# Tag Info

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 ...

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 ...

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. ...

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 ...

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 ...

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 ...

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.

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 ...

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 ...

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 ...

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 ...

2

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 ...

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 $$\mathbf{v}_{cm} = \mathbf{v}_A + \mathbf{\omega} \times ... 1 Consider your arms as pendulums. The period of a pendulum is determined by its length and by gravity. Though the period is affected neither by the weights nor by the amplitude of your arms, these two quantities affect your balance and efficiency. When you run you get into a rhythm of arms and legs. Your legs are do the work; your arms are along for the ... 1 The ball will fall out of the tube, but I think this is precisely because there is no friction to supply the centripetal acceleration. There is still a normal force however this is not towards the point of rotation. Friction would supply a force along the direction of the tube. Since you cannot supply an adequate centripetal acceleration, the ball should ... 1 Maybe this thought experiment? Suppose you have a frictionless wheel and surface, in a vacuum, etc., and you spin up the wheel and push it forward so that its linear speed just matches the rotational speed and it moves along the surface with no slippage, i.e. zero net velocity at the point of contact. In this scenario, there are no forces of any kind, so ... 1 There might be normal friction acting on the rolling wheel, namely static friction. The static friction force, F_f, is often written as,$$ 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

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

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