Exploring Uniform Circular Motion Step 1: Go to https://www.explorelearning.com Step 2: Open the activity called Uniform Circular Motion ____________________________________________________________ Prior Knowledge Questions (Do these BEFORE using the Gizmo.) 1. Circle the appropriate italicized response. Speed is a scalar / vector quantity that measures magnitude only / both magnitude and direction. Velocity is a scalar / vector quantity that measures magnitude only / both magnitude and direction. Acceleration is a scalar / vector quantity that measures how an object’s speed / velocity is changing over time. 2. A boy is whirling a yo-yo above his head in a counterclockwise direction. At the exact moment shown at left, he lets go of the string. In which direction will the yo-yo travel? Draw an arrow on the image to show the yo-yo’s direction. 3. Do you think the released yo-yo’s path will be straight or curved? Explain. _________________________________ __________________________________ Gizmo Warm-up The Uniform Circular Motion Gizmo™ shows a pink puck that is floating above a circular air table. The puck is held to the center of the table by a string so that it travels in a circle at a constant speed. Check that the radius is 8.0 m and the mass is 5.0 kg. Set the velocity to 5.0 m/s. 1. Select the BAR CHART tab and select Velocity from the dropdown menu. The three bars represent the magnitude (speed), and the x component and y component of the puck’s velocity. A. Click Play. Does the magnitude of the velocity (i.e. the speed) change over time? _______________ B. Do the x and y components of the velocity change over time? _______________ C. What part of the object’s velocity is changing, its speed or its direction? _______________ D. Can something be accelerating even if its speed is constant? How so? Velocity, acceleration, and force Get the Gizmo ready: Click Reset. Select the DESCRIPTION tab. Turn on Show velocity and acceleration vectors. Introduction: Velocity is a vector quantity that describes both the speed and direction of an object’s motion. Vectors are represented by arrows. While the speed of the puck is constant, its direction changes continually as it travels in a circle. Because its direction is changing, the puck undergoes acceleration even though its speed is constant. Question: How is the velocity of a revolving body related to its acceleration? 1. Observe: On the SIMULATION pane, observe the directions of the velocity (green) and acceleration (purple) vectors. A. What do you notice? __________________________________________________ ___________________________________________________________________ B. Click Play. What do you notice about the vectors as the puck moves in a circle? ___________________________________________________________________ 2. On the picture below showing the ball at five separate locations, draw and label the direction of the instantaneous velocity at these locations on your drawing. 1 5 2 3 4 3. As the ball revolves, what direction is the velocity vector always pointing relative to: a) The circumference of the circle? b) The radius of the circle? Based on your diagram in question 2, determine the change in velocity, Δv, between each of the four instantaneous velocity vectors. Positions: Instantaneous velocity vectors: Sketch of Change in velocity vector, Δv [hint: Δv = vn + (-vn-1)] Sketch of v1 Sketch of v2 Position #1 to position #2 Sketch of v2 Sketch of v3 Sketch of v3 Sketch of v4 Position #2 to position #3 Position #3 to position #4 Sketch of v4 Sketch of v5 Position #4 to position #5 What do you notice about the general direction of Δv? 4. On the picture below showing the ball at five separate locations, draw and label the instantaneous velocity vector, and the acceleration vector at these locations on your drawing. 1 5 2 4 3 Discovering the formula for Centripetal Acceleration Get the Gizmo ready: Click Reset. Set the radius to 2.0 m, the mass to 1.0 kg, and the velocity to 10.0 m/s. Introduction: The acceleration toward the center that keeps objects in uniform circular motion (circular motion at a constant speed) is called centripetal acceleration. An understanding of centripetal acceleration was one of the key elements that led to Newton’s formulation of the law of universal gravitation. Question: How is centripetal acceleration related to radius, mass, and velocity? 1. Record: Select the BAR CHART tab, select Acceleration from the dropdown menu, and turn on Show numerical values. The bar on the left shows the magnitude of the acceleration, or |a|. 2. Holding Speed Constant: Keeping the mass at 1.0 kg and the speed at 10.0 m/s, record the magnitude of centripetal acceleration for each given radius value. Radius: 2.0 m 4.0 m 6.0 m 8.0 m 10.0 m Acceleration (m/s2): 3. Calculate: To calculate the radius factor, divide each radius by the original radius (2.0 m). To calculate the acceleration factor, divide each acceleration value by the original acceleration (50.00 m/s2). Radius: 2.0 m Radius factor: 1 Acceleration factor: 1 4.0 m 6.0 m 8.0 m 10.0 m 4. Holding Radius Constant: Set the radius to 2.0 m and the velocity to 1.0 m/s. Keeping the radius constant, record the magnitude of centripetal acceleration for each speed indicated in the table below: Speed: 1.0 m/s 2.0 m/s 3.0 m/s 4.0 m/s 5.0 m/s Acceleration (m/s2): 5. Calculate: To calculate the speed factor, divide each speed by the original speed (1.0 m/s). To calculate the acceleration factor, divide each acceleration value by the original acceleration (0.50 m/s2). Speed: 1.0 m/s Speed factor: 1 Acceleration factor: 1 2.0 m/s 3.0 m/s 4.0 m/s 5.0 m/s 6. Make a mathematical rule: Using your above results, predict a mathematical equation for the relationship between speed (v) , magnitude of centripetal acceleration ( ac ) and radius ( r ) (i.e. find the equation). Use two of your trials data to test your result. Predicted Formula for Centripetal Acceleration Test using two trials r(m) ac = After testing your formula, write your final version of your formula. Final Version of Formula for Centripetal Acceleration ac = v (m/s) Calculated acceleration, from formula (m/s2) Actual value for acceleration, from data (m/s2) Extension: Equation for Centripetal Force Newton’s first law states that an object will continue at the same velocity (speed and direction) unless acted upon by an unbalanced force. Click Pause when the puck is in approximately the position shown at right. Imagine at this moment the string connecting the puck to the center is cut. A. Is any force acting on the puck now? ______________ B. Draw an arrow on the image to represent the direction of the puck’s motion. C. Would the path of the puck be straight or curved? ___________________________ If you are sitting in the back seat of a car that makes a hard left turn, you will feel pushed toward the right side of the car. Why does your body move to the right? _________________________________________________________________________ _________________________________________________________________________ _________________________________________________________________________ _________________________________________________________________________ Newton’s second law states that a force will cause objects to accelerate in the direction of the force. A. Given the fact that the puck is accelerating, what can you conclude? ___________________________________________________________________ B. What is the direction of the force on the puck as it travels in a circle? ___________________________________________________________________ Newton’s second law states that net force is equal to the product of mass and acceleration: Σ� = m�. Based on Newton’s second law, what is the formula for the net centripetal force on an object in uniform circular motion with mass m and orbital speed v? Formula for Centripetal Force: ΣFc = Formula for Centripetal Acceleration: ac = CIRCULAR MOTION SUMMARY & INTRODUCTORY PROBLEMS All objects that follow a curved path must have force acting towards the centre of that curve. We call this force the centripetal force. (Greek: “Centre-seeking”). Some examples of circular motion and the associated centripetal forces are shown below: Example Force Responsible Circular planetary orbits Gravity Electron orbits Electrostatic force on electron Car cornering Static friction between road and tires Car cornering on a banked turn Component of normal force Aircraft banking Horizontal component of lift on the wings Diagram Centripetal Acceleration Examples: Centripetal Acceleration Example #1 What is the centripetal acceleration of a point on the perimeter of a bicycle wheel of diameter 70. cm when a bike is moving at 8.0 m/s? Sketch: Centripetal Acceleration Example #2 A race car makes one lap around a track of radius 50. m in 9.0 s. Sketch: a) What is the speed? b) What was the car’s centripetal acceleration? Centripetal Acceleration Example #3 Norman swings a rubber ball attached to a string over his head in a horizontal, circular path. The piece of string is 1.5 m long and the ball makes 120 complete rotations each minute. Sketch: a) What is the speed of the ball? b) What is the ball’s centripetal acceleration? Centripetal Force Problems: 1) Calculate the centripetal force acting on a 925 kg car as it rounds an unbanked curve with a radius of 75 m at a speed of 22 m/s. (6.0x103 N) Diagram: Force(s) acting to hold object into circular motion: Reasoning - How the force(s) contribute to the net centripetal force: Solution: FBD of Car: 2) An electron (m = 9.11x10–31 kg) moves in a circle whose radius is 2.00 x 10–2 m. If the force acting on the electron is 4.60x10–14 N, what is its speed? (3.18x107 m/s) Diagram: Force(s) acting to hold object into circular motion: Reasoning - How the force(s) contribute to the net centripetal force: Solution: FBD: EXAMPLES OF VERTICAL CIRCULAR MOTION Example #1: Roller Coaster Consider a roller coaster car passing through a vertical loop. On the diagram below, draw the force vectors for the normal force, and the weight, at both the top and bottom positions of the loop. Top of Circle Select the direction pointing toward the centre of the circle as positive. Force Name Radial component Net force, ΣFc Now, write the centripetal force statement for the top of the circle: F c = ( ______ + _________) = 2 mv r Now, rearrange your equation above for the normal force: Bottom of Circle Select the direction pointing toward the centre of the circle as positive. Force Name Radial component (Towards the centre of the circle) Net force, ΣFc Write the centripetal force statement for the bottom of the circle: Now, rearrange your equation above for the normal force: Compare your two equations for the normal force. Use it to explain where you will feel lighter, and where you will feel heavier: Consider the mechanical energy of the coaster as it travels along the loop. Assuming a frictionless surface, use conservation of mechanical energy to show that the coaster’s speed is NOT uniform as it moves from the bottom of the loop to the top: Kinetic Energy Expression: Gravitational Potential Energy Expression: Mechanical Energy Expression: Bottom of Loop Top of Loop Based on the analysis of forces & speeds of the coaster at the top & bottom of the loop: A) where is the coaster at the GREATEST risk of falling? Why? B) where is the coaster at the LEAST risk of falling? Why? C) In the WORST-CASE SCENARIO, if the coaster isn’t travelling FAST ENOUGH, the coaster will briefly lose contact with the track’s surface at the top of the loop. What will happen to the NORMAL force from the track acting on the coaster at this point? Modify the centripetal force equations from the previous page to show the relationship between the force(s) acting on the coaster if this happens, and the minimum speed the coaster must have to maintain circular motion: D) Explain, using Newton’s 1st Law, how the coaster can continue to complete the loop despite briefly losing contact with the track at the top, as long as it’s traveling at the minimum speed specified in part above: Example #2: Driving along a hill Consider a car traveling along a vertical semi-circular hill. On the diagram below, draw the force vectors for the normal force and the weight at the top of the hill. Top of Hill Select the direction pointing toward the centre of the circle as positive. Force Name Radial component (Towards the centre of the circle) Net force, ΣFc Now, write the centripetal force statement for the top of the circle: F c = ( ______ + _________) = 2 mv r A) If the car travels too fast, what can happen to it at the top of the hill? Explain why, using circular motion concepts. B) In the WORST-CASE SCENARIO, if the car is travelling TOO FAST, the car will briefly lose contact with the road’s surface at the top of the hill. What will happen to the NORMAL force from the road acting on the car at this point? Modify the centripetal force equations from the previous page to show the relationship between the force(s) acting on the coaster if this happens, and the maximum speed the car can have to maintain circular motion (instead of projectile motion): C) Explain, using Newton’s 1st Law, how the car can continue to complete the hill despite briefly losing contact with the road at the top of the hill, as long as it’s traveling no faster than the maximum speed specified in part B above: More Vertical Loop Practice 1) You are riding your bike on a track that forms a vertical circular loop. If the diameter of the loop is 10.0 m, what is the minimum speed required for you to make it around the loop? (7.00 m/s) Diagram of Situation: FBD at point of MINIMUM force: Forces providing ΣFc: , Solution: Select the direction pointing toward the centre of the circle as positive. Force Name Net force, ΣFc Radial component (Towards the centre of the circle) 3) You are swinging a 2.5-kg bucket of water in a vertical circle. Assuming that the radius of the rotation of the water is 0.95 m, and the maximum force of tension that the string can provide before breaking is 100N: A) What is the maximum speed the bucket can have without breaking the string? Diagram: FBD at point of MAXIMUM tension: Forces providing ΣFc: , Solution: Select the direction pointing toward the centre of the circle as positive. Force Name Net force, ΣFc Radial component (Towards the centre of the circle) B) Assume the bucket is travelling at precisely the maximum speed calculated in part A) when it’s at the bottom of its arc. Using conservation of mechanical energy (assuming no friction/air resistance), determine the speed the bucket will have at the top of the arc. Use the chart below to help: Kinetic Energy Expression: Gravitational Potential Energy Expression: Mechanical Energy Expression: Bottom of Arc Top of Arc C) Determine, using centripetal force concepts, if the speed of the bucket at the top of the arc found in part B) is sufficient for it to complete its circular orbit: FBD at top of loop: Forces providing ΣFc: , Solution: Select the direction pointing toward the centre of the circle as positive. Force Name Radial component (Towards the centre of the circle) L.S. R.S. Check: L.S. = ΣFc (from component chart) = R.S. mv 2 r (formula for Net force, ΣFc ΣFc) Friction and Uniform Circular Motion A Car rounding a bend It turns out that understanding circular motion is useful for: Let’s consider a car taking a curve, by now it should be clear there must be a ___________________ present to keep the car on the curve or, more precisely, in uniform circular motion. For a flat road, this force actually comes from: Don’t be misled by the outward force against the door you feel as a passenger: Skidding The limit of steering in a curve occurs when the centripetal acceleration equals the maximum static friction. Ex. A curve on a dry road (µs = 1.0) is safe at a speed of 90 km/h. What is the safe speed on the same curve with ice (µs = 0.2)? General expression for radius of curvature: Radius of curve: Maximum safe speed for icy pavement: Example: Negotiating a Curve A 1000 kg car rounds a curve on a flat road of radius 50 m at a speed of 50 km/hr (14 m/s). Set-up the component chart and develop a general equation relating friction force, mass, and speed: Vector Y-component Radial component (Towards the centre) b) Will the car make the turn, or will it skid, if: (i) (ii) the pavement is dry and the coefficient of static friction is s = 0.60? the pavement is icy and s = 0.25? Level Curve – Final Example A 1500 kg car moving on a flat, horizontal road negotiates a curve as shown. The radius of the curve is 35.0 m and the coefficient of static friction between the tires and dry pavement is 0.523. Find the maximum speed the car can have and still make the turn successfully. Banked Curves On a frictionless banked curve, the centripetal force is the ______________________ of the normal force. The _______________________ of the normal force balances the car’s ____________________. The Theory of Banked Curves Race cars (and street cars for that matter) require some help negotiating curves. By banking a curve, the car’s own ________________, through a component of the _____________, can be used to provide the centripetal force needed to stay on the road. In fact for a given angle there is a _________________ speed for which no _______ is required at all. Note that we’ve picked an unusual _____________________ system. Not down the inclined plane, but aligned with the _________________ direction. That’s because we want to determine the _______________ of any force or forces that may act as a __________________________. Important Points to Remember: When solving a problem in which an object is accelerating ALONG an incline, DO tilt the Cartesian plane to reduce the number of force components. When solving a problem in which an object is REVOLVING AROUND a banked curve, DON’T tilt the Cartesian plane, since the x-axis should point to the center of the circle around which the car/object is turning. Example: Banking Angle For a car traveling with a speed, v, around a curve of radius, r. a. Derive the general equation for the angle at which a road should be banked so that no friction is required to keep the car on the road. Vector b. Y-component Radial component (Towards the centre) What is this angle for a curve of radius 50 m at a designed speed of 50 km/h? Summary: Determining Banking Angle