Presentation on theme: "7.2 Applications of Circular Motion"— Presentation transcript:
17.2 Applications of Circular Motion Newton's Second law - RevisitedAmusement Park Physics
2The Centripetal Force is not a new type of force. It is the NET force
3Examples of centripetal force As the moon orbits the Earth, the force of gravity acting upon the moon provides the centripetal force required for circular motion.As a car makes a turn, the force of friction acting upon the turned wheels of the car provides centripetal force required for circular motion.As a bucket of water is tied to a string and spun in a circle, the tension force acting upon the bucket provides the centripetal force required for circular motion.
4Newton's Second Law - Revisited Where Fnet is the sum (the resultant) of all forces acting on the object.Newton's second law was used in combination of circular motion equations to analyze a variety of physical situations.Note: centripetal force is the net force!
5Steps in solving problems involving forces Drawing Free-Body DiagramsDetermining the Net Force from Knowledge of Individual Force ValuesDetermining Acceleration from Knowledge of Individual Force ValuesOr Determining Individual Force Values from Knowledge of the Acceleration
6Example 1 Ff = μFNorm; FNorm = mg; Ff = μmg 3780 N = μ(9270 N); A 945-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 25.0 m. Determine the force of friction and the coefficient of friction acting upon the car.FNormGiven: m = 945 kg; v = 10.0 m/s; R = 25.0 mFind: Ffrict = ? μ = ?FfFf = Fnet = m*v2/RFf = (945 kg)*(10m/s)2/25mFf = 3780 NFgravFf = μFNorm; FNorm = mg;Ff = μmg3780 N = μ(9270 N);μ = 0.41
7Example 2The coefficient of friction acting upon a 945-kg car is The car is making a 180-degree turn around a curve with a radius of 35.0 m. Determine the maximum speed with which the car can make the turn.Given: m = 945 kg; μ = 0.85; R = 35.0 mFind: v = ? (the minimum speed would be the speed achieved with the given friction coefficient)FNormFf = FnetFfFf = μFN = μFN ;Ff = (0.85)(9270N) = 7880 NFgravFnet = m*v2/R7880 N = (945 kg)(v2) / (35.0 m); v = 17.1 m/s
8Example 3A 1.50-kg bucket of water is tied by a rope and whirled in a circle with a radius of 1.00 m. At the top of the circular loop, the speed of the bucket is 4.00 m/s. Determine the acceleration, the net force and the individual force values when the bucket is at the top of the circular loop.Fgrav = mg = 14.7 Na = v2 / R = 16 m/s2Fnet = ma = 24 N, downFnet = Fgrav + Ftens24 N = 14.7 N + FtenFtens = 24 N N = 9.3 Nm = 1.5 kga = ________ m/s/sFnet = _________ N
9Example 4A 1.50-kg bucket of water is tied by a rope and whirled in a circle with a radius of 1.00 m. At the bottom of the circular loop, the speed of the bucket is 6.00 m/s. Determine the acceleration, the net force and the individual force values when the bucket is at the bottom of the circular loop.Fgrav = m • g = 14.7 Na = v2 / R = 36 m/s2Fnet = ma = 54 N, upFnet = Ften - Fgrav54 N = Ften – 14.7 NFten = 54 N N = 68.7 Nm = 1.5 kga = ________ m/s/sFnet = _________ N
10Example 5Anna Litical is riding on The Demon at Great America. Anna experiences a downwards acceleration of 15.6 m/s2 at the top of the loop and an upwards acceleration of 26.3 m/s2 at the bottom of the loop. Use Newton's second law to determine the normal force acting upon Anna's 864 kg roller coaster car at the top and the bottom of the loop.Given:m = 864 kgatop = 15.6 m/s2 , downabottom = 26.3 m/s2 , upFgrav = m•g = 8476 N.
11Example 6Anna Litical is riding on The American Eagle at Great America. Anna is moving at 18.9 m/s over the top of a hill which has a radius of curvature of 12.7 m. Use Newton's second law to determine the magnitude of the applied force of the track pulling down upon Anna's 621 kg roller coaster car.Given:m = 621 kg; v = 18.9 m/s; R=12.7 mFgrav = m * g = 6086 Na = v2 / R = (18.9 m/s)2 / (12.7 m) = 28.1 m/s2Find:Fapp at top of hill