Specifically, if you are in a box with no windows ... 0 It’s all about engine's efficiency. According to wikipedia: gasoline engine's efficiency = 1/(BSFC × 0.0122225) (Also Actual efficiency can be lower or higher than the engine’s average due to varying operating conditions.) To calculate BSFC (Brake specific fuel consumption) use the formula: BSFC=r/P (where: r is the fuel consumption rate in grams per ... 2 The first figure shows the parallel and normal components of a vector$\mathbf{r}$relatively to a direction$\mathbf{n}$. Based of this, the second figure shows the centripetal acceleration. In case of plane circular motion$\omega R = v$. 1 The answer is two arcs. One arc with a constant gee loading in one direction and then flipping to the opposite direction. This is called the bang-bang method, and it is no very smooth, but the gee forces never exceed the specified maximum. Given a path$y(x)the instantaneous radius of curvature at each x is $$\rho = \frac{ \left(1+ \left(\frac{{\rm ... 1 Hint: Since the car begins from rest and keeps accelerating, the equation d=vt is no longer valid, because velocity is not constant. Note that you can, however, say that d=v_0 t + {}^1/_2 a t^2 = {}^1/_2 a t^2 and the acceleration will be the same for both time intervals. Hope that gives you a nudge in the right direction! Also, these equations may be ... 0 Whenever a pendulum moves, it accelerates since the velocity vector is constantly changing. So any pendulum that 'pendulates' forever will 'accelerate' forever. However, I think the acceleration you are talking about is better described as angular acceleration. Perhaps a better way to describe it is if it will complete a full rotation with increasing ... 0 I don't quite understand what you are asking. In your diagram you have$$ \begin{aligned} a_{tangential} & = \dot{\omega} \ell \\ a_{centripetal} & = \omega^2 \ell -g \end{aligned} what else to do want to know? If the angle is changing (and therefore your acceleration are not aligned with X and Y) then use \begin{aligned} a_X & = ... 0 If the pivot is accelerating horizontally (together with the body) at a rate ofa_{pivot}$then the angular acceleration of the pendulum is $$\ddot{\theta} = - \frac{m c (a_{pivot} \cos\theta + g \sin \theta)}{I_{zz} + m c^2}$$ where$c$is the distance from the pivot to the center of mass,$m$the total swinging mass and$I_{zz}$the mass moment of ... 0 As it happens, auto manufacturers (and the NHTSA I believe) have rules about maximum allowable jerk during cruise control, automatic braking, and similar machine-based events. (Trust me -- I work for an automotive active safety product company) Certainly airbags are required to have a maximum 'acceleration' as they open up, to reduce (and sadly, not ... 0 It is probably the case that your swing is too long because the increase in the amplitude of a swing is achieved by changing the position of the centre of mass of the person on the swing relative to the position of the pivot. For a long swing the percentage change in the position of the centre of mass in relation to the length of the swing will be smaller ... 0 You must pull the swing correspond to its natural frequent. In the case of yours (swing), natural frequent of swing is$2\pi\sqrt{\frac{g}{l}}$and for your swing, it is approximately 8 seconds. This means you must pull the swing once per each 8 seconds. For starting, it is better that you release the swing from a height. 2 Let's look at the hodograph of a constant radius & constant velocity motion. Left: trajectory of one of the masses. Right: hodograph, i.e. locus of the velocity vectors. Now, let's look closer at how the velocity changes during a small time interval$\mathrm dt$. A force is needed to rotate it (difference between the brown and red arrows). If you ... 1 Take a free particle moving on a plane in polar coordinates $$\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} r \cos \theta \\ r \sin \theta \end{pmatrix}$$ The velocity is found from the chain rule, with clear separation for radial and tangential components: $$\begin{pmatrix} \dot{x} \\ \dot{y} \end{pmatrix} = \begin{vmatrix} \cos \theta & ... 3 Here is one way of looking at it via a velocity-dependent potential.^1 The Coriolis potential is$$\tag{1} U_{\rm cor} ~=~ -m({\bf v} \times {\bf \Omega})\cdot{\bf r} ~=~-{\bf v}\cdot ({\bf \Omega}\times{\bf r} ),$$cf. Ref. 1. The factor 2 comes from two different terms in the corresponding force formula$$\tag{2} {\bf F}~=~\frac{\mathrm d}{\mathrm ... 0 If acceleration is linear, use the 'kinematic equations'. If not, use differentiation/integration (in this case, you will probably be given a formula which expresses acceleration). 0 The acceleration due to gravity on the surface of Earth can change in various ways as the earth shrinks. If you look up at the rate of change of g with respect to decrease in R the radius of Earth, it will vary as 1/Radius of the old value ; so the acceleration due to gravity increases with reduction in radius. If you also consider the change in spin ... 0 The value of acceleration due to gravity varies inverse squarely as the distance from the center of earth to the center of gravity of the object. Suppose we are concerned about the value of acceleration due to gravity on the earth's surface. Then the distance between the object and the center of earth will be the radius of the earth (assuming the earth to be ... 3 The strength of the gravitational field depends on how far you are from the center of the mass, assuming the mass density is radially dependent or constant. It also depends on the total mass inside your position: $$g=\frac{GM_E}{r^2},$$ where$r$is the distance from the center of the mass distribution. So if you stay where you are now, the gravitational ... 4 If you stayed at the same radius while the earth shrinks then nothing will change other than you'll start to fall. If you stand on the surface of the new smaller radius then you will feel the increased gravitation. 2 First draw a circular path. In order for an object to follow that path at a constant speed, there must be a force acting on it towards the centre of the circle. At any instant, that force has no component in the direction of motion, and so, if that's the only force acting on the object, its speed will stay constant. Now draw a spiral path (i.e. one in which ... 0 I have tried to illustrate the error which has been made about the second term. Assume that a mass at position$A$($\vec r_1$) is under the influence of an attractive central force which originates from point$C$. The velocity of the mass is$\vec v_1$with radial and tangential components of velocity$\vec v_{1r}$and$\vec v_{1\theta}$. A little ... 0 The reason for the two behavior of the same objects accounts for the forces acting on it. When you drop a 5 kg body and another 25 tonne body vertically downwards, they will fall under the influence of gravity alone and both will fall with the same acceleration g (free fall). In such a case the weight of the body is zero. You know that fact, which you may ... 2 All other things being equal, if a heavier object will roll at a higher speed down hill than a lighter one With the qualification "All other things being equal" your statement is not correct. The falling acceleration is the same because doubling the mass of an object doubles the force causing the acceleration (the object's weight) which means that ... 1 if a heavier object will roll at a higher speed down hill Free fall and rolling are two different behaviors of objects. It is correct that for free fall all objects get the same acceleration ( minus friction and drag) but free fall is not the same as rolling. For going down a hill free fall can be compared to sliding, as was pointed out in the comments ... 1 In physics we recognize two different kinds of mass: inertial and gravitational. Inertial mass tells us how much an object resists a change in motion - or how much force is needed to effect an acceleration. Gravitational mass describes how much attraction (due to gravity) an object experiences as a result of gravity. Now despite very careful experiments, ... 0 The position-time graph you've sketched looks roughly right (though be careful to make the scale on the vertical axis consistent). The data don't look at all linear and don't justify a straight line. The second graph has a couple problems. Is it a speed vs. time graph or a velocity vs. time graph? In what you've written you refer to velocity so, if ... 1 As a supplement to Farcher's answer, pointing out that the constant acceleration equations fail here, we can still use$\vec{F} = m \vec{a}$to solve for the speed of the bullet - by means of integration. We have$- k \vec{x} = m \vec{a}$, but we can rewrite$\vec{a}$.${a} = \frac{d^2 x}{d t^2} = \frac{ d v}{d t} = \frac{dv}{dx} \frac{dx}{dt} = v ...

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Your first method is using the conservation of energy for the spring bullet system. You second method fails because the acceleration is not constant and so you cannot use a constant acceleration kinematic equation. On seeing $ma = -kx$ you might reorganise that the acceleration is proportional to the displacement from a point (the position of the free ...

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Parametrize the particle's worldline w.r.t. $t$: $$~x^\mu (t) = (t,1)$$ Its four-velocity is: $$u^\mu =\frac{d x^\mu}{d \tau}$$ To evaluate this we use the fact that: $$d \tau^2 = coshx~dt^2-dx^2$$ Also use $x=1$ and $dx=0$: $$d \tau^2 = cosh(1) dt^2$$ Therefore: u^\mu = \frac{d x^\mu}{d \tau} = \frac{d t}{d \tau} \frac{d x^\mu}{d t} = ...

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