Advanced Circular Motion

Advanced Circular Motion UNDERSTANDING CENTRIPETAL ACCELERATION AND FORCE ON FLAT SURFACES, AND INCLINED SURFACES

Circular Motion on Flat Plane On a flat, level curve, the friction between the tires and the road supplies the centripetal force. If the tires are worn smooth or the road is icy or oily, this friction force may not be available. The car might not be able to move in a circle, and will keep going in a straight line off the road.

Flat Plane Curve cont., The force due to gravity is balanced by the support force of the road while the frictional force between the tires and the surface is unbalanced. Unbalanced forces lead to acceleration in the direction of the net force. ??= ????????? ??2 ? = ??? Or, ?2 ?= ??

Motion On A Banked Curve Some curves are banked to compensate for slippery conditions. In addition to any friction forces that may or may not be present, the road exerts a normal force perpendicular to its surface. The downward force of the car s weight is also present. These two forces are no longer balanced and influence the overall net force (????) directed toward the center of the circle. The centripetal force is directed toward the center of the circle, not parallel to the banked road.

Banked Curve cont., The effect of banking is to tilt the normal force (??) toward the center of curvature of the road so that the inward radial component (??sin?) can supply the required centripetal force. Vehicles can make a sharp turn more safely if the road is banked. If the vehicle maintains the speed for which the curve is designed, no frictional force is needed to keep the vehicle on the road.

Banked Curve cont., There is no acceleration along the y axis, so the sum of the forces in the y plane is zero. ??cos? ?? = 0 Therefore, ?? cos ? ??= The horizontal component of the normal force (??) is provided by the force the road exerts against the car. The horizontal component of the normal force is unbalanced and results in the centripetal acceleration directed towards the center of the turn.

Banked Curve cont., Engineers can utilize the centripetal force equation to determine the banking angle that allows a car to travel in a curve of radius ? with constant speed ? and require no friction force. ??= ??sin? Or, ??2 ? = ??sin? ?? cos ?for the normal force (??) Substitute ??2 ? ?? cos ? sin? =

Banked Curve cont., Sine dividing by cosine is also known as the tangent ??2 ? = ?? tan? Solving for the angle (?) ?2 ?? ? = tan 1

Banked Curve cont., A banked curve is designed for one specific speed, called the design speed . If the banked curve is icy so that there is no friction force at all, then traveling at a speed higher than the design speed means the car will slide out, up, and over the edge. Traveling at a lower speed than the design speed means the car will slide in, down, and off the bank. When a banked curve has to be negotiated at a speed above the design speed (or if you are asked to find the maximum speed), friction is required. The frictional force (?????????) acts parallel to the road surface.

Banked Curve With Friction Resolve the normal force FNand friction FFinto horizontal and vertical components. ??= ??cos? ?????????sin? ??= ??sin? + ?????????cos? The horizontal components of ??and ?????????are both directed inward toward the center of the curve, therefore, these two force components combine to determine the centripetal force ??2 ? = ??sin? + ?????????cos?

Banked Curve With Friction cont., Because there are no unbalanced forces in the vertical direction, the upward forces must equal the downward forces, therefore: ??cos? = ?????????sin? + ?? Substitute ?????????= ??? into both equations. ??2 ? = ??sin? + ???cos? ??cos? = ???sin? + ??

Banked Curve With Friction cont., Solve both equations for ??. =??2 ??sin? + ?cos? ? ??cos? ?sin? = ?? and set these equations equal to each other (because there is only one force acting normal to the surface). ??2 ?? ??= = ? sin ?+ ? cos ? cos ? ? sin ? From this equation the unknown variable for mass, velocity, and radius of turn can be determined.

Banked Curve With Friction For problems involving a minimum speed for the vehicle to travel around the curve without skidding, the frictional force is directed up the incline to keep the vehicle from sliding down to the bottom of the banked curve. Resolve the normal force FNand friction FFinto horizontal and vertical components. ??= ??cos? + ?????????sin? ??= ??sin? ?????????cos?

Banked Curve With Friction The horizontal component of ??is inward toward the center of the curve and is positive; ?????????is directed outward away from the center of the curve and is negative. These two force components combine to determine the centripetal force ??2 ? = ??sin? ?????????cos?

Banked Curve With Friction Because there are no unbalanced forces in the vertical direction, the upward forces must equal the downward forces, therefore: ??cos? + ?????????sin? = ?? Substitute ?????????= ??? into both equations. ??2 ? = ??sin? ???cos? ??cos? + ???sin? = ??

Banked Curve With Friction cont., Solve both equations for ??. =??2 ??sin? ?cos? ? ??cos? + ?sin? = ?? and set these equations equal to each other (because there is only one force acting normal to the surface). ??2 ?? ??= = ? sin ? ? cos ? cos ?+ ? sin ? From this equation the unknown variable for mass, velocity, and radius of turn can be determined.

Normal Force Difference In an inclined plane problem such as a skier on a hill or a block on a ramp, the normal force is given by ??= ??cos? In every banked curve problem the normal force is given by ?? cos ? ??= Both equations are correct!

??Difference cont., The difference is due to the direction of the acceleration. The direction of the acceleration in the inclined plane problem is down toward the bottom of the incline. There is no component of the acceleration in the normal direction. the direction of the acceleration in the banked curve is horizontally toward the center of the circle. This produces a component of acceleration in the normal direction. In the inclined plane problems, I taught you to rotate the x- and y- axes so that the x-axis is parallel to the surface and the y-axis is perpendicular to the surface. We did not do this with the banked turn problems. This results in the acceleration being zero on the y axis for both cases. In the inclined plane problems, the acceleration is zero along the normal perpendicular to the incline; thus the normal force equals a component of gravity.)

??Difference cont., In the frictionless banked turn, the acceleration is zero vertically; thus the force of gravity equals a component of the normal force. In the inclined plane problems, one component of gravity causes the acceleration (the other component cancels out the normal force). In the banked turn, one component of the normal force causes the acceleration (the other component cancels out the force of gravity).

Uniform Circular Motion Velocity ? =? ? n = number of revolutions (rotations) ?=?2?? r = radius t = time Period = time for one revolution (rotation)