# Tag Info

3

The energy produced by the Hiroshima detonation is estimated to be $63\text{ TJ}$ ( $63 \times 10^{12}\text{ Joules}$). Later detonations were much much larger. If an individual is going to excite the resonances of the wind instrument with that much energy, conservation of energy requires that the same amount of energy must be put into the instrument by ...

1

Usually, to simplify things, the pulleys are assumed to be massless/frictionless. In this case, if the tensions were different. we would have a net torque. However, I=0 so this is impossible without a persistent infinite angular acceleration (which is also not possible). So we take the tensions to be the same. Similarly, in the frictionless case, there is ...

2

$\def\vm{{\bf m}}$If the magnets are free to rotate they will find the lowest energy state. This occurs when all the magnets are aligned head to tail along the line. $$\begin{equation*} \rightarrow\rightarrow\rightarrow\cdots\rightarrow\tag{1} \end{equation*}$$ If now the magnet at one end is rotated a quarter turn and then fixed, it does not follow that ...

6

The conservation laws provide just a few equations so if there are more degrees of freedom you can find trajectories that obey all conservation laws but which do not obey the dynamics. E.g. two particles of the same mass with no forces acting on them or between them. If they travel on opposite sides of a circle at constant speed about their fixed centre of ...

1

You don't get the same amount of work, N, from each coil. Units further away give you less work. Therefore, total work recovered by the coils is not L*N. (by the way L and N have very specific meanings in electromagnetism so you should avoid using those variable names) In fact, total energy of the system is preserved at all time because the potential energy ...

2

I think the first part of the curve (the part between your hand and the stationary "node" where the curve crosses back over the vertical axis) would be described by the Troposkein Curve, or at least be extremely similar. This curve would, I imagine, either fully describe, or at least be related to, any other segments between nodes if you had more than one. ...

4

Hanging my iPhone with case (0.16 kg), on a lever arm 12 cm from the center of the door knob just starts turning it. Therefore, $$m\cdot g\cdot L = 0.19 \mathrm{\,Nm}.$$

0

There is so many different variables in play to really give any answer, how old is the door noob, was there a bit of paint in the door nob assembly from the last time it was painted, has the components worn out or have started to oxidize, just to many different variables that I want to be able to give you an answer.

2

How about 5 inch-pounds max (0.56 Nm). It would take a weight of 1 pound hung at a distance of 5 inches to turn some doorknobs. Of course it really varies with the strength of the return spring, the friction and the amount of travel the latch needs.

0

There are two separate sources of friction you need to worry about there. First is just the bearing friction -- the force due to the bearing used on your axles. This is so complicated that there's no way to really calculate it; you should just measure it. And it's quite possible that it will change as the bearings wear or lose lubrication. Second is ...

0

You have to know where the center of gravity is. If $a$ is the % distance along the wheelbase for the center of gravity (50% = center, 0% = front, 100%=back), and $b$ the % distance along the track for the center of gravity (50% = center, 0% = left, 100% right) then the weight fractions for each wheel are:  (\mbox{front-left}) = (\mbox{Weight}) ...

6

I agree with @ja72 answer, except for what seems to me a minor slip (the kind I do all too often). I think he meant more energy to start or stop rather than more force, the same force for a longer time, or more force for the same acceleration. Also the energy increase is only due to the fact that a large wheel is likely to have a greater mass. If the mass ...

6

Larger are better because they can roll over stuff better (like gravel or sticks), but have more inertia and so require more force to start or stop. Also the larger wheel requires less friction to roll because the ratio of wheel diameter to axle diameter is larger. There are some trade-offs as you try to make the wheels as thin as possible to reduce ...

0

The forces acting on the toboggan are: 1) Vertical force of gravity downward; since there is no vertical velocity, this force does no work. 2) Vertical reaction force from the surface, upward; see 1) above 3) Vertical component of parent force upward; see 1) above 4) Horizontal friction force (snow on toboggan) backward; since this force is ...

0

I usually take the dot product of the force vector with the instantaneous velocity vector at the point where the force is applied. If $\vec{F}\cdot \vec{v} = 0$ then the force does no work.

1

Let's adopt polar coordinates. Fix the rotating body at the origin. Everything is happening in a plane. I assume that both springs rotate with the rotating body, i.e the follow the movement. Thus, they have the same angular velocity $\theta$. Let's take the moment of inertia of the rotating body at the origin as J. I also assume a symmetry of the system in ...

0

As far as I understood, there are two cases: Rotor: If there is a rotor at the end of A1 axis which rotates with constant rotation rate, then obviously the angular velocity is conserved. However, in this case energy is not conserved at all times(its average over long periods is though); and the rotor will receive some energy at times and it will do work on ...

0

Work is the component of force parallel to the direction of motion times the displacement. That component of force could of course point in the opposite direction of motion (anti-parallel). The work done by the force is positive in the first case and negative in the second. For instance, the direction of the force of gravity on a freely falling body (dropped ...

5

In the context of classical mechanics as you describe, negative work is performed by a force on an object roughly whenever the motion of the object is in the opposite direction as the force. This "opposition" is what causes the negative sign in the work. Such a negative work indicates that the force is tending to slow the object down i.e. decrease its ...

Top 50 recent answers are included