# Uniform Circular Motion

## Presentation on theme: "Uniform Circular Motion"— Presentation transcript:

Uniform Circular Motion

Circular Motion ACT 1 B C A v Use this to motivate circular motion involves acceleration. Answer: B A ball is going around in a circle attached to a string. If the string breaks at the instant shown, which path will the ball follow?

Circular Motion

Uniform Circular Motion
For a particle in uniform circular motion, the velocity vector v remains constant in magnitude, but it continuously changes its direction.

Acceleration in Uniform Circular Motion
v2 v1 R R v1 v2 v centripetal acceleration aave= Dv / Dt Acceleration toward center Acceleration is due to change in direction, not speed. Object turns “toward” center  must be a force on object toward center.

Centripetal Acceleration

Centripetal Acceleration ACT
Which motion has the largest centripetal acceleration?

Preflight “Gravity, Normal Force, Friction”
Consider the following situation: You are driving a car with constant speed around a horizontal circular track. On a piece of paper, draw a Free Body Diagram (FBD) for the car. How many forces are acting on the car? A) B) C) D) 4 W FN SF = ma = mv2/R a=v2/R R 0% 28% 44% f correct “Gravity, Normal Force, Friction” “Fn = Normal Force, W = Weight, the force of gravity, f = centripetal force.” 16

Common Incorrect Responses
Acceleration: SF = ma Centripetal Acceleration Force of Motion (Inertia not a force) Forward Force, Force of velocity Speed Centrifugal Force (No such thing!) Centripetal (really acceleration) Inward force (really friction) Internal Forces (don’t count, cancel) Car Engine

Circular Motion Requires Net Force
Net force may be provided by the tension in a string, the normal force, or friction, among other sources – as with any net force.

Preflights Consider the following situation: You are driving a car with constant speed around a horizontal circular track. On a piece of paper, draw a Free Body Diagram (FBD) for the car. The net force on the car is W FN A. Zero B. Pointing radially inward C. Pointing radially outward SF = ma = mv2/R a=v2/R R f 22% 67%11% correct “This is why many racetracks have banked roadways” “Centripetal force is always pointing inward" “Ever spin stuff in a bowl? It gravitates outward.” 16

Dip ACT Suppose you are driving through a valley whose bottom has a circular shape. If your mass is m, what is the magnitude of the normal force FN exerted on you by the car seat as you drive past the bottom of the hill A. FN < mg B. FN = mg C. FN > mg a=v2/R R correct FN v SF = ma FN - mg = mv2/R FN = mg + mv2/R mg 20

Dip Example While driving on a country road at a constant speed of 17.0 m/s, you encounter a dip in the road. The dip can be approximated by a circular arc with a radius of 65.0 m. What is the normal force exerted by the car seat on an 80.0 kg passenger at the bottom of the dip?

Merry-Go-Round ACT Bonnie sits on the outer rim of a merry-go-round with radius 3 meters, and Klyde sits midway between the center and the rim. The merry-go-round makes one complete revolution every two seconds. Klyde’s speed is: Klyde Bonnie (a) the same as Bonnie’s (b) twice Bonnie’s (c) half Bonnie’s Bonnie travels 2 p R in 2 seconds vB = 2 p R / 2 = 9.42 m/s Klyde travels 2 p (R/2) in 2 seconds vK = 2 p (R/2) / 2 = 4.71 m/s 33

Rounding a Corner A 1,200 kg car rounds a corner of radius r = 45.0 m. If the coefficient of friction between the tires and the road is ms = 0.82, what is the maximum speed the car can have on the curve without skidding? Question: How does this result depend on the mass of the car?

Banked Curves

Revolving in a Circle Vertical: Radial:
An energetic father places his 20 kg child in a 5.0 kg cart to which is attached a 2.0 m long rope. He then holds the end of the rope and spins the cart and child in a circle, keeping the rope parallel to the ground. If the tension in the rope is 100 N, what is the cart’s tangential speed? Vertical: m = 25 kg r = 2 m T = 100 N v = ? Radial:

50 •• Driving in your car with a constant speed of 12 m/s, you encounter a bump in the road that has a circular cross-section, as indicated in Figure 6–30. If the radius of curvature of the bump is 35 m, find the apparent weight of a 67-kg person in your car as you pass over the top of the bump. w N

51•• Referring to Problem 50, at what speed must you go over the bump if people in your car are to feel “weightless?”

Centrifuge A centrifuge rotates at a rate such that the bottom of a test tube travels at a speed of 89.3 m/s. The bottom of the test tube is 8.50 cm from the axis of rotation. What is the centripetal acceleration acp at the bottom of the test tube in m/s and in g (where 1 g = 9.81 m/s2)?