explain why a centripetal force does not work on a circularly moving object?
816881
Work is commonly defined as the dot product of force and displacement.  Essentially this means that you are moving in the same or exact opposite direction as your field lines for your given force.  Example: Friction does work to stop a moving box as the force is directed anti-paralell to your direction of travel.\n\nWhen you move in a circle, centripetal force is always directed toward the inside of the circle, which is exactly perpendicular to your direction of motion in the circle.  Since the force is perpendicular to the direction of motion, there is no work done.
336809
On a flat curve, the centripetal force on the car is merely the friction between the tires and the road. Note, however, that this friction is STATIC friction, which has a maximum value. \n\nF(radial)=f(static)=mv^2/r\n\nIf the velocity exceeds a certain value, the car will slide. \n\nOn a banked road, however, one component of the normal force will point inwards, adding to the centripetal force. This allows the car to take the curve at higher velocities without sliding:\n\nF(radial)=f(static)+normal*sin(angle)=mv^2/r
198674
There is no such thing as "centrifugal" force.\n\nThe mud stuck on a rotating wheel is undergoing circular motion.  At any instant in time as the wheel rotates, the mud stuck on the side has a velocity vector pointed straight outward tangential to the wheel...and yet it still goes around in a circle.  Since the velocity is constantly changine, since the direction is constantly changing ("speed" may or may not be constant), there must be some type of force acting on the mud to cause it to accelerate.  This force is called centripetal force, the force causing the mud to contantly accelerate toward the inside (axis of rotation) of the wheel.\n\nThis foce is not magic, something must be applying the force on the mud.  You could call it the "stickyness" factor of the mud or a type of friction adhearing the mud on the wheel.  In reality, it is due to thesruface tension of water [but that is a topic for another day].\nThis centripetal force is applied on the mud so that is experiences centripetal acceleration causing the musts velocity to constantly change as it goes around the wheel, without this force, the mud would just fly off in a  straight line.\n\nAt some point, when the wheel is turning fast enoug, not enough centripetal force is being applied to the mud in order to keep it adhearing to the wheel and continue going around in a circle.\nCentripetal force = m(v^2 / r),\nwhere m is the mass [of the mud], v is the tangential speed [of the mud], and r is the radius of the rotation.\nAs you can see, a larger centripetal force is required if the wheel is spinning very fast, so if the wheel is spining so fast that the mud does not get enough centripetal force, it will fly off.  Another fact would be a smaller radius of rotation, the smaller the radius, the higher the force needed.
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Think of the wheels as spinning masses or revolving weights, in fact gyroscopes. \nEach wants to turn left or right while spinning but is constrained by it's axis,\nHowever both drift overtime.\nNow add another pair further back and wider apart, and this pair are not on the same line of travel.\nWith four wheels all wanting to drift the resultant direction becomes a straighter line.
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You need to look at the normal acceleration versus the acceleration due to gravity at that angle.\n\nThe normal acceleration = v^2/R (where R is the radius and v is the velocity) and if this is greater than the component of gravity pointed towards the center of circle then airborne. So as it slides down the circle the velocity, v, is increasing, at the same time the component of gravity gets smaller.\n\nAt 0 degrees  (on top) v is zero but all the gravity is pointed towards the center of the circle. At 90 degrees the velocity is equal to square root of (2 * g * R) but the gravity component passing through the center of circle is 0. So somewhere between 0 and 90 degrees it takes off.\n\nI will let you do the math, but I would check 48 degrees as your instructor implied. (wink wink), use conservation of energy to find v.
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You say "when" like you think it is going to happen any minute.\n\nThe acceleration due to gravity does not depend on the rotation of the Earth, it only depends on the mass of the Earth and the distance from it.  No matter how fast you are traveling or the Earth is spinning, there will alwasy be a constant gravitational attraction between the Earth and some mass at a constant distance.\n\nHowever, the fact that the Earth is spinning has the effect of lowering the percieved influence of gravity in a way.\nGravity can be through of in this case as the centripetal force which is causing you travel in circular motion.\nWithout going into too much detail (and I fear that I am not explaining it clearly), the force of gravity can only cause so much of a centripetal acceleration, beyond that and you cannot continue in the same circular path.\n\nThere is a point where the force of gravity would not be enough to "hold you down" if the Earth was spinning fast enough.  You set your weight (weight = force of gravity) eqaul to the centripetal force (the force causing you to moving in circular motion), then you can solve for the tangential speed of the Earth's rotation which you would be "weightless".\nHowever, the force of gravity in and of itself is in no way effected by the Earth's rotation.\n\nI hope I explained it clearly enough, this topic lends itself to people using in correct terms like "centrifugal force" which is really not correct.
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The basic formula that we know in circular motion is:\n\nF=mv^2/r, \n\nwhere F is the centripetal force, m the mass, v is the linear velocity, and r is the radius.\n\nWe also know that v=wr\n\nwhere v is the linear speed, w (supposed to be omega) is the angular velocity, and r is the radius.\n\nSubstitute the value of v in our first formula:\n\nF=m(wr)^2/r\n F =mw^2r      Formula 1)\n\nIn this problem, we also know that w which is the angular velocity is the same for m1 as for m2 because they are located along the same line which is rotating about one end of the rod.\n\nWe are also given the value of the tension in the inner section as 3x that of the outer section. Let's draw a simple free body diagram:\n                  \nT1=3F2<----0-->T2=F2<---0\n\nThe first 0 is m1 and the second 0 is m2.\n\nThe diagram might get distorted when I send it, but let's just cross our fingers it comes out clear enough.\n\nThe net force acting on m1 is (3F2-T2). But T2=F2. \nTherefore, net force on m1 =3F2-F2=2F2. Let's call\nthis net force as F1. Thus F1=2F2. Now we also know\nthat \n\nF1=m1w^2r    \nbased on formula 1) above.  Now substitute the value of F1=2F2, and we get:\n\n2F2=m1w^2r   Equation 1)\n\nAnd based on the same formula F=mw^2r:\n\nF2=m2w^2(2r)    Equation 2)\n\nRemember that for m2 the radius is twice that of m1, which is a given condition in this problem.\n\nNow divide equation 2 by equation 1:\n\nF2=m2w^2(2r)\n------------------\n2F2=m1w^2r\n\nF2,w^2, and r will cancel out leaving:\n\n1/2=2m2/m1\n\nDivide both sides by 2:\n\n1/4=m2/m1\n\nm2/m1=1/4\n          =0.25
792058
you should visit the following link\nhttp://en.wikipedia.org/wiki/Centripetal_acceleration
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Centrifugal force is a fact, but we call it a fictional force. We interpret what we observe as a force because we want to apply the laws of inertia and so on that appropriate for a stationary, inertial frame to a nonstationary, noninertial frame. \n\nIt has the same root as the word "fugitive", a latin word meaning to flee. So if you spin a rock on a rope, it seems to you that the rock wants to flee away from your hand, and the string is holding it back with its centripetal force (or center seeking force, again from the latin). The fleeing force or centrifugal force is not really there, it is just our interpretation of what we observing from the application of Newton's laws in a rotating frame.
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Whenever a force is acting perpendicular to the motion of an object and if the force is pointed to ward a fixed point then the path of the object will be circular.
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I haven't done any physics problems for ages, but I'll give it a try.\n\nThe centripetal acceleration = V^2/R \n\nBut, because the acceleration due to gravity is about 9.8m/sec^2 and this is important to the answer, you first want to express V in m/sec;\n\nV = 360km/hr = 360,000m/3,600sec = 100m/sec.\n\nSo the centripetal acc. = (100m/sec)^2 /500m = 10000/500 = 20m/sec^2, which is about 20/9.8 = 2.04 g-forces. This force is always directed towards the center of rotation. \n(Note: if you were on a spinning wheel space station where gravitational forces were negligible, and it had a radius of 500m and had a tangential V of 100m/sec, and you were walking around the inside of the station at the 100m extremity, you would have the apparent weight of 2.04 gs)   \n\nAs far as his "weight" is concerned: if the seat of the plane measured his weight at different points around the loop then it would vary. \n\nAt the top of the loop: his recorded weight would be 70kg x (20-9.8)/9.8 = 72.86kg. This is because gravity is reducing the  weight. \n\nAt the bottom: it would be 70x(20+9.8)/9.8 = 212.86 kg\nThis is because gravity is adding to the weight.\n\nAt the mid-points gravity and the centripetal acc. would be perpendicular to each other and the recorded weight would simply be 70x20/9.8 = 142.86kg\n\nI hope that helps.
826674
use conservation of angular momentum in these kind of problems\nNOT energy conservation\nif you dont know how to use the conservation equation, better read up on examples
9163
There is a force, called centripetal force, that exerts on every body in rotational movement. If that force would not exist the body would go in straight line.\nThe magnitude of that force is\n\nF=m ω^2 r\n\nwhere m [kg] is the mass of the body, ω [rad/s] the angular velocity of the body and r [m] the radius of the trajectory.\n\nWe have the relation between rpm and angular velocity:\n\nrpm=60ω/2π => ω=πrpm/30\n\nHence, F=m π^2 rpm^2 r/900\n\nAccording to Newton: F=m a  (where a is the acceleration in m/s^2)\n\nThus a=π^2 rpm^2 r/900\n\nto get the expression in g you divide both sides of the equation by gravity's acceleration g (~9,81 m/s^2)\n\nSo RCF=(π^2 rpm^2 r) / (900 g) \n(RCF=relative centrifugal force)\nRCF=1.118*(10^-3) * rpm^2 * r        in SI units
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e, because of the Newton's 1st law: "An object at rest will remain at rest unless acted upon by an external and unbalanced force . An object in motion will remain in motion unless acted upon by an external and unbalanced force". Assuming inertial frame of reference.\nThe object can still have angular acceleration if there are two forces cancelling each other but are applied to an object at different points. \n\nIn real life there is always something that will slow down the object even in space.
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This is very similar to your other question (see m responce there).\n\nTo help you figure this out, it would help if you though just exactly how the velocity changes are the plane goes around the circle.  If you cant visualize it, draw a diagram.\nAs the plane travels around the circle, its velocity vector is always pointed straight out, tangent to the edge of the circle.\nBy the time the plane gets 1/4 way around the circle, it no longer has any of its velocity in the original "positive" direction, it is all directed perpendicular to the original direction.
728073
At faster speeds you have turbulent flow, with boundary layer separation on the trailing edge. At slower speeds you have laminar flow with no boundary layer separation. Around the transition funny stuff happens to the flow where it can't quite decide what to do ( e.g. in flow around a circular cylinder, there is the Strouhal effect).  So - and I'm guessing here - that as the ball slows and the boundary layer starts to reattach, that it doesn't happen quite symmetrically, and thus imparts a torque to the ball, making it spin.\n\nThe ball is round - why wouldn't it happen symmetrically? I'd guess that some small imperfection in the air flow, possibly due to the seams in ball, gets amplified in a nonlinear manner in the air flow.
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The Coriolis effect works due to the fact that the earth (different from what is commonly taught in school) is NOT an inertial frame, because of the rotation the earth has; inertial frames are created when an object is moving at constant velocity, and the earth has no lineal movement, the tangential velocity (as vector) is ever changing although it's magnitude is always kept (like in any other rotating frame), then this velocity is not constant it has an acceleration (known as centripetal acceleration) and as a result pseudo forces (like Coriolis force) are created (like when you're in a car and you accelerate, this is called the inertial force).\nThe Coriolis force term can be obtained theoretically (If you want to profoundest I recommend you the book named "Classical dynamics of particles and systems" by Stephen T. Thornton & Jerry B. Marion), and, as seen in wikipedia, this force depends on the velocity mainly (as vector) and the angular velocity (whose cross product changes sign depending on you're on the Northern hemisphere or the Southern hemisphere, and it's cero on the Ecuador, so that this force is equal in magnitude but different in direction in the Northern and the Southern hemispheres at the Tropic of Cancer and the Tropic of Capricorn, for example)\nI did an experiment about the Coriolis effect in my mechanics lab. You can use or build an scaled Foucault pendulum (because an usual Foucault pendulum is just impractical, by the time you need for the effect to be noted and by the size predicted by some theory you need the pendulum to be) and observe how are felt the fictitious forces. As you may see in the theory given by some books, online articles or even the wikipedia for the effect to be seen (precession of the pendulum) it is needed some parameters to have certain values (Rossby number) and by creating and small rotating frame with a relative high angular velocity (for example 34 rpm) you can see, according to the rotating frame, the precession that shows the Coriolis effect.\n\nEdited:\n\nIf you know spanish you can get more detailed information on the Coriolis effect and non intertial frames in http://www.sc.ehu.es/sbweb/fisica/cinematica/coriolis/coriolis.htm\n\nOtherwise, you can even use a translator, this site is worth and for many spanish speaking students in physics from high school to university it's been a good resource...