# University Physics: Mechanics

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University Physics: Mechanics
Lecture 11 Ch6. Friction and Centripetal Force University Physics: Mechanics Dr.-Ing. Erwin Sitompul

Uniform Circular Motion
Let us recall that when a body moves in a circle (or a circular arc) with the radius R at constant speed v, it is said to be in uniform circular motion. The body has a centripetal acceleration (directed toward the center of the circle) of constant magnitude given by: (centripetal acceleration) As shown here, a hockey puck moves around in a circle at constant speed v while tied to a string looped around a central peg. The centripetal force is the radially inward pull T on the puck from the string. Without that force, the puck would slide off in a straight line instead of moving in a circle.

Uniform Circular Motion
From Newton’s second law, we can write the magnitude F of a centripetal force as: (magnitude of centripetal force) Because the speed v is constant, the magnitudes of the acceleration and the force are also constant. However, the directions of the centripetal acceleration and force are not constant; they vary continuously and always point toward the center of the circle (radially inward). The positive direction of the axis is chosen to be radially outward.

Checkpoint When you ride in a Ferris wheel at constant speed, what are the directions of your acceleration a and the normal force FN on you (from the upright seat) as you pass through the highest point of the ride? the lowest point of the ride? The highest point a downward, FN upward The lowest point a and FN upward

Example: “Allo Diavolo”
In a 1901 circus performance, “Allo Diavolo“ introduced the stunt of riding a bicycle in a loop-the-loop. Assuming that the loop is a circle with radius R = 2.7 m, what is the least speed v Diavolo could have at the top of the loop to remain in contact with it there? By the time Diavolo remain in contact with the loop, FN = 0.

Example: Riding the Rotor
A Rotor is a large, hollow cylinder that is rotated rapidly around its central axis. When the rider’s speed reaches some predetermined value, the floor abruptly falls away. The rider does not fall with it but instead is pinned to the wall while the cylinder rotates. Suppose that the coefficient of static friction μs between the rider’s clothing and the wall is 0.40 and that the cylinder’s radius R is 2.1 m.

Example: Riding the Rotor
(a) What minimum speed v must the cylinder and rider have if the rider is not to fall when the floor drops? (R = 2.1 m) Vertical calculations Radial calculations

Example: Riding the Rotor
(b) If the rider’s mass is 49 kg, what is the magnitude of the centripetal force on her? If the Rotor initially moves at the minimum required speed for the rider not to fall (7.173 m/s) and then its speed is increased step by step, what will happen with (a) the magnitude of fs; (b) the magnitude of FN; (c) the value of fs,max? Increase, decrease, or remain the same?

Negative lift exerted by passing air
Example: F1 Car A modern F1 car is designed so that the passing air pushes down on it, allowing the car to travel much faster through a flat turn in a Grand Prix without friction failing. This downward push is called negative lift or down force. The following figure represents an F1 car of mass m = 600 kg as it travels on a flat track in a circular arc of radius R = 100 m. Because of the shape of the car and the wings on it, the passing air exerts a negative lift FL downward on the car. Take μs = 0.75 and assume that the forces on the four tires are identical. Negative lift exerted by passing air

Free body diagram for the car
Example: F1 Car (a) If the car is on the verge of sliding out of the turn when its speed is 28.6 m/s, what is the magnitude of FL? Vertical calculations Free body diagram for the car Radial calculations On the verge of sliding

Example: F1 Car (b) The magnitude FL of the negative lift on the car depends on the square of the car’s speed v2. Thus, the negative lift on the car here is greater when the car travels faster, as it does on a straight section of track. What is the magnitude of the negative lift for a speed of 90 m/s? By the time the speed is 90 m/s, the negative lift is already greater than the gravitational force. Thus, the car “could have” run on a long ceiling with that velocity (324 km/h).

Example: Banked Curved Highways
Curved portions of highways are always banked to prevent cars from sliding off the highway. When the highway is wet, the frictional force maybe significantly low, and banking is then essential. The figure below represents a normal car of mass m as it moves at a constant speed v of 20 m/s around a banked circular track of radius R = 190 m. If the frictional force from the track is negligible, what bank angle θ prevents sliding?

Example: Banked Curved Highways