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22

Shift the upper configuration to the left a short distance at equilibrium. Result: the left wheel goes a little up, the right goes a little down, the train tilts clockwise, the center of mass is to the right of the centerline between the wheels, and therefore the center of mass provides a restorative force to push the train back to the right. Shift the ...

13

In both diagrams in the question, the left wheel has a smaller radius at the contact point than the right wheel. Because they're fixed to a common axle, in any given amount of time, the right wheel will travel a greater distance than the left, so the axle as a whole will rotate anti-clockwise (when viewed from above) about a vertical axis. As it does this, ...

5

The equilibrium position in this case is not where the spring is not stretched, it is actually stretched by a $\Delta x$ amount with $F_{spring}(0) = k\Delta x$. So the spring force on point A is a little smaller than in point -A, since $F_{spring}(A) = -k(A-\Delta x)$ and $F_{spring}(-A) = k(A+\Delta x)$ so it compensates the "extra" force. You have ...

4

I think you have misunderstood the meaning of the equation $$\Delta X \Delta P \geq \hbar / 2 \, .$$ This is not surprising given that the notation used here is really, really misleading. It should be written like this $$\sigma_X \sigma_P \geq \hbar / 2 \, .$$ To understand this we have to explain what $\sigma_X$ and $\sigma_P$ mean. Suppose you have a ...

3

Torque is a mathematical object called a bivector, produced by taking the wedge product of $\mathbf{r}$ and $\mathbf{F}$. Bivectors can be thought of as area elements of planes; the magnitude is equal to $rF \sin \theta$ and the plane itself contains $\mathbf{r}$ and $\mathbf{F}$. By a great coincidence, in three dimensions, there are three distinct planes ...

3

The contact with the rail creates a kinematic center of rotation where the reaction forces meet. The rail car will tend to rotate about this center as a result of side loads. If the center is above the center of mass, the rail car acts like a hanging pendulum. A small deflection will cause a restoring torque opposing the swing. If the center is below the ...

3

Moments of inertia are additive. Suppose you have particle $A$ with a moment of inertia $I_A$ and particle $B$ with a moment of inertia $I_B$. Then the total moment of inertia of both particles is just $I_A + I_B$. You can imagine a ring as being made up from lots of point particles, all at a distance $R$ from the central axis. In that case the moment of ...

2

In accordance with Hooke's law the force is linear with distance. Incorporating gravity only means that the equillibirum position of the spring has changed, the "zero" around which it oscillates. The gravitational pull is already compensated by the spring. Thus the magnitude of the force is euqal at $-A$ and $+A$. Edit: When the gravitational pull on the ...

2

Sorry, but the currently accepted answer is not precise enough. In the situation you discussed, there is no applied force acting on the object, but there is still the normal force $N$, which prevents the object from "falling through" the surface due to gravity. The most common model of kinetic friction is $F_k = \mu N$, where $\mu$ is the coefficient of ...

2

Torque is a vector whose direction is always out of plane. The same with angular velocity $$\vec{\tau} = \vec{r} \times \vec{F}$$ $$(0,0,\tau) = (x,y,0) \times (F_x,F_y,0) = (0,0,x F_y - y F_x)$$ $$\vec{v} = \vec{\omega} \times \vec{r}$$ $$(v_x,v_y,0) = (0,0,\omega) \times (x,y,0) = (-y\, \omega,x \,\omega,0)$$ I always think of planar quantities as a ...

2

This answer may not look like a typical answer, but I am attempting to instill a key concept, so please bear with me. For this type of problem, where you are investigating a possible solution, UNITS are EXTREMELY important. What are the units of momentum? What are the units of force? Note that if units do not match across an equal sign, the answer is ...

2

If $m$ < $n$, then your $F$ is a function from $m$-dimensional flat space to $n$ dimensional flat space, which maps $\mathbb{R}^m$ onto a $m$ dimensional surface in $\mathbb{R}^n$. If the motion of a particle in $\mathbb{R}^n$ is restricted to this surface, then obviously a force is required to keep it on that surface. Without such a force, it follows a ...

2

Firstly, you must qualify moment of inertia by the axis it is taken about. If you translate this point, the inertia also changes as described by the parallel axis theorem. So you question relates to the moment of inertia of a ring about the ring's axis of symmetry normal to the ring's plane, and to a point on this plane at the same distance from this ...

1

I assume that you want both particles to be free to move, each in a circle. That circle should be about the center of mass of the system, which will be at $$\frac{m_1\vec{r}_1+m_2\vec{r}_2}{m_1+m_2}.$$ The radius, $R$ of each circle will be the distance from the mass to the center-of-mass point. Calculate the Newtonian gravity force on a particle, set it ...

1

For common center (barycenter) orbits, the velocities will be \begin{align} v_1&=\sqrt{\frac{Gm_2r_2}{\left(r_1+r_2\right)^2}}\\ v_2&=\sqrt{\frac{Gm_1r_1}{\left(r_1+r_2\right)^2}} \end{align} which, since $m_1=m_2$ and $r_1=r_2$, will be the same value, $v\approx0.22$ for your values of $G,\,m,\,r$. Since you've placed the two objects along the $x$ ...

1

"It will soon fall off" - indeed. Hence the part of the question that says "at what value of $\theta$ does the particle leave the surface" By first seeing what force normal to the surface is needed to keep the particle "on the surface" and following the curvature, and then finding when this force is less than gravity, you break the problem into two ...

1

Here is a method using energy. Assume the mass starts a distance $A$ from its equilibrium point and it moves past the equilibrium point a distance $B$ before turning around. The initial spring energy is equal to the final spring energy plus the energy lost due to the work by friction: \begin{align} \frac12 kA^2 &= \frac12 kB^2 + FA + FB \\ 0 &= ...

1

Draw a FBD for a small angular section of the rope (subtending an angle $\delta\theta$), showing the normal reaction from the pulley and the friction force when the system is about to slip and then let $\delta\theta\to0$. You should get: $$\frac{\mathrm{d}\,T}{\mathrm{d}\,\theta} = -\mu\,T$$ subject to $T(0) = 10^6{\rm N}$ where $\theta$ is the angular ...

1

Model 1: Kick is an impulse A common approach for this problem would be to model the football player's kick as a sudden impulse. In this case, the kick happens in an instant and, immediately after the kick, the ball has a finite speed. This speed does not depend on the spring. So, initially, no, the spring has no effect: the speed of the ball is ...

1

Let us assume the initial conditions are $x(0)=A$ and $\dot x(0)=0$. The equation of motion as long as $\dot x(t)<0$ is given by $$m\ddot x=-kx+F$$ What happens at $t=0$? We should distinguish $F < kA$ and $F \geq kA$. In the case $F > kA$, suppose that the mass starts to move with $\dot x(0)<0$, the friction will immediately start pulling the ...

1

Well, it's not energy, its power $P$. $~~~~~~~~~~~~P = \int F \cdot dv$ And since power is the derivative of energy $P = \dot E$, your world makes sense again ;). Regarding your problem I agree with Yanping Cai, the kinetic energy of the car $E_{kin}$ must be converted into potential energy of the spring $E_{spring}$. $~~~~~~~~~~~~\frac{1}{2} m_{max} ... 1 You'll find a one parameter family of solutions, because you have 4 independent quantities while in this problem you have the 3 independent units for mass, length and time. In your solution, you can see that the freedom to choose the parameter b comes from the fact that the density of air divided by the density if the object is dimensionless. To fix b ... 1 The way I like to phrase pretty much all of dimensional analysis is, "you can only take an arbitrary mathematical function of dimensionless parameters: mathematics doesn't directly deal in any other sorts of functions." When you see$[[R]] = \text m, ~~[[M]] = \text {kg},~~[[v]] = \text{m/s},~~[[\rho]] = \text{kg}/{\text m^3}$your first question needs to ... 1 Since drag is the resistance force of an object moving through a fluid (a fluid friction term), then you can make the physical argument for the value of$b$. This frictional force exists only at the boundary between the object and the fluid itself (a surface force), this means that the drag force must be independent of the mass of the object, thus$b=0$. ... 1 No. Even if you include some additional things for angular momentum there are still many things not in Newton's laws of motion. For instance, the fact that mass doesn't change from one value on Tuesday to a different value on Wednesday. Newton talks about mass in the Principia but it isn't about motion or forces so it isn't part of his Laws of Motion. ... 1 Friction comes into play whenever there is relative motion between the surfaces in contact or a tendency of motion between the same.There need not necessarily be an externally applied force on either of the bodies,that is,there need not necessarily be a relative acceleration initially,merely relative motion or a tendency for the same. It is the frictional ... 1 Moment of inertia depends on how far away mass is from the axis. In a ring of radius$R$, all the mass is$R$from the axis. For a single particle$R$away from the axis... well, all the mass is$R\$ from the axis.

1

Suppose someone suggests that following a perfectly elastic collision, two billiard balls are each traveling twice as fast as they were before (and opposite to their original directions). You can't prove him wrong using conservation of momentum, but you can prove him wrong using conservation of energy. Therefore conservation of energy has implications that ...

1

Because the pulley possesses mass, you need to apply a non-zero net torque to it to increase its angular acceleration (assuming that is the goal here). If the tensions were the same on both sides of the contact point between the string and the pulley, there would be no angular acceleration.

1

Problems that depict situations where the tensions are same on ropes on both sides of the pulley are ideal situations.It is stated so in order to minimize any complexities that may arise if the pulley was to rotate.Now, if the tensions were not equal on both sides, the pulley would experience a net non-zero torque and hence a net angular acceleration and ...

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