Circular Motion Part 2 By: Heather Britton

Circular Motion Part 2 According to Newton’s 2nd Law, an accelerating body must have a force acting on it F = ma

Circular Motion Part 2 Substituting for a we get ΣF = (mv 2 )/r The net force is directed towards the center just as a goes towards the center

Circular Motion Part 2 This force is called centripetal (center seeking) force It is measured in Newtons like all forces It must also be applied by another object

Circular Motion Part 2 If the plane of motion is horizontal the force is uniform If the plane of motion is at all vertical the force is not uniform due to gravity There is more force at the bottom There is less force at the top

Circular Motion Part 2 Example 3 A model airplane (m = 0.9 kg) is swung in a circle by a guide wire and moves with a constant speed parallel to the ground. Find the tension in the guide wire (length = 17 m) for speeds of 19 m/s and 38 m/s.

Circular Motion Part 2 When a car travels around a corner it experiences centripetal acceleration in the direction toward the center of the turn The tires static friction provide the force There is a maximum speed for each curve that is independent of mass That is why there are posted speed signs

Circular Motion Part 2 Example 4 Compare the maximum speeds at which a car can safely negotiate an unbanked turn (radius 50.0 m) in dry weather (μ = 0.900) and icy weather (μ = 0.100).

Circular Motion Part 2 Turns can be negotiated at a faster speed if the turn is banked at an angle Like before we can break down forces at an angle into components

Circular Motion Part 2 Now we will break down the normal force into components to show that part of the normal force points to the center thus increasing the centripetal force

Circular Motion Part 2 Looking at these components we can derive an equation for a banked curve relating angle, velocity, and radius. The centripetal force is the sin component of the normal force tan θ = (v 2 ) / (rg)

Circular Motion Part 2 Example 5 The oval track at the Daytona 500 has a maximum radius of 316 m and are banked at a 31°. If the track were frictionless what is the maximum velocity the cars can negotiate the turn with?