Physics 12 Unit 5 Circular Motion and Gravitation Part 1
1. Nonlinear motions According to the Newton s first law, an object remains its tendency of motion as long as there is no external force acting on it (i.e., the inertia of an object). When an unbalanced force is applied to it, then its inertia will be changed. Depending on the direction in which the force is applied, the object might be speeded up or slowed down. The amount of change is determined by the Newton s second law: F net = ma Most of our previous discussions focussed on only the linear motion in which the applied force and the motion of an object are aligned. Unit 5 - Circular Motion and Gravitation (Part 1) 2
Indeed we have seen examples where the motion of an object is deflected in the presence of an applied force. For instance, in an inelastic, oblique collision, two masses which are initially approaching one another are bounced off in the directions different from their original paths. (Refer to Unit 4!) Unit 5 - Circular Motion and Gravitation (Part 1) 3
Projectile motion is another type of examples in which objects move on a curved trajectory. It is obvious, therefore, that an object, under the influence of an applied force, could travel along a nonlinear path. Unit 5 - Circular Motion and Gravitation (Part 1) 4
2. Linear and angular speed When a force is applied in a proper direction, an object is able to move in a circular path about a point (i.e., the center of the circular path). In this situation, the object is said to be undergoing a circular motion. There are two speeds associated with any object moving in a circular path: (1) Linear speed (2) Angular speed What is the difference between them, and how are they related to one another? Unit 5 - Circular Motion and Gravitation (Part 1) 5
Consider the following diagram: Assume an object moves counter-clockwise at a constant speed v. If the distance it travels in time t is s, then we can define the linear speed as: v = s t Unit 5 - Circular Motion and Gravitation (Part 1) 6
In the meantime, the object has moved along the path that constitutes an angle θ of a complete circle. Hence, we can define another quantity, angular speed, as: ω = θ t These two speeds are related to one another due to the fact that θ 2π = This implies s = rθ. Note that here θ is an angle in radians. s 2πr Unit 5 - Circular Motion and Gravitation (Part 1) 7
Substituting this expression into the linear speed yields v = s t = rθ t = rω Hence, the linear speed is the angular speed times the radius of the circular path. Unit 5 - Circular Motion and Gravitation (Part 1) 8
Example: Find the speed of a particle which is moving on a circular path of radius 0.30 m with an angular speed of 4 rad/s. Solution: Using v = rω, we have v = 0.30 4 = 1.2 m/s Example: A particle moving on a circular path of radius 4 m has a speed of 36 m/s. Find its angular speed. Solution: Using ω = v/r, we have ω = 36 4 = 9 rad/s Unit 5 - Circular Motion and Gravitation (Part 1) 9
3. Motion on a circular path In order for an object to move on a circular path, an unbalanced force must be applied. This force is called the centripetal force, F c, which points towards the center of the path. This force gives rise to the centripetal acceleration, a c, which also points towards the center. Note that both the centripetal force and centripetal acceleration are perpendicular to the velocity, or the motion, of the object. Mathematically, a c = v2 r F c = ma c = mv2 r Unit 5 - Circular Motion and Gravitation (Part 1) 10
Consider an object that moves from A to B in a time interval t. Recall that velocity is a vector. Hence, by definition the acceleration is: a = v t = v 2 v 1 t The change of velocity, v, is not the numerical difference of v 1 and v 2, but their vector difference. F D E Unit 5 - Circular Motion and Gravitation (Part 1) 11
Note that the triangles ABC and DEF are similar (you can try to prove it); therefore, v v = AB r When θ is small, AB l. Hence, v Here we drop the absolute signs for simplicity: v = l r v = v r l Unit 5 - Circular Motion and Gravitation (Part 1) 12
Dividing both sides by t gives v t = v r l t Note that a c = v/ t and v = l/ t. Thus, a c = v r v = v2 r This acceleration is pointing towards the center of the circular path, and is therefore called the radial acceleration or centripetal acceleration. Unit 5 - Circular Motion and Gravitation (Part 1) 13
Usually circular motions are described in terms of frequency f, which is the number of revolutions per second. A related quantity, called period T, is also used commonly. Period refers to the time taken for an object to complete one revolution. Symbolically, T = 1 f Using these, the velocity and centripetal acceleration of an object undergoing a circular motion can be rewritten as: v = 2πr T a c = 4π2 r T 2 Unit 5 - Circular Motion and Gravitation (Part 1) 14
Example: A 150-g ball at the end of a string is revolving uniformly in a horizontal circle of radius 0.600 m. The ball makes 2.00 revolutions in a second. What is its centripetal acceleration? Solution: The linear speed of the ball is v = 2πr 2π 0.600 = = 7.54 m/s T 0.500 Therefore, the centripetal acceleration is a c = v2 r = 7.54 2 0.600 = 94.7 m/s2 Unit 5 - Circular Motion and Gravitation (Part 1) 15
Example: The Moon s nearly circular orbit around the Earth has a radius of about 384,000 km and a period of 27.3 days. Determine the acceleration of the Moon toward the Earth in terms of g. [2.77 10-4 g] Unit 5 - Circular Motion and Gravitation (Part 1) 16
In Physics 12, we will look at several typical examples of circular motion. Depending on the degree of complication, we classify them into the following categories: 1. Horizontal circular motion (e.g. ball-and-string) 2. Vehicles rounding a sharp turn 3. Conical pendulum 4. Vertical circular motion (e.g. Ferris wheel, roller coaster) 5. Banked curves (e.g. car racing) Unit 5 - Circular Motion and Gravitation (Part 1) 17
3. Horizontal circular motion This is the most simple type of circular motion because only the centripetal force comes entirely from the string connected to the object. Gravity, on the other hand, can be ignored as it always remains perpendicular to the plane of the circular motion, and therefore has no contribution to the resulting centripetal force. Unit 5 - Circular Motion and Gravitation (Part 1) 18
The free body diagram for such a system is very simple. Mathematically: F c = T Unit 5 - Circular Motion and Gravitation (Part 1) 19
Example: Estimate the force a person must exert on a string attached to a 0.20 kg ball to make the ball revolve in a horizontal circle of radius 0.800 m. The ball makes 4 revolutions per second. Ignore the string s mass. Solution: The centripetal force is F c = mv2 r = m 4π2 r T 2 = 0.20 4π2 0.800 0.25 2 = 101 N Unit 5 - Circular Motion and Gravitation (Part 1) 20
Another example of horizontal circular motion involves a vehicle rounding a turn on a flat road. In this case, there exists no string which provides the necessary centripetal force. Instead, it is exerted by the friction between the tires of the car and the road. Unit 5 - Circular Motion and Gravitation (Part 1) 21
The free body diagram associated with this case is given below: Note that F c = F f = μf N = μmg Unit 5 - Circular Motion and Gravitation (Part 1) 22
Therefore, in order for the vehicle to round the curve safely, the static friction on the road must be strong enough to provide sufficient centripetal acceleration. However, it is not always the case. Due to different road conditions, the friction may be too small to afford enough friction. If the car rounds the curve too fast, it will skid. The situation gets worse if the wheels are locked when the brakes are applied too hard. Now the wheels stop rotating and the kinetic friction exists, causing the tires to slide. That s why modern vehicles are equipped with the antilock braking system (ABS) which can avoid sliding. Unit 5 - Circular Motion and Gravitation (Part 1) 23
Example: A 1000-kg car rounds a curve on a flat road of radius 50 m at a speed of 15 m/s. Will the car follow the curve, or will it skid? Assume: a) the pavement is dry and the coefficient of static friction is 0.60. b) the pavement is icy and the coefficient of static friction is 0.25. Solution: The centripetal force required is given by F c = mv2 r = 1000 15 2 50 = 4500 N If the static friction is equal or larger than this, then the car will round successfully. Otherwise it will skid. Unit 5 - Circular Motion and Gravitation (Part 1) 24
(a) The static friction exerted by a dry road is F f = μmg = 0.600 1000 9.8 = 5880 N Since F f > F c, the car will follow the curve. (b) The static friction exerted by an icy road is F f = μmg = 0.250 1000 9.8 = 2450 N Since F f < F c, the car will skid. Unit 5 - Circular Motion and Gravitation (Part 1) 25
Example: A 750 kg car travelling at 18.0 m/s comes to a sharp turn in the road where the radius of the curve is 136 m. (a) Find the centripetal acceleration and force acting on the car as it begins the turn. (b) If the coefficient of static friction between tires and road is 0.254, will the car be able to complete the turn at this speed without sliding off the road? Unit 5 - Circular Motion and Gravitation (Part 1) 26
4. Conical pendulum A particle is attached to the lower end of a light inextensible string, the upper end of which is fixed. When the particle describes a horizontal circle, the string describes the curved surface of a cone. This arrangement is known as a conical pendulum. Unit 5 - Circular Motion and Gravitation (Part 1) 27
The free body diagram for a conical pendulum looks like the following: The vertical component of F T balances F g. F g = F T cos θ The horizontal component of F T gives rise to the centripetal force F c. F c = F T sin θ The radius of the path is given by: r = l sin θ Unit 5 - Circular Motion and Gravitation (Part 1) 28
Example: A 140 g ball is fastened to one end of a 0.24 m string, and the other end is whirled in a horizontal conical pendulum. Find: a. the speed of the ball in its circular path; b. the tension in the string that makes an angle of 30 to the vertical. Unit 5 - Circular Motion and Gravitation (Part 1) 29
Example: A conical pendulum consists of a light inextensible string AB, fixed at A and carrying a particle of mass 50 g at B. The particle moves in a horizontal circle of radius 3 m and center vertically below A. If the angle between the string and the vertical is 30, find the tension in the string and the angular speed of the particle. Unit 5 - Circular Motion and Gravitation (Part 1) 30
5. Motion on a banked curve Recall that for a car travelling around a curve on a horizontal road, the friction between the tires and the road provides the necessary centripetal force. That is F f = μmg = mv2 r If the speed of the car exceeds the maximum possible value dictated by the frictional force, then the car will slip. This problem can be avoided by banking the road. That means, the level of the road on the inside of the bend is made lower than the level of the road on the outside of the bend. Unit 5 - Circular Motion and Gravitation (Part 1) 31
A banked curve looks like the following: Unit 5 - Circular Motion and Gravitation (Part 1) 32
What happens when a car is rounding a banked curve? Look at the following free body diagram for a banked curve: The vertical component of F N counteracts the gravity The normal force from the road The horizontal component of F N provides the centripetal force Unit 5 - Circular Motion and Gravitation (Part 1) 33
In an ideal scenario where the car has no tendency to slip either up or down the slope, there exists no friction between the tires and the road perpendicular to the direction of the motion, and thus F N sin θ = mv2 r The speed at which the car travels without the tendency to slip is called the null speed. It is given by v = rg tan θ (Try to derive this equation!) Unit 5 - Circular Motion and Gravitation (Part 1) 34
Example: A car travels around a bend of radius 400 m on a road which is banked at an angle θ to the horizontal. If the car has no tendency to slip when travelling at 35 m/s, find θ. Solution: Since there is no tendency for the car to slip, the null speed of the bend is 35 m/s. Hence, θ = tan 1 tan θ = v2 rg v2 rg = 35 2 tan 1 400 9.8 θ = 17.35 Unit 5 - Circular Motion and Gravitation (Part 1) 35
Example: A train of mass 50 tonnes travels around a bend of radius 900 m. The outer rail of the track is raised 7 cm above the inner rail and the distance between the rails is 1.4 m. Calculate the speed at which the train should travel so that there is no force acting between the flanges on the wheels and the rails. [21 m/s] Unit 5 - Circular Motion and Gravitation (Part 1) 36
If a car is travelling on a banked curve at a speed other than the null speed, it may skid either up or down the slope depending on its actual speed. (1) Actual speed > null speed There will be a tendency for the car to slip up the slope. Therefore, a friction will act between the tires and the road. The friction points down the slope. Unit 5 - Circular Motion and Gravitation (Part 1) 37
Based on the free body diagram, we can deduce that mv 2 = F r N sin θ + μf N cos θ F N cos θ = μf N sin θ + mg Combining these two equations, we can find the maximum speed that the car can travel without sliding: v max = rg sin θ + μ cos θ cos θ μ sin θ (Again, try to derive this equation!) Unit 5 - Circular Motion and Gravitation (Part 1) 38
Example: The radius of a velodrome curve is 40 m and the banked angle is 15. If μ = 0.20, what is the maximum speed at which a cyclist can take this curve without slipping? Unit 5 - Circular Motion and Gravitation (Part 1) 39
(2) Actual speed < null speed When the car enters into a banked curve with the speed below the null speed, it does not have enough inertia to maintain the circular path. Consequently it gradually slides down the slope. In this case, the road exerts a friction to keep it from slipping. This frictional force is pointing up along the road. F f Unit 5 - Circular Motion and Gravitation (Part 1) 40
In this situation: mv 2 r = F N sin θ μf N cos θ F N cos θ + μf N sin θ = mg Combining these two equations we can work out the minimum speed at which the car can travel without slipping: v min = rg sin θ μ cos θ cos θ + μ sin θ Unit 5 - Circular Motion and Gravitation (Part 1) 41
Example: A car travels around a bend in a road which is a circular arc of radius 62.5 m. The road is banked at an angle tan 1 5/12 to the horizontal. If the coefficient of friction between the tires of the car and the road surface is 0.4, find the range of speeds in which the car can be driven around the bend without slipping occurring. [2.96 m/s to 24.5 m/s] Unit 5 - Circular Motion and Gravitation (Part 1) 42
6. Vertical circular motion A circular motion with an object moving at a constant speed is a uniform circular motion because the net force on the object is exerted towards the center of the circle always. In some situations, on the other hand, the net force on the moving object is not always centripetal. Many examples: Unit 5 - Circular Motion and Gravitation (Part 1) 43
If the net force acting on the object is not directed toward the center but at an angle, then there will be two components of this force: (1) The one pointing toward the center of the circle gives rise to the centripetal force that keeps the object moving in a circle. (2) The one pointing parallel to the motion of the object gives rise to the tangential acceleration of the object. This is the reason why the object has a varying acceleration. Unit 5 - Circular Motion and Gravitation (Part 1) 44
The centripetal acceleration arises from the change of direction of the velocity of the object. Thus a c = v2 r On the other hand, the tangential acceleration arises from the change of magnitude of the velocity of the object; that means a t = v t The net force on the object is the vector sum of them, and its magnitude is given by: a = a c 2 + a t 2 Unit 5 - Circular Motion and Gravitation (Part 1) 45
Example: A race car starts from rest in the pit area and accelerates at a uniform rate to a speed of 35 m/s in 11 s, moving on a circular track of radius 500 m. Assuming constant tangential acceleration, find a. the tangential acceleration; and b. the radial acceleration at the instant when the speed is 15 m/s. Unit 5 - Circular Motion and Gravitation (Part 1) 46
It is very complicated to analyze vertical circular motion in general because the net force is not always pointing towards the center of the circle. Therefore, we will only investigate the objects, undergoing a vertical circular motion, at some specific positions where forces are interacting with the net centripetal force. Specifically, we will look at 3 systems: 1. Object attached to a string 2. Vertical loop of track 3. The Ferris wheel Unit 5 - Circular Motion and Gravitation (Part 1) 47
(1) Mass attached to a string The associated free body diagram is depicted below: Unit 5 - Circular Motion and Gravitation (Part 1) 48
We can conclude: At the bottom At the side At the top F T = F c + F g F T = F c F T = F c F g The minimum speed at which the mass can go around the top without falling is called the critical velocity. At this speed, no tension is exerted. Hence, F c = F g The centripetal force is provided by the gravity on the mass. Unit 5 - Circular Motion and Gravitation (Part 1) 49
Example: A 0.90 kg mass attached to a cord is whirled in a vertical circle of radius 2.5 m. a. Find the tension in the cord at the top of the circle if the speed of the mass is 8.7 m/s. b. Find the tension in the cord at the bottom of the circle if the speed is maintained at 8.7 m/s. c. What minimum speed must it have at the top of the circle so as not to fall from the circular path? d. At the speed in (c) and neglecting any friction, how fast will the object be going at the bottom of the circle? e. What is the tension in the cord at the bottom at this speed? Unit 5 - Circular Motion and Gravitation (Part 1) 50
(2) Vertical loop of track When a vehicle is moving inside a vertical loop of track, it experiences the normal force acting on it from the track. This normal force is always pointing towards the center of the loop. It is therefore obvious that the resulting free body diagram for such a system is equivalent to the one for a mass attached to a string, except that the tension F T is replaced by the normal force F N. Hence F c = F N + F g (top) F c = F N F g (bottom) Unit 5 - Circular Motion and Gravitation (Part 1) 51
Example: A 20.0 g steel ball-bearing on a rail rolls from rest at point A. Assuming negligible friction, if h = 0.25 m and R = 0.050 m, a. what is the speed of the bearing at point B? b. what normal force must the rail exert on the bearing at B? Unit 5 - Circular Motion and Gravitation (Part 1) 52
(3) The Ferris wheel The free body diagram for the Ferris wheel is different from the other two; the cars are mounted on the outside of the rim, thereby having the normal force point upward always. Unit 5 - Circular Motion and Gravitation (Part 1) 53
Schematically, At top: At bottom: F c = F g F N F c = F N F g Unit 5 - Circular Motion and Gravitation (Part 1) 54
Similar to the case of swing a mass vertically, there exists a critical speed at the top of a Ferris wheel below which a person will remain in his seat without flying off. In this situation, F N = 0; therefore F c = F g If the Ferris wheel is moving faster than the critical speed, the rim will exerts a negative normal force to the compartment. It implies that the compartment will become apparently weightless and fly off the rim. Unit 5 - Circular Motion and Gravitation (Part 1) 55
The same idea can be applied to the case of a vehicle going over a round hill or a curved surface. The free body diagram is as follows: The centripetal force that keeps the car going on the road is F c = F g F N. Unit 5 - Circular Motion and Gravitation (Part 1) 56
Since r, the radius of the round hill, is fixed, the maximum centripetal force that a vehicle could receive is determined by F g. This leads to the critical speed: v = rg In this case, the centripetal force comes entirely from gravity. If the actual speed of the vehicle is larger than v, then gravity will not be sufficient to hold it on the road, and it will fly off (i.e., airborne). Unit 5 - Circular Motion and Gravitation (Part 1) 57
Example: A 62 kg student drives his 450 kg car at 25 m/s up towards the top of a hill of radius 70 m. a. What normal force will the driver s seat exert on him at the top of the hill? b. How fast can he drive his car over the hill without being airborne? Unit 5 - Circular Motion and Gravitation (Part 1) 58
7. More practice on circular motion Question: In a rotor-ride in a carnival people are rotated in a cylindrically walled room. (See image below). The room has a radius of 4.6 m and the rotational frequency of 0.50 Hz when the floor drops. What is the minimum coefficient of friction between the walls and the people so the people stay pressed to the wall? Unit 5 - Circular Motion and Gravitation (Part 1) 59
Question: A frictionless 3.0 kg cart rolls down an incline, and then enters into a loop of track as shown below. a. What is the minimum speed at which the cart can pass the top of the loop without falling off? b. What is the speed of the cart at the bottom of the loop? c. From what minimum height should the cart be released so that it does not fall off the track? Unit 5 - Circular Motion and Gravitation (Part 1) 60