# Tag Info

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There is critical angle $\mu = \tan( \theta_c)$ for the leg where if exceeded the foot would slip. The less the available friction $\mu$ the smaller the critical angle. Even without ice, try to walk on a dirt path using a really long stride and when your foot pressed down when the leg is at a high enough angle away from vertical it will slide. It is the ...

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If you project the velocity of a water parcel traveling up the channel into radial coordinates in both the inertial frame (space frame) and the rotating frame of reference (earth frame) then I think the necessary effects will become apparent. In the earth's frame, the velocity is (use a ' to symbolize the rotating frame): $$\vec{v} = 0\cdot\hat{r}' + ... 17 I often wondered about these things - then I came up with a simple experiment that works for me because I have a simple bike computer (thing with a magnetic pickup on the spokes that updates my speed every second). I find a flat piece of road, and ride at a certain speed (say 20 mph on my road bike, or 15 mph on my mountain bike). I then stop pedaling at a ... 5 The general rule-of-thumb http://bicycles.stackexchange.com/questions/1505/how-do-on-road-mountain-bike-speeds-translate-to-road-bike-speeds is that by switching from a mountain bike with knobbies to a road bike, one increases the speed achieved with the same "human wattage" by 15-20 percent. That's of order 5 km/h increase of the speed. I would guess ... 1 Let me start at the end. Indeed, the attractive forces between atoms may make it harder to lift an object. The actual reason is adhesion or cohesion – if the surfaces are "sticky", it's hard to separate them. Adhesion is only counted in friction if we study the effect of the adhesion on the sliding motion when the surfaces remain in contact. If the surfaces ... 0 I get the same answer as others but in a different way. First I look at the slip speed v_s(t) = \omega(t) r + v(t) and find the time needed to get v_s(t)=0. The time functions of the motion depend on the constant friction (until rolling starts) with equations$$ v(t) = v_0 - \mu g t \\ \omega(t) = \omega_0 - \frac{\mu m g r }{I} t With the general ... 0 Your expressions are all correct, except for your work due to torque. Because the cylinder isn't rolling, \theta \neq \frac{d}{R}. Torque is constant though, so we can write \theta = \omega_0 t -\frac{1}{2}\alpha t^2. Furthermore, the work due to torque is negative: W = \int F \cdot ds + \int \tau \cdot d\theta = \mu_k mg d - \mu_kmgR\theta And then ... 0 I found a couple definitions of impulsive force, one of which states The force that two colliding bodies exert on one another acts only for a short time, giving a brief but strong push. This force is called an impulsive force. Frictional forces don't typically satisfy the criterion of strong in magnitude. The idea behind this definition is that a ... 4 You seem to be saying that friction couldn't speed it up, because nothing else is moving that fast. Well, how fast is it moving? We can imagine the gyroscope axis parallel to the z axis, and the casing to be aligned such that the x axis goes through it. If the casing is tipped slightly, the gyroscope resists that turning and one side of the shaft has firm ... 1 I'm going to just use 32^∘ below; it doesn't make a difference. Your equation isn't correct. You should have F_x−f_k = 0 or Fcos32∘−f_k = 0. The x-component of F is F_x = F.cos32^∘. Writing F_x.cos32^∘ doesn't make logical sense. Why? Shouldn't I be able to use ∑F=0 in this problem to find the answer? Yes, you can. 2 There is, effectively, only gravitation and friction acting on the pack of gum. However, the friction is not that strong (it is mostly independent of the velocity of the book, and dynamic friction is weaker than static friction) and it doesn't have that much time to act. Hence it doesn't affect the momentum of the gum noticeably. This is very related to the ... 0 If the shape is described with polar coordinates such that the radial distance from some center is R(t) and t is the angle from some datum, then you can define the pressure angle \alpha at each location t as \tan( \alpha) = - \frac{1}{R(t)} \left( \frac{{\rm d}}{{\rm d}t} R(t) \right)  The pressure angle will give the direction of the normal ...

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Backspin! Those shots in which the cue ball "draws" backwards after hitting the target ball involve backspin. Without backspin, the cue ball cannot reverse direction. Consider what happens when the cue ball is not spinning at all when it hits the target ball. The cue ball will come to a dead stop if it hits the target ball straight on. Think of Newton's ...

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The direction the ball will take depends on the angular momentum. The velocity with which the ball moves or bounces backwards but the chief determinant is the spinning effect of the incoming ball.

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First of all, if the collision is elastic, the distribution of momentum in between the components is completely determined by momentum and energy conservation! This statement is most obvious in the center-of-mass frame where the total momentum is zero and the two objects are moving in opposite directions. The momentum conservation (the total momentum is ...

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Many students confuse the term work in physics with the conventional term of work. Your body wastes energy when you push something, and when that something doesn't move... 100% is wasted in the biological efficiency. 1st step: forget the concept of how hard it would be for you to do it. How much work is a table doing by holding up a 1kg weight? zero. It ...

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It has been engineered, based on observations hair patterns of insects (Droplet slides down when substrate is oriented so that the hairs point downwards, while it was attached in the first two orientations)

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To get the initial conditions you guess and look at whether the guesses fit the physical situation. For example, suppose that you have a mass on a spring and you are holding mass so that the spring is slightly stretched and then release it. The mass is not moving at t=0 so the initial condition is that the velocity is zero at t=0. But you could imagine other ...

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Your intituition is totally different because ennumerous forces change the situation from the ideal situation predicted by work enrgy theorem, some of them are: Air Drag/Friction, Rotational Friction due to differences in size of tyres, different aerodynamic effects due to different body design.. etc.

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Something moving in a circle at steady speed is experiencing constant inward accelleration. From F = mA, we know that this requires a force on a object to accellerate it, and that force is proportional to the mass of the object and the accelleration. In the case of a car going around at circle at steady speed, that force comes from the ground pushing on ...

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Friction forces act as a response, and opposite, to velocity, not force (that would be normal forces). The car has a liner velocity in the forward direction, and it keeps moving indefinitely, ignoring any residual friction. Then, if the steering wheel is turned left, the front tires are rotated to the left, thus there appears a frictional force ...

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The simple answer is that the chemical properties of oil and water result in them having smaller coefficients of friction than other surfaces like concrete. The frictional force exerted on a skidding body is given by $F=\mu_k N$ where $N=mg$ is the normal force of the body, and $\mu_k$ is the coefficient of kinetic friction. Assuming the same object skids on ...

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One word really: friction. Surfaces that are "slippery" or difficult to walk on have much smaller coefficients of static and kinetic friction. Static friction is why its hard to get things moving and kinetic friction is what slows things down once they start moving.

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Unlike other substances, water is denser in its liquid form. I'd guess that the pressure that you exert when you walk on the ice melts some of it into water, making for a slippery surface

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Short answer: No, Long answer: let's say your plane is moving to the right. If the plane were purely horizontal (like a cube) and friction less, the block placed on top won't even move (there is no force on it except gravity and the normal force from the block underneath it). Now lets look at an inclined plane. The block placed on top again has only two ...

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