# Tag Info

49

The same reason objects which are heavier on one side tend to fall with the heavy side down: the tip of the arrow is denser than the rest of the arrow. The center of gravity is offset from its geometrical center, so the air drag, which is based on the object's geometry, causes a torque together with gravity as seen in this very professional picture of a body ...

46

Air. Conservation of angular momentum does infact dictate that whatever rotation it starts with it should end with, provided nothing else acts on it. Air allows its forward momentum to act on it. Consider a weather vane, a wind sock, or a flag. They rotate when not facing into the wind because one side presents more wind resistance than the other. Once ...

12

In the movies, arrows shot into the air rotate so that during the descent, the arrow head hits ground first. What is the source of this angular momentum? An arrow shot on the Moon would not do that. Air and the geometry of the arrow are key. An arrow flying through the air is subject to two forces, gravity and aerodynamic drag. Gravitation will not make an ...

9

The zig-zag strategy seems trivially obvious - but it might not be the better strategy in a particular situation. I suggest rather than asking under what conditions this strategy is preferable, you ask under what conditions the counter-intuitive straight-line strategy is preferable. The advantage of zig-zagging is that it presents a smaller "collision ...

8

As pointed out by dmckee in his comment, anyone (including myself) that has practised bow and arrow knows the arrow by weight and fletch inspection. As my English corrector points out, fletch doesn't seem to be a very commom word: it means that feather at the end of the dart/arrow. Everything resumes to how good is the fletching. When shot the arrow wants, ...

5

Newton's third law tells us that the momentum imparted on one body is equal and opposite to the momentum imparted on another if they interact. We then have $$\Delta \vec p_1~=~-\Delta\vec p_2.$$ The change in momentum is $\Delta \vec p_i~=~m\vec a_i\Delta t$, $i~=~1,~2$. The change in momentum is with Newton's second law due to a force so that $$\vec F_1~... 3 Your reasoning is essentially correct, apart from the last paragraph. To conclude, note that Newton's equation:$$\ddot {\mathbf r}(t) = \mathbf f(\mathbf r (t)),$$with initial condition \mathbf x (0)=(x_0,0,0), \dot {\mathbf x} (0)=(0,0,0) can be solved by puttin y(t)=z(t)\equiv 0, thus reducing to a one dimensional problem:$$\ddot x (t)=f(x(t)),$$... 3 What should be the minimum value of v_0 in order to hit the monkey while it's in air? Minimum value for v_0 is when arrow hits to the monkey just before it (monkey) reaches to the ground. Or, minimum value for v_0 is when arrow's range is equal to horizontal distance between hunter and monkey. So you need to find the time of monkey's fall. Or, you ... 3 the bee makes movement of air that causes air pressure, and this pressure is distributed around, if the box is closed, the amount of lift above because low pressure is equal to the weight below caused by air pressure, so no change on weight but if the box is open, i think that the air outside the box will flow in and the air below will 'hit' the box, ... 2 The flaw is your assumption that In this [accelerating] frame we don't see any force so the first law of dynamics is respected. In the accelerating reference frame you do see evidence of a force, even though you don't see the effect you are expecting (acceleration of the object). Like an observer standing on the surface of the Earth, acted on by the ... 2 if the body is rolling on a plane, then its degree of freedom is 2: one for rotation about the body's axis and one for translation of its center of gravity in forward and backward direction. if there is no slipping, the translation can be calculated from the rotation and the radius of the block. Thus the degree of freedom is degenerated to 1. 2 No this is not a physically valid equation. It is a mathematical description of the pendulum in which the variables do not have units. The corresponding physical equation would be:$$I \frac{d^2\theta}{dt^2} + b \frac{d\theta}{dt} + c \sin(\theta) = T \sin(2\pi ft),$$where$$[I]=kgm^2, [b] = Nms, [c] = [T] = Nm$$For small angles$$ sin(\theta) = \...

2

Newton's second law, force f is $$f=m\frac{d^2 x}{d t^2}$$ x is position vector of the particle. $$f=-\frac{d v}{dx}$$v is the potential energy. $$m\frac{d^2 x}{d t^2}=-\frac{d v}{dx}$$ Multiply both sides with $\dot x$ $$\frac{m}{2} \frac{d\dot x^2}{dt}=-\frac{dv}{dt}$$ $$\frac{d}{dt}(\frac{1}{2}m\dot x^2+v)=0$$ i.e., $$\frac{dE}{dt}=0$$Energy is ...

2

It is not true that the same force has to create the same change in kinetic energy. For instance, if two equal forces of opposite directions are applied on a body, the body does not change its energy. Thus each force makes zero work, or zero change in kinetic energy. You could tell that both forces create kinetic energies in different directions and that ...

2

Energy is force times distance. You haven't specified a distance, so you can't say how much energy that force delivers. If you allow the force to rotate the sphere, the application point will accelerate faster than it would if the sphere does not rotate. So a constant force will deliver energy to the sphere faster than it would in the non-rotating case. ...

2

When you jump from a height, you gather momentum. Absorbing this momentum at landing reduces the size of the maximum force, and thus the "pain". Let us assume that the distance over which a person can absorb the momentum of the fall is proportional to their height (proportional to the length of their legs). In that case, the taller person can absorb the ...

2

If we are at the equator on the surface of an atmosphere-free, non-rotating, perfectly spherical planet (no mountains or trees, etc) then the optimal angle should be zero (launch towards the horizon, tangent to the planet's surface). This assumes the initial velocity is the orbital speed at the surface of the planet, which is approximately the escape ...

2

Mass removal due to wear relates to sliding distance. This gives to wear rate relates to sliding speed. However, wear rate is not only only a function of sliding speed. Surface hardness also plays a role. AL-SI, according to this paper, will be hardened in the beginning session. With the surface hardness increased, the wear rate decreases. When it can no ...

2

It seems to me that the key to this trick is to build up enough elastic energy in the rope - which requires you to "build tension" by riding the edge hard, as explained in this video. If you do the trick too close to point A, there is limited lateral motion needed to build tension - but as you take off, the force you are looking for will disappear as the ...

2

The drone flies by its propellers exercising force against the air inside the airplane. It flies with respect to the air inside the airplane. Since the air is being carried by the airplane, the drone will fly with respect to the airplane. It'll fly forward or in whichever direction you point it, with respect to the X. Assuming the airplane is airtight.

1

Relative motion between block and belt will stop when speeds of block and belt become same. So, after finding the acceleration of the block due to friction, you should find the time taken for it to reach the speed of the belt. Note that, speed of the belt, is constant.

1

Remember that static friction is a constraint force: it enforces the rule "no motion between these surfaces" as long as the force needed to do so does not exceed the maximum. The force of static friction will have the value needed to prevent relative motion unless that value exceeds the maximum. So there are two possibilities to what happens here. ...

1

When $F=2N$, the a friction of $2N$ would also act in the opposite direction on $A$ This is wrong. Friction force isn't equal to force $F$. It is equal to $F_f=m_Ba_B$ Free body diagrams of blocks are as below: When $F=2\;\mathrm N$ then we have: $$F-F_f=m_Aa\Longrightarrow2-F_f=2a$$ $$F_f=m_Ba\Longrightarrow F_f=4a$$ Hence $$a=\frac 13\;\mathrm{m/s^2}... 1 We know$$f_n=\frac1{2\pi}\sqrt{\frac km} We then ask ourselves what is k and m. For this case, mass is $980kg$. Is k stiffness of 1 spring or 4 spring? The 980kg mass is not sitting on 1 spring. So it should be 4 spring. The tricky part is we don't use 4k. Instead we use k for stiffness equivalent to 4 springs. With 80kg, we get 1.2cm deflection. So it ...

1

You need the effective spring constant of the four springs in parallel and then use the whole mass of the car to consider the motion of the car or use the spring constant of one spring and use a quarter of the mass of the car.

1

For a pulley that has mass and moment of inertia, there must be a net torque on the pulley for the pulley to demonstrate an angular acceleration. Assuming that mass "M" is greater than mass "m", a net torque necessarily requires that the counterclockwise torque from mass "M", given by the equation Torque1 = T1(R), is larger than the clockwise torque from ...

1

Yes. Tension can vary if external forces are acting between the ends of the string - such as gravity (if the string has mass) and friction where the string makes contact with other objects (such as the pulley). For example, suppose you attach one end A of a uniform massless string to a support and the other end C to a vertically hanging mass M. This ...

1

The tension in a string is uniform Not always. The tension of the string is uniform in some cases: If String is mass-less and its particles don't move with respect to each other (i.e. string is inextensible or if it is extensible, it reach to its final tension). If String is mass-less and there is no friction between string and pulley. For string with ...

1

We must all keep in mind that for average atmospheric pressure, and assuming Zhang could pull a hard vacuum with his abdomen (which is probably not feasible, but serves to provide us with a bound), the maximum (negative) pressure he could achieve is only about 14.7 psia. Given 36 tonnes, you can back calculate what the diameter of the bowl would have to be. ...

1

The exact mechanism you describe for how suction cups work is how the rice bowl work. Instead of the bowl being flexible, though, it's his body (skin and muscles) that are providing the change in volume necessary for the suction. So, instead of the suction cup creating the volume change, it's the surface the suction cup (the bowl) is attached to, Mr. Zhang'...

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