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3

When it is half full, however, the water bottle rotates for one half-spin, and then it stops rotating. Why is this? This is why you don't want to ship oil across the ocean in a half-full oil tanker. If you do, you had better equip that tanker with some very good anti-slosh mechanisms. The same goes for trucks, trains, and spacecraft carrying fluid. ...


1

I guess, The most important question is wheter you will accelerate the fluid when throwing the bottle. This and conservation of angular momentum will be much more important than dissipation of energy in the fluid. If it is full, the water, due to the constraints on the system, will be accelerated with the bottle when throwing. So it will just continue to ...


0

The ball will fall out of the tube, but I think this is precisely because there is no friction to supply the centripetal acceleration. There is still a normal force however this is not towards the point of rotation. Friction would supply a force along the direction of the tube. Since you cannot supply an adequate centripetal acceleration, the ball should ...


1

EDIT: The "old" explanation below is quite informal, as gravity cannot produce a torque around the center of mass. See the new explanation below it. Humans lean forwards slightly as they run (more so if they run faster, and have to counter larger drag forces). This makes gravity exert a counter-torque that exactly balances the torque produced by action of ...


0

This is the equation that solves the problem: $$\frac{d\vec{L}}{dt}=\vec{r}\wedge\vec{F}$$ In this case $\frac{d(IV_{wheel})}{dt}=rF\rightarrow IV_{Wheel}\Omega_{prec}=RMg$ from this equation you can get $\Omega_{prec}$ since it is the only unknown, the weight is unimportant because it is canceled


0

There is no "precession force" acting at the point where the arm holds up the shaft. The force from the arm must balance Mg, and hence the vertically upwards force is just Mg. However, the wheel has angular momentum, let's call it L (not to be confused with the length of the shaft) which equals $I\omega$, where $I$ is the moment of inertia around the axis ...


-1

It is because there aren't any forces acting on the mud keeping turning with the tire that it flies off. At whatever point the mud comes off, it will travel tangent to the tire at first and the follow a parabola due to earths gravity. It is most likely the more loose mud will come off first and at that point the tangential direction of the tire points ...


3

You were probably expected to note that the path of any point of the tire is a cycloid. At the point of contact with the ground, the tire is not moving. As it rises from the ground it is moving faster, with a speed that is $v(1+\sin \theta)$ where $v$ is the bike speed and $\theta$ is measured counterclockwise from horizontal. The centrifugal force is ...


0

Let's consider two ships passing each other. When they pass, a rope is thrown from a ship to the other ship. Then then rope is pulled sharply. That causes the ships to collide, the rears of the ships hit each other and the ships start to spin. In the previous scenario part of the energy used to pull the rope became rotational energy of the ships, that ...


3

How does the kinetic energy of a ballerina increase? Conservation of angular momentum: $$L_1=L_2 \implies I_1\omega_1=I_2\omega_2\quad\quad (1)$$ Pulling in your arms reduces moment of inertia $I$, since the same mass is now distributed over a volume closer to the spin centre, $I=\sum mr^2$. As you say, reducing $I$, so $I_2<I_1$, implies ...


0

Think about what the parameters are that determine the period of the pendulum $T = 2 \pi \sqrt{ \frac{l}{g}}$. And is there a change in any of these variables.


0

The equation from the link is an approximation. In your analysis you do not include the mass of the car when considering linear acceleration. Nor do you consider the reaction force of the wheel/axle accelerating the car. You can split the driving torque into two portions, the portion that rotationally accelerates the wheel, and the portion that is ballanced ...


7

Ernie is close to the correct answer, but the fundamental thing that needs to be considered is how the internal energy of the body flows. I researched this in a very interesting book I''m still reading, Principles of Animal Locomotion . Chapter 7 addresses running and section 7.5 discusses Internal Kinetic Energy . Limb accelerations can store kinetic energy ...


6

Speculating here... I suspect that for light weights the answer is yes - with the right technique. Your center of mass moves up and down which requires energy being absorbed and expended by your legs. Moving your arms with small weights should allow you to even out the motion, lowering the peak stress on your legs so they tire more slowly. In a sense you ...


1

Consider your arms as pendulums. The period of a pendulum is determined by its length and by gravity. Though the period is affected neither by the weights nor by the amplitude of your arms, these two quantities affect your balance and efficiency. When you run you get into a rhythm of arms and legs. Your legs are do the work; your arms are along for the ...


1

I think you are overcomplicating this. Consider an arbitrary point P moving with linear speed $\mathbf{v}_A$. Linear momentum is $$\mathbf{P} = m \mathbf{v}_{cm}$$ Angular momentum at the center of mass is $$\mathbf{L}_{cm} = I_{cm} \mathbf{\omega}$$ Linear velocity of the center of mass is $$\mathbf{v}_{cm} = \mathbf{v}_A + \mathbf{\omega} \times ...


2

If we look at $CP_i X m_i (\omega \times CP_i)$ we can say that the cross product in the parenthesis gives the component of the vector $CP_i$ along the direction of $\omega$. Let us call that component $R_i$. (note: $R_i$ is the perpendicular distance between the particle in the system of particles in which we are interested in and the axis of rotation of ...


0

I have read that the formula for angular velocity: $$\dot {\vec r}=\vec \omega \times\vec r \tag{1}$$ does not hold in some situations, but the book does not specify what situation so please could you produce a list of when this formula does not hold. That expression is only true in the case of circular motion. It fails whenever the radial component ...


2

I just want to formulate John Rennie's answer in a slightly different and more general way. For any object of angular momentum $I$, the angular acceleration $\dot \omega$ is given by $$\dot\omega = \frac{\Gamma}{I}$$ where $\Gamma$ is the torque. Now in the case of an object with masses $m_i$ distributed along the length at distance $\ell_i$ from the ...


1

There might be normal friction acting on the rolling wheel, namely static friction. The static friction force, $F_f$, is often written as, $$ F_f \leq \mu_s F_n, $$ where $\mu_s$ is the coefficient of static friction and $F_n$ the normal force, in this case the weight of the wheel. Note the less-than-or-equal-to sign. The magnitude of this friction force ...


1

Maybe this thought experiment? Suppose you have a frictionless wheel and surface, in a vacuum, etc., and you spin up the wheel and push it forward so that its linear speed just matches the rotational speed and it moves along the surface with no slippage, i.e. zero net velocity at the point of contact. In this scenario, there are no forces of any kind, so ...


2

The force doesn't get "used up" by creating a torque. The torque and the force exist simultaneously. You correctly computed the torque due to that force. If that is the only force on this object, then the linear acceleration of the object is given by Newton's 2nd law: $$a = \frac{100~\rm N}{m}$$ where $m$ is the mass of the object. That is the linear ...


47

I'm going to assume the bottom end of the rod is fixed, so the rod rotates around it. I think this is what you have in mind - shout if it isn't. So at some point during its fall the rod looks like: The mass of the rod is $m$ and the mass oif the weight on the end is $M$, and I've drawn in the forces due to gravity. To write down the equation of motion ...


7

I assume that you're imagining two sticks whose bases stay stuck still to the ground as they tip over. In this case, what we should do to calculate the tipping rate (I wouldn't exactly say "falling"-- I think it's misleading) is consider the torque applied to each stick as it tips. What the answer really comes down to is the struggle between torque (the ...


-3

Yes you are right. There is a resist in gravity because of an applied weight. Therefore the acceleration of the beam falls slightly slower depending on the mass of the object. Unless nothing are holding them together, both beams would fall at the same rate except the applied weight on the top would fall slower. For an example, If we are in an elevator and ...


3

The author of the question may be expecting you to recognize or assume that the axle of the platform is supported by the floor (probably by way of some kind of superstructure), in which case linear momentum would not be conserved for a system defined as the student and the platform because of the reaction between the platform and the floor. It is, in my ...


3

Your answer set is wrong. Linear momentum is always conserved in closed systems, so your initial guess was correct. Good job! Apparently you know more about physics than the people who write the text books. Also, you should read the policies on posting homework related questions on this site.


0

The translational component is already taken into account by your use of the parallel axes theorem to derive the moment of inertia for a disk rotating about its edge. But if you just use translation you are not accounting for the rotation of the disk... At the lowest point in the arc the points are not all moving in the same direction: some move faster, some ...


3

I looked up leap second in Wikipedia. It is a second added (usually) to clocks to keep them in sync with the atomic clock. Civilian clocks use Coordinated Universal Time (UTC), sometimes erroneously called Greenwich Mean Time (which no longer exists). Atomic clocks use International Atomic Time (TAI). UTC and TAI are in sync. Civilian clocks tick at the ...


0

Without a guideline on what you're wanting to optimize for specifically, this question is hard to answer. Different types of tires/wheels are better for different things, and differ in several different ways. Some general considerations to take into account, then: When riding on a hard surface, a narrower tire is preferred due to its lighter weight, more ...


-1

No. You have to take moments about the center. Taking moments clockwise, 252Nm + 240Nm - 246Nm - 198Nm = 48Nm Therefore it has a clockwise moment of 48Nm which implies it moves in the clockwise direction.


1

You have a compound pendulum so you need to consider its moment of inertia, by using the parallel axis theorem on its parts


3

You're mistaken. If I visualize , the speed of rotation increases as the person's hands fold inwards, this indicates that the angular velocity increases. The net angular momentum however, remains conserved. (If we consider the system to be isolated, that is. For every isolated system, the angular and linear momentum is always conserved.) Also, for the given ...


2

For disc $I=(2m*r*r)/2$ for small body $I=m*r*r$ Hence net Moment of inertia is their sum.


0

To get to a final rotational velocity $\Omega$ is takes the same amount of energy regadless of where you push. Why? Well the final kinetic energy is $K=\frac{1}{2} I \Omega^2$ (where $I$ is the mass moment of inertia about the hinge) and this value does not depend on where you push. This is kind of a boring result. What does differ is how much you need to ...


4

Let's figure it out. First let's figure out the force you would need to slide the object. In order to slide it you'd need a force that overcomes friction. $$ F_{\text{slide}} > \mu m g $$ where $\mu$ is the coefficient of friction. In order to tip it, you'd need to cause a torque that will cause the can to rotate about its far bottom corner. If you ...


2

If there is no resistance then there will be no net torque applied about the center of mass. So any initial rotational speed will remain. The rotation center is going to be the center of mass. The effect of gravity will be to accelerate the center of mass, and it will have no effect on the rotational motion of the body. See this accepted answer for a ...



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