45

I've confirmed the experiment, using a McD_n_lds paper drinks cup and a beer can hollow plastic ball of about $5\mathrm{g}$, of about the same diameter as a ping pong ball (PPB): The observed effect depends largely on the cup being soft and permanently deformable (like an object made of blutack or playdough), so its collision with Earth is inelastic. A ...


20

I don't know if there's a beautiful solution for this. I'd love to see it, if it exists. What I can do is show you how I slogged my way through it. All praise to the mighty Mathematica. Part I: Obtaining the Equations of Motion First, we can dispense with the cylinder and rod and consider only a point mass $M$ on a ring of mass $m$ and radius $R$. Define $...


17

As mentioned in the comments above, the ball in the cup is similar to Galilean Cannon. The maximum height to which the ball can bounce $h_{max}$ can be estimated using the law of energy conservation: $$(m+M)gH=mgh+E_{cup}+E_{water}+E_{heat},$$ where $m$ is the mass of the ball, $M$ is the mass of cup+water, $H$ is the initial height from which the ball was ...


14

Potential energy is still a scalar quantity even in more than one dimension. This is because it only has a magnitude, there is no direction of potential energy. You can think of it as similar to the temperature in a room. Even though it varies with position, it still does not have a direction.


11

Turns out the tube does jump if the rod is no less than 13 times the mass of the tube. My previous answer had a couple of mistakes that yielded the wrong result, here is the updated one. Let $M$ be the mass of the tube, $m$ the mass of the rod and $R$ the radius of the tube. Let $\theta$ be the angle between the vertical and the direction of the rod from the ...


9

No, just because a value changes over space doesn't mean it is a vector quantity. Potential energy is not a vector. It is a scalar quantity related to its corresponding conservative force by $$\mathbf F=-\nabla U=-\left(\frac{\partial U}{\partial x}\hat x+\frac{\partial U}{\partial y}\hat y+\frac{\partial U}{\partial z}\hat z\right)$$ So the force components ...


8

Recall that if a ball normally hits a wall elastically, its velocity will be exactly reversed. Suppose the whole system hits the ground with speed $v$. Now, as the cup and the water hits the soft mat, their speed quickly reduces, and may start moving upward (depending on how soft the mat is) before the ping-pong ball is affected by a reaction force. Suppose ...


6

Since potential energy is function of position, ... hence can be considered as vector quantity? You make a conceptual mistake: "$n$-dimensional vector quantity" does not mean that a quantity depends on the position in an $n$-dimensional coordinate system. "$n$-dimensional vector quantity" means that a quantity requires $n$ different real ...


6

My hypothesis why the ping pong ball receives a large upward impulse: The floating ping pong ball is displacing some water. The amount of displacement does not change much during the fall. As the cup hits the floor the deceleration of the quantity of water gives a short pressure peak. Because of that pressure peak the water that is in contact with the ping ...


5

The force on object $1$ due to object $2$ can be computed by doing a six-dimensional integral, $$-G\int_{V_1}d^3\mathbf{r}_1\rho_1(\mathbf{r}_1)\int_{V_2}d^3\mathbf{r}_2\rho_2(\mathbf{r}_2)\frac{\mathbf{r}_1-\mathbf{r}_2}{|\mathbf{r}_1-\mathbf{r}_2|^3},$$ where $\rho_1(\mathbf{r}_1)$ is the mass density of object $1$ and $\rho_2(\mathbf{r}_2)$ that of object ...


4

Definition: The change in potential energy of the system is defined as the negative of work done by the internal conservative forces of the system. Potential energy may vary with space just like mass of a non uniform rod which may be represented as $f(x,y,z)$. After all potential energy is basically negative of work done by internal conservative forces which ...


4

The answer is that the moment of inertia is changing not only due to the instantaneous rotation about the contact point, but also because of the horizontal motion of the cylinder. The position of the rod (I won't write the z component of vectors) is $$r=R(\sin \theta, 1+\cos\theta)+\int_{t_0}^t (R\omega(t'),0)dt'$$ where at $t=t_0$, the cylinder is above the ...


3

Mathematically, moving between inertial and non-inertial frames correspond to moving terms from one side of Newton's second law to the other side. So, in your non-inertial frame accelerating with the incline you have for Newton's second law along the incline (using your notation) $$Mg\sin\theta+Ma_0\sin\theta=Ma_\text{net}$$ Moving to the inertial frame we ...


3

I don't understand why reaction force would decrease at the start of a countermovement jump. The reaction force changes because of changes in the vertical acceleration of the center of gravity of the persons body throughout the counter movement, as follows. At position A, before the person body starts dropping, the net force on the person is zero, so that $...


3

The article specifies the equation dealing with kinetic energy is looking at the relative kinetic energy. For a perfectly inelastic collision, the bodies are not moving relative to each other, so the relative kinetic energy is $0$. Thus there is no contradiction. To add more detail to this, the best thing to do is to work in the center of momentum frame, ...


3

Torque here is not external, you can tell because the total angular momentum in the system is the sum of the angular momentum of the two disks. Therefore the two disks are what makes up the system, neither of them are an external object. They only exchange momentum between each other, as they have both applied torques to each other. It is the same concept as ...


3

Both. Suppose that a particle of mass $m$ is in uniform circular motion with radius $r$ and tangential velocity $v_T$. We know that there must be a centripetal force maintaining this motion $$F_c = m \frac{v_T^2}{r}.$$ We also know that the system has an associated angular momentum whose magnitude is $$L = rmv_T.$$ Now if we only increase the centripetal ...


2

The slowing-down of your car is a function of all the resistive forces acting upon it: that is, all forces that are trying to dissipate the car's kinetic energy relative to the surface of the Earth. Hence, depending on the exact nature of those forces, the slow-down time, and the deceleration profile, can vary. That said, we can nonetheless come up with some ...


2

If the system is the two discs then the frictional forces apply internal torques which have a net value of zero - the internal torques are opposite in direction and equal in magnitude. If no external torques are applied then angular momentum is conserved.


2

Let's assume that tension increases down the rope then for this section of rope to be in equilibrium $$T-(T+\Delta T)=\Delta mg$$ As rope is massless, $\Delta m=0$ So, $\Delta T=0$ Therefore the magnitude of tension is constant throughout the massless rope.


2

Why is the tension dependent on the acceleration of the trolley? The acceleration of the large block $M$ and the attached pulley does put a force on the string because the string must accelerate with the pulley. This increases the tension of the string and so creates a larger force on both blocks $m_1$ and $m_2$. How can acceleration of the trolley prevent ...


2

To calculate the equation of motion we obtain the sum of the torques about point A, because we don't have to take care about the contact force. first I obtain the vector u from point B to A $$\vec{u}=R\,\begin{bmatrix} 0 \\ -1 \\ 0 \\ \end{bmatrix}- R\,\begin{bmatrix} \sin{\theta} \\ \cos(\theta) \\ 0 \\ \end{bmatrix}=-R\,\begin{bmatrix} \...


2

Assume that the water in the cup is compressible and inviscid, experiencing one-dimensional flow and thereby satisfying the one-dimensional Euler equations. Initial conditions, velocity =$\sqrt{gh}$ downward and pressure =1 atm, are both uniform. The bottom of the cup is struck from below in such a way that the velocity of the water is reduced and the ...


2

In general, you can only apply $ \tau_C = \tfrac{\rm d}{{\rm d}t} L_C $ about the center of mass C. The expression about a different point is quite more complex. You can see that taking the torque about another point A (not the center of mass C), and the derivative of angular momentum about A isn't enough to solve the problem. Using the standard ...


2

If you define the gravitational potential energy between two bodies to be zero when the bodies are infinitely far apart, then naturally the gravitational potential is always negative. However, the potential energy can increase in a closed system (while remaining negative). Consider two bodies that are receding from each other. As they get farther apart, ...


2

The up and down movement of each individual point visualised only the fact that the wave does not transfer mass. However, the points are not independent, but coupled: If the position of a specific point $x_i$ is $x_i - x_0$, where $x_0$ is the equilibrium position, the neighbouring points have similar positions. The energy is transferred due to this ...


1

The law of conservation of angular momentum states that when no external torque acts on an object, no change of angular momentum will occur. Yes there is friction between the discs,when they come into contact . Consider the resultant of the friction forces acting on the discs to be F. As shown above they are an action-reaction pair.They are internal forces. ...


1

The EOM's with $x'=v(\tau)\,\tau+x$ where $v(\tau)$ is the velocity between $x'$ and $x$ you obtain the kinetic energy and the potential energy of a pendulum that move in the prime system. Pendulum position vector $$\vec{R}=\left[ \begin {array}{c} v \left( \tau \right) \tau+L\sin \left( \varphi \right) \\L\cos \left( \varphi \right) \end {array} \right] ...


1

loose sand has no shear strength. This is why bike wheels skid on sand: the sand adheres to the tire, but that sand shears loose from the rest of the sand. Then you fall down go BOOM.


1

It seems I've figured out why the tube will jump if the mass of the rod is large enough, but I can't calculate the exact threshold. My proof of the jump is below. Let's suppose that the mass of the original tube (i.e., without the rod) is infinitesimal, whilst the mass of the rod is finite. Let's also assume that the tube won't jump. I'm going to prove the ...


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