Ryan Hyon Chapter 6: Circular Motion and Other Application of Newton’s Laws Background: This chapter covers the application of Newton’s laws to uniform circular motion. • Uniform circular motion o When an object’s speed around a circular path remains constant o Though speed may be constant, velocity is not This is an example of vertical uniform circular motion. Roller-­‐
coasters are typical examples in vertical uniform circular motion AP problems. constant because a circular path implies that the direction of the velocity is always changing o Since the velocity is always changing, this means that there must be acceleration – this acceleration does not change the speed of the object but rather the direction. •
Applying Newton’s Laws to Uniform Circular Motion o Fc = ma o
Fc =
§
This is an example of horizontal uniform circular motion. This classic example – a ball tied to a string – is frequently used as an AP problem. mv 2
r
Setting these two centripetal force equations equal to each other reveals that centripetal acceleration is equal to the velocity of the object squared divided by the radius of the circular path. Easy problem An object of mass 5 kg moves at a constant speed of 6 m/s in a circular path of radius 2 m. Find the object’s acceleration and the net force responsible for its motion. 2 1 Solution: By definition, an object moving at constant speed in circular path is undergoing uniform circular motion. Therefore, it experiences a centripetal acceleration of magnitude v2/r, always directed towards the center of the circle: v 2 (6m/s)2
ac = =
= 18m/s2 r
2m
Newton’s Second Law, coupled with the equation for centripetal acceleration, gives the net force that produces the centripetal acceleration: Fc = mac The equation gives the magnitude of the force. As for direction, recall that because F = ma, the directions of F and a are always the same. Since centripetal acceleration points towards the center of the path, so does the force that produces it. Therefore, it’s called centripetal force. Centripetal force if provided by everyday forces such as tension, friction, gravity, or normal forces. Fc = (5kg)(18m/s2 ) = 90N The centripetal force acting on this object has a magnitude of 90N. v
v
Medium problem A 10 kg mass is attached to a string that has a breaking strength of 200 N. If the mass is whirled in a horizontal circle of radius 80 cm, what maximum speed can it have? Assume the string is horizontal. Solution: The first thing to do in problems like this is to identify what force(s) provide the centripetal force. In this example, the tension in the string provides the centripetal force: FT provides Fc ,
mv 2
FT =
r
rFT
v=
m
vmax =
rFT max
m
vmax =
(0.80m)(200N)
10kg
vmax = 4m/s
The maximum speed it can have is 4 meters per second. Hard problem A roller-­‐coaster car enters the circular-­‐loop portion of the ride. At the very top of the circle (when the people in the car are upside down), the speed of the car is 25 m/s, and the acceleration points straight down. If the diameter of the loop is 50 m and the total mass of the car (plus passengers) is 1200 kg, find the magnitude of the normal force exerted by the track on the car at this point. Solution: When analyzing circular motion, consider all forces pointing towards the center to be positive and all forces pointing away to be negative. There are two forces acting on the car at its topmost point: the normal force exerted by the track and the gravitational force, both of which point downward. The combination of these two forces, FN + Fg, provides the centripetal force: mv 2
FN + Fg =
r
2
mv
FN =
− Fg
r
mv 2
FN =
− mg
r
v2
FN = m( − g)
r
"
%
(25m/s)2
FN = (1200kg)$
−10m/s'
$# 1 (50m)
'&
2
FN = 1.8x10 4 N