1. By: Morgan Henkelman Elma Sakian Lauren Cotter Stephanie Wood

2. In “linear” motion, the whole object moves at a uniform speed and distance. (In other words, they all go to the same place, within the same amount of time) However, in a circular motion, two points on the same object can be moving at different speeds and different distances.

3. Rotational motion measures how long it takes a point on an object to make a revolution. We describe rotational motion using equations that are very similar to those used in linear motion. The main difference is that they are now accompanied with the word, “angular.” Example: acceleration becomes angular acceleration, etc.

4. Angular Velocity of an object: w = ΔΘ÷ΔΤ Angular Acceleration of an object: a = Δ w ÷ΔΤ Torque: T =Fr sin Θ Newton’s Second Law: a= T net ÷ I Moment of Inertia of a point mass: I= mr ²

5. A 900 kg car makes a 180 degree turn with a speed of 10.0 m/s. The radius of the circle through which the car is turning is 25 meters. Determine the force of friction, and the coefficient of friction acting upon the car.

6. Known Information: m = 900 kg v = 10.0 m/s R = 25.0 m Requested Information:F frict = ??? mu = ???? ("mu" - coefficient of friction) The mass of the object can be used to determine the force of gravity acting in the downward direction. Use the equation Fgrav = m * g where g can be approximated as 10 m/s/s. Knowing that there is no vertical acceleration of the car, it can be concluded that the vertical forces balance each other. Thus, Fgrav = Fnorm= 9000 N. This allows us to determine two of the three forces identified in the free-body diagram. Only the friction force remains unknown Since the force of friction is the only horizontal force, it must be equal to the net force acting upon the object. So if the net force can be determined, then the friction force is known. To determine the net force, the mass and the kinematic information (speed and radius) must be substituted into the following equation: Substituting the given values yields a net force of 3600 Newtons. Thus, the force of friction is 3600 N. Finally the coefficient of friction ("mu") can be determined using the equation which relates the coefficient of friction to the force of friction and the normal force. Substituting 3600 N for F frict and 9000 N for F norm yields a coefficient of friction of 0.400.

7. Amusement park rides that spin are designed to thrill riders using the physics of rotational motion. The thrill is produced by a “force” that is present only when the ride spins.