Circular Motion. - The velocity is tangent to the path and perpendicular to the radius of the circle

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1 Circular Motion Level : Physics Teacher : Kim 1. Uniform Circular Motion - According to Newton s 1 st law, an object in motion will move in a straight line at a constant speed unless an unbalance force acts upon it - In other words, to change the speed of an object we need to apply a force. But a force needs to be applied to change the direction of the motion. - If there is an unbalance force, continuously acting toward a common center, then the motion of that object is called circular motion. If the speed is constant, then it is called uniform circular motion - The velocity is tangent to the path and perpendicular to the radius of the circle If a ball is tied to a string and spun over your head, the ball is being pulled toward the center continuously, resulting in a circular motion. direction of the force direction of the speed - In uniform circular motion, though the speed is constant throughout the circular motion, the direction is constantly changing. The acceleration of this type is called centripetal acceleration 2. Newton s Second Law Applied to Uniform Circular Motion - Some examples of centripetal forces *~ Centripetal force is not another type of force. Any inward unbalance force that contributes to circular motion becomes a centripetal force~* i) Tension force : if a ball tied to a string is spun in a circle, the string is continuously pulling the ball inward, forcing the ball to move in a circular motion ii) Gravitional force : our moon is constantly pulled towards the Earth, forcing the moon to move in a circular motion iii) Static frictional force : an object place on the surface of a spinning desk is forced to move in circular motion due to frictional force between the object and the surface of the disk iv) Normal force: if an object is placed in a empty bucket and spun, the object is continuously pushed inwards by the inner surface(the bottom of the bucket), forcing the object to move in a circular motion

2 *~Observe video Rotor in an Amusement Park~* *~Observe video Cars driving on Banked wall~* 3. Centripetal Acceleration - The acceleration in uniform circular motion is always perpendicular to the path and always points toward the center of the circle. - The acceleration of a uniform circular motion is called a centripetal acceleration, and its magnitude is a r = - If we apply Newton s 2 nd law along the radial direction(inward direction), we find that the value of the net force causing the centripetal acceleration can be expressed as F r = ma r = Q1) A 4kg mass on the end of a string rotates in a circular motion on a horizontal frictionless table. The mass has a constant speed of 2m/s and the radius of the circle is 0.8m. ii) Find the magnitude of the centripetal force. a) 10N b) 12N c) 15N d) 20N Q2) A highway curve has a radius of 140m and is unbanked. A car of mass kg goes around the curve at a speed of 24m/s without slipping. Top View v ii)what is the magnitude of the centripetal force of the road on the car? a) 5038N b) 4035N c) N d) Rear View!! F c F c

3 f =µf N F r = ma r = Q3) A 0.1kg weight is placed 0.3m from the center of a rotating, horizontal turntable. It slips when its speed is 0.5m/s. ii) Find the magnitude of that force? a) 0.054N b) 0.083N c) 0.102N d) 0.156N Top View Side View turntable! weight! weight! turntable! iii) What is the coefficient of static friction between the coin and turntable? a) b) c) d) Engineering a highway curve. see p.138 in the textbook!! **Q4) If a car goes through a curve too fast, the car tends to slide out of the curve. Especially when the road is wet or icy, the friction between the tires and road cannot be reliable. To prevent this, highways are banked when the roads are curved. If a car is expected to move around a curve of radius 200m at 25m/s, what should be the value of the banking angle if no dependence is to be placed on friction? a) 12.1º b) 17.7º c) 25.2º d) 35.3º F r i) Draw free-body diagram to represent the forces. Remember that the x-axis must be parallel to the plane of the motion! ii) Divide any forces into components if necessary iii)write the equations for X&Y components and solve for θ hint : tanθ=sinθ / cosθ, and use inverse tangent to get the angle (tan -1 ).

4 f =µf N F r = ma r = Q5) A small ball is fastened to a string L=0.24m long and suspended from a fixed point P to make a conical pendulum. The ball describes a horizontal circle about a center and the string makes an angle of 15 with the vertical. Find the speed of the ball. See solution in the next page! a) 0.404m/s b) 0.612m/s c) 0.943m/s d) 1.21m/s θ Q6) At an amusement park there is a ride in which cylindrical shaped chambers spin around a central axis. People are standing on the floor facing the axis, their backs against the wall. When the chamber reaches the speed of 3.2m/s, the floor opens but the people stay on the wall. An 83kg person feels a 560N force pressing against his back. i) What is the radius of a chamber? See solution in the next page! a) 1.5m b) 2.2m c) 3.4m d) 4.1m ii) What is the coefficient of static friction? a) 0.32 b) 0.67 c) 1.10 d) 1.45

5 sol for Q5) T θ Y Tcosθ The x-axis must be parallel to the plane of the motion According to our x-y axis, tension for T needs to be divided into components. Using Newton's 2nd law F r =Tsinθ = F y =Tcosθ =mg Tsinθ X If we divide the two equations above, then 'T' and 'm' will cancel and sinθ/cosθ=tanθ. So, mg tanθ = To find the speed v, we can v = However, the radius of the circle 'r' is not 0.24m. The length of the string is 0.24m. That is, L=0.24m. L and r has the relationship r=lsinθ. So v = = = 0.404m/s sol for Q6) Y The x-axis must be parallel to the plane of the motion Using Newton's 2nd law F r =F N = (1) f s F y = f s =mg (2) i) Finding the radius of the chamber F N X From (1), F N =, we can rewrite as r = = =1.5m mg ii) Finding the coefficient of static friction µ s From (2), f s =mg. And since f s = µ s F N µ s F N = mg So µ s = = =1.45