With centripetal acceleration, when would the normal force of motorcyclist be zero?
925006
Since the magnitude of the acceleration is v^2/r, and the radius of the Earth is approximately 4000 miles (21120000 ft), you would have to solve the following equation: v^2/21120000=32\nv^2=675840000\nv=25997feet per second  or\nv=4.92 miles per second
1011658
Ummmm, I think your talking about Centripetal and Centrifugal force....
437948
On a motorcycle, hit on the front brake, not too hard, but enough to slow down the front wheel but still be able to harness the momentum shift when applying the brake, which tilts the motorcycle at an angle, usually about 60 degrees, and the motorcycle keeps rolling.\n\nAn endo is when the front tire is still on the ground and the back is in the air.  It's the opposite of a wheelie.  Rolling just means you're still in motion when you're doing it
850378
Centripetal acceleration a=r*w^2, r=0.35 m, w=2500*2pi/60 rad/s is angular speed, v=w*r is linear speed;
792607
0 km/hr^2 .... because your speed doesn't change with time due to constant as you mentioned
726238
let mass of car = M\n\n0.6 = Force of Friction / Normal Force , Normal Force = 9.81xM\n\nForce of Friction = 5.886xM\n\nthis means that\n\ncentri force must not exceed 5.886xM\n\na=v^2 / 10, F_centri = (v^2 / 10)xM\n\n(v^2 / 10)xM=5.886xM , M's cancel out\n\nv=7.67
660806
centripetal force and centrifugal force\ncentripetal force and centrifugal force, action-reaction force pair associated with circular motion. According to Newton's first law of motion, a moving body travels along a straight path with constant speed (i.e., has constant velocity) unless it is acted on by an outside force. For circular motion to occur there must be a constant force acting on a body, pushing it toward the center of the circular path. This force is the centripetal (“center-seeking”) force. For a planet orbiting the sun, the force is gravitational; for an object twirled on a string, the force is mechanical; for an electron orbiting an atom, it is electrical. The magnitude F of the centripetal force is equal to the mass m of the body times its velocity squared v 2 divided by the radius r of its path: F=mv2/r. According to Newton's third law of motion, for every action there is an equal and opposite reaction. The centripetal force, the action, is balanced by a reaction force, the centrifugal (“center-fleeing”) force. The two forces are equal in magnitude and opposite in direction. The centrifugal force does not act on the body in motion; the only force acting on the body in motion is the centripetal force. The centrifugal force acts on the source of the centripetal force to displace it radially from the center of the path. Thus, in twirling a mass on a string, the centripetal force transmitted by the string pulls in on the mass to keep it in its circular path, while the centrifugal force transmitted by the string pulls outward on its point of attachment at the center of the path. The centrifugal force is often mistakenly thought to cause a body to fly out of its circular path when it is released; rather, it is the removal of the centripetal force that allows the body to travel in a straight line as required by Newton's first law. If there were in fact a force acting to force the body out of its circular path, its path when released would not be the straight tangential course that is always observed.
144016
The net force needed is zero.  Force is mass * acceleration, and acceleration suggests a change in velocity.  For constant speed and direction, the acceleration is zero, and hence the net force is zero.  When moving at constant velocity, the frictional and drag forces (which work to slow the car) and the drive force of the engine are exactly equal and opposite.  The net force is the sum of these two opposite forces, and hence the net force is Zero.
708184
The centripetal acceleration is v^2/r and must equal the acceleration due to gravity if the people are to stay in the car.\n\nv^2/r = g\nv^2 = gr\n\nv = sqrt(gr) = sqrt(9.8*12) = sqrt(117.6) = 10.8 meters/sec
93119
Force = mass x acceleration\n\nacceleration = force/mass.\n\nWith a given force, as mass increases, acceleration decreases.
295799
k ill try to help u  even though i dont condone cheating. i just know how it is to not understand something and its due in a few hours and it sucks. just remember, nobody will be able to help u on the exam.\n\nso 1. \na. acceleration is change in velocity divided by time. \nsince it started from rest, initial velocity is 0 and then final velocity is given as 25.3 m/s. and the time is given as 0.59s. so to get acceleration   you would do  \na =  Vf-Vi / t\n  =  25.3 - 0 / 0.59\n  = 42.88......\n\nb. the force is equal to mass x acceleration. since you know the mass and acceleration, you just multiply them and you get F = 38421.69... N. \n\nc. you have to find the normal force in this case, which is the force that opposes gravity. like upward force. so in this case, gravity is acting on the driver and force of gravity is equal to mass x 9.8 = 666.4 N. and that means the normal force is 666.4N (since he is not moving up or down the total vertical force = 0)\n\n2. ok well in this case, you know the initial velocity is 0, the distance is 404m and the time is 4.944s. if its constant acceleration, you could assume that the average velocity is just Vi+Vf/2. so if thatz average velocity, then distance = average velocity x time\n     404 = 4.944s  x  (0+Vf)/2\n 404 x 2 / 4.944 = Vf\nVf = 163.43........m/s\n\nthe acceleration would just be Vf-Vi/t. and since you know Vf now  you just do   a = 163.43-0/4.944\na = 33.056....m/s^2\n\n3. ok well acceleration is 9.8m/s^2 (gravity)     distance is 9.7m. initial velocity is 0m/s. so this time, you have to solve 2 equations. the first one is the acceleration equation. a = Vf-Vi / t = Vf / t = 9.8\n9.8 = Vf/t\nVf = 9.8t\nand the other one is the distance = t  x  Vi+Vf/2   average velocity one. \n9.7 = t x Vf / 2\n19.4 = Vf x t\n19.4 = (9.8t) x t\n19.4 = 9.8t^2\nt^2 = 1.979...\nt = 1.406....s\n\nnow that you know the time, you find the Vf by putting it in any of those 2 equations. Vf = 9.8t = 9.8 (1.406)\nVf = 13.788... m/s\n\npart b. so in this case, you use the initial velocity as the one u just found (the velocity of the swimmer at the surface of the water)   so Vi = 13.788m/s\nand Vf = 0    since they come to a stop. and d = 2m\nso you have to find the acceleration. first you need to find the time by using the average velocity equation.\nd = t x Vi+Vf/2\n2m = t x (0+13.788)/2\n4 = 13.788t\nt = 0.29s\nnow that you know the time, you have to find acceleration\na = Vf-Vi/t\n  = 0-13.788 / 0.29\na = -47.5286m/s^2    its negative because its downwards and up is positive in our case. so the force = mass x acceleration and you just got acceleration and her mass is 65kg so you just multiply by 65kg to get F = 3089.36N\n\n\n4. well F = ma and you have the force. when they say it weighs this much newtons, then it means  use gravitational acceleration which is 9.8. so 2350N = m x 9.8m/s^2\nm = 2350/9.8\nm = 239.79kg\n\ni hope you actually understand this stuff. try to do your homework by yourself   dont take advantage of this site. youll just end up screwing yourself over. and if u dont get anythin, let me know and ill try to help
262728
Simple.\nThe car tyres rotate thus the force is called centripetel force(f).It is equal to v^2/r where r is the radius of tyre.\n\nforce at 15m/s = 15*15/r\nforce at 30m/s = 30*30/r\n\nWhen we divide them we get 1:4 the answer\nit is assumed that car tyre is of same radius in both cases.
870312
a)For rigid body, center of mass accelerates in the direction of applied external forces. Just Newton's second law and therefore true.\nb)As in Newton's second law F=ma, force is dependent on acceleration and not directly to velocity only. So, there is explicit time dependence. hence, it really matters how fast the object gets to 10m/s or 20m/s velocity starting from rest. Therefore this statement is inconclusive with unknow time.
912606
The simple answer is that the vector sum of the forces acting on an object must be zero if the object is to remain in "equilibrium" or in other words to have constant velocity.
981896
The upward reaction force (at the top of the bump) is NOT mg + mv^2/r.\n\nPer unit mass, it is g - v^2/r. Note the sign.\n\nAlso, the relevant mass is not the whole mass of the car m but, say, about *half* of it---the rear wheels are not even in contact with the bump unless it's a really huge one (was this overlooked when the question was set?)\n\nThe previous answers are *almost* correct---except for the sign mistake (+ instead of - in the centripetal term). Taking the upward direction as +ve, the gravity force is -ve (-g per unit mass), the bump's reaction n is +ve, and the centripetal force is -ve (c = -v^2/r per unit mass), with c = -g + n (gravity and reaction add up to provide the req'd centripetal acceleration, which is negative, i.e. downward). So, per unit mass, the reaction is \n\nn = -gravity + centripetal = g - v^2/r.\n\nIOW, at that point, the bump's curvature actually lessens the load on the wheels (it's a different story at the point of initial contact of course). \n\nThe relevant mass for the front wheels is m/2, so the numerical answer is:\n\nReaction on the *front* wheels is \n(g - v^2/r)m/2 = (9.8 - (16.6)^2/42.1)1720/2\n= 2800 (newtons) = 286 kG (kilograms-force)\n\n[An answer that forgets to halve the mass would be 5600 N.]\n\nReaction on the *rear* wheels is: gm/2 = 9.8*1720/2 = 8428 (newtons) = 860 kG (kilograms-force)\n\n\nPS: I see now that the correct sign did also occur to gp4rts in his last edit.
784822
According to Newton's first law of motion, a moving body travels along a straight path with constant speed (i.e., has constant velocity) unless it is acted on by an outside force.\n\nFor circular motion to occur there must be a constant force acting on a body, pushing it toward the centre of the circular path. This force is the centripetal (“centre-seeking”) force. The centrifugal (centre-fleeing) force will, therefore, be the opposing force which pushes away from the centre\n\nThe two forces are equal in magnitude and opposite in direction.
12014
Though I like most answers to this question, posted here, and also enjoy this source http://howthingswork.virginia.edu/print.php?title=Bicycles&startdate=0&enddate=99999999&topic=bicycles, I would actually answer this question with Newton's Three Laws of Motion (http://csep10.phys.utk.edu/astr161/lect/history/newton3laws.html), especially the Second Law of Motion: F = ma.\n\nIn this case, a few forces come into play: gravitational force, gyroscopic precessional force, as well as the force applied by the rider of the bike, to name a few.  If the net force (F) is non-zero, the bike's center of gravity would move with an acceleration (a.)\n\nWhen a bike is stationary, there is only gravitational force.  If the bike's center of gravity goes off its base of support, the bike would move toward the ground in that off direction.  Now if someone would apply a force whose vertical component cancels or is greater than the gravitational force, then the bike would not drop to the ground, for the same reason: no F, no a.\n\nWhen a bike is moving, the forces generated by the gyroscopic precession and the steering of the rider would cancel out the gravitational force, keeping the bike moving and not falling to the ground.  At a slower speed (hence a smaller gyroscopic precession force) or with a novice rider (not knowing how to steer the wheel) or a combination of both, these forces would be less than the gravitational force, and the bike would still fall.  Any kid learning how to ride a bike would tell you that :-).  Similarly if we push a bike forward, it would eventually fall, once the gravitational force would dominate and a (acceleration to the ground) would be greater than 0.