* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Download Answers - hrsbstaff.ednet.ns.ca
Survey
Document related concepts
Modified Newtonian dynamics wikipedia , lookup
Equations of motion wikipedia , lookup
Newton's theorem of revolving orbits wikipedia , lookup
Hunting oscillation wikipedia , lookup
Coriolis force wikipedia , lookup
Fictitious force wikipedia , lookup
Jerk (physics) wikipedia , lookup
Newton's laws of motion wikipedia , lookup
Centrifugal force wikipedia , lookup
Seismometer wikipedia , lookup
Transcript
2.2.2 In Class or Homework Exercise 1. Sue whirls a yo-yo in a horizontal circle. The yo-yo has a mass of 0.20 kg and is attached to a string 0.80 m long. a. If the yo-yo makes 1.0 complete revolution each second, what force does the string exert on it? m 0.20kg r 0.80m f 1.0 Hz FT ? If we draw a free body diagram, we see that the string cannot be perfectly horizontal since there is no force other than the string to balance the force of gravity: However, for this and other problems involving horizontal circles we will assume (unless told otherwise) that the string is horizontal and only exerts a horizontal force. 1 f 1 1.0 1.0 s T 2 r v T 2 (0.80) 1.0 5.03m / s v2 r (5.03) 2 0.80 31.6m / s 2 ac Fc mac FT mac (0.20)(31.6) 6.3 N b. If Sue increases the speed of the yo-yo to 2.0 revolutions per second, what force does the string now exert? Since she is doubling the frequency, she is cutting the period in half. This will double the speed and will quadruple the acceleration (since ac v 2 ). This will require a force 4 times as large FT 4(6.3) 25 N UNIT 2 2D Motion RRHS PHYSICS Page 59 of 136 2. A car rounds a curve on a flat road of radius 50.0 m at a speed of 50.0 km/h. Will the car make the turn if (a) the pavement is dry and the coefficient of static friction is 0.60, (b) the pavement is icy and 0.20 ? There are a number of ways in which this problem may be approached. We will first calculate the required coefficient of friction to make the turn at the given speed. r 50.0m v 50.0km / h 13.9m / s ? The free body diagram looks like this: So Fc mac F f mac FN mac mg m v2 r v2 gr (13.9) 2 (9.80)(50.0) 0.39 This tells us the required (or minimum) coefficient of friction. a. Since the dry coefficient of friction is 0.60, there will be enough friction to make the turn. b. Since the wet coefficient of friction is only 0.20, there will not be enough friction to make the turn UNIT 2 2D Motion RRHS PHYSICS Page 60 of 136 3. A gravitron circus ride consists of a cylindrical room with vertical walls (as shown below) that rotates about its axis People are positioned against the circular wall. When the room is rotating at a certain speed, the floor drops out and the people in the room remain suspended against the outer wall. Consider a gravitron circus ride that has an 8.0 m radius and rotates 0.36 times per second. a. Draw a free body diagram indicating all of the forces involved. Notice that the normal force is horizontal (perpendicular to the wall) and friction is up (opposing the person trying to slide down the wall). There is no outward force on the person. b. What coefficient of friction is necessary to prevent the people from sliding down the wall? UNIT 2 2D Motion RRHS PHYSICS Page 61 of 136 r 8.0m f 0.36 Hz ? T 1 f 1 0.36 2.78s 2 r T 2 (8.0) 2.78 18.1m / s v In order for people to not slide down the wall, the force of friction must balance the force of gravity: Vertical Ff Fg FN mg Horizontal The centripetal acceleration (and therefore the centripetal force) in the diagram is to the left. It must therefore be provided by the normal force exerted by the wall: Fc mac FN mac If we combine the horizontal and vertical equations, we get FN mg ( mac ) mg v2 g r gr 2 v (9.80)(8.0) (18.1) 2 0.24 c. People often describe the ride by saying that they were being pressed against the wall. Is this true? Not really. There body has a natural tendency to continue travelling in a straight line (Newton’s First Law), but the wall is in the way. The wall is exerting an inward force to make them travel in a circular path and this is what they are feeling. UNIT 2 2D Motion RRHS PHYSICS Page 62 of 136 4. What minimum speed must a roller coaster be travelling when upside down at the top of a circle if the passengers are not to fall out. Assume a radius of curvature of 8.00 m. In general for the people in the roller coaster, there would be two forces exerted downward at the top of the path, the force of gravity and the normal force of the roller coaster cart; however, the slower the cart is travelling the lower the magnitude of the normal force. At the minimum speed, the normal force is zero, so the only force acting on the person is the force of gravity. This means that the person has a downward acceleration of 9.80 m/s2. Since there is no upward force, the person is falling at 9.80m/s 2 at this point. For the passengers to no fall out of the car, it must also have a downward acceleration of 9.80m/s2. v2 ac r v2 9.80 8.0 v 8.85m / s If the cart goes slower than 8.85 m/s, its downward centripetal acceleration is less than the 9.80 m/s2 that the person is accelerating at and the person will fall out of the cart. 5. A coin is placed 18.0 cm from the axis of a rotating turntable of variable speed. When the speed of the turntable is slowly increased, the coin remains fixed on the turntable until a rate of 58 rpm is reached. What is the coefficient of static friction between the coin and the turntable? r 18.0cm 0.180m f 58rpm 0.97 Hz ? T 1 f 1 0.97 1.03s UNIT 2 2D Motion 2 r T 2 (0.180) 1.03 1.10m / s v RRHS PHYSICS v2 r (1.10) 2 0.180 6.72m / s 2 ac Page 63 of 136 Fc mac F f mac FN mac mg mac (9.80) 6.72 0.69 6. A ball on a string is revolving at a uniform rate in a vertical circle of radius 96.5 cm. If its speed is 3.15 m/s and its mass is 0.335 kg, calculate the tension in the string a. at the top of its path It should be noted that although we are assuming a constant speed throughout the vertical circle in this problem, this is not a realistic situation. The ball’s speed must change in order to satisfy conservation of energy (this will be addressed in the next problem). r 96.5cm 0.965m v 3.15m / s m 0.335kg FT ? Fg mg (0.335)(9.80) 3.28 N v2 r (3.15) 2 0.965 10.3m / s 2 ac The acceleration is downward. Using down as positive, UNIT 2 2D Motion RRHS PHYSICS Page 64 of 136 Fc mac mac F mac FT Fg mac FT Fg (0.335)(10.3) FT 3.28 FT 0.17 N b. at the bottom of its path m 0.335kg ac 10.3m / s 2 Fg 3.28 N FT ? Since we are now at the bottom of the path, the acceleration is upward. We will use up as positive. Fc mac mac F mac FT Fg mac FT Fg (0.335)(10.3) FT 3.28 FT 6.73 N c. at the middle of its path (halfway between top and bottom) m 0.335kg ac 10.3m / s 2 Fg 3.28 N FT ? UNIT 2 2D Motion RRHS PHYSICS Page 65 of 136 Since the centripetal acceleration is horizontal at this point, only the tension in the string contributes to (and therefore provides) the centripetal force Fc mac FT mac (0.335)(10.3) 3.45 N d. Assuming a constant speed throughout the entire vertical circle, where would the string be most likely to break and in what direction would the ball travel immediately after the string broke? Since the tension is greatest at the bottom of the path, the string would be most likely to break at the bottom. Since the ball is travelling horizontally at the bottom of the path, it would continue to move horizontally (either left or right) immediately after the string breaks. 7. A 5.0 kg mass is being swung in a vertical circle on a 3.0 m rope. a. What is the critical speed (i.e. the minimum speed at which the ball will maintain a circular path) for this mass? If the ball is to fail to maintain a circular path, this will occur at the top of the path. Consider a mass on a string being swung in a vertical circle; as the mass is swung faster, the string must exert more force to maintain a circular path. As the mass is swung more slowly, the string exerts less force. If you slow the mass down too much, it falls out of the circle at the top. The critical speed occurs at the exact moment that the tension goes to zero so that the entire centripetal force at this time is being provided by the force of gravity. The mass must therefore have an acceleration equal to the acceleration due to gravity: v2 ac r v2 g r v gr (9.80)(3.0) 5.4m / s . UNIT 2 2D Motion RRHS PHYSICS Page 66 of 136 b. Calculate the tension in the rope at the ball's lowest point, assuming that the ball is travelling at its critical speed at the top of the circle. Assume no change in energy for the system. Fg mg m 5.0kg r 3.0m FT ? (5.0)(9.80) 49 N Since we are assuming no change in energy for the system, the ball must be travelling faster at the bottom since it is at a lower height and will have less potential energy. If we use the lowest point in the circle as the reference level, Ei E f E pi Eki Ekf mghi 12 mvi2 12 mv 2f (9.80)(6.0) 12 (5.4) 2 12 v 2f v f 12.1m / s v2 ac r (12.1) 2 3.0 48.8m / s 2 Since we are now at the bottom of the path, the acceleration is upward. We will use up as positive. UNIT 2 2D Motion RRHS PHYSICS Page 67 of 136 Fc mac mac F mac FT Fg mac FT Fg (5.0)(48.8) FT 49 FT 290 N 8. Tarzan plans to cross a gorge by swinging in an arc from a hanging vine. If his arms are capable of exerting a force of 1500. N on the vine, what is the maximum speed he can tolerate at the lowest point of his swing? His mass is 85.0 kg; the vine is 4.0 m long. Fa 1500.N Fg mg m 85.0kg (85.0)(9.80) r 4.0m 833 N v? Since Tarzan is at the bottom of the path, the acceleration is upward. We will use up as positive. Fc mac v2 ac mac F r mac Fa Fg v2 7.85 4.0 mac Fa Fg v 5.6m / s (85.0)a 1500. 833 c ac 7.85m / s 2 9. How many revolutions per minute would a 15 m diameter Ferris wheel need to make for the passengers to feel weightless at the top? UNIT 2 2D Motion RRHS PHYSICS Page 68 of 136 If the passengers are to feel weightless, the must be in free fall which means that their downward acceleration must be 9.80m/s2. This must be the centripetal acceleration. ac 9.80m / s 2 r 7.5m f ? v2 r v2 9.80 7.5 v 8.6m / s ac 2 r T 2 (7.5) 8.6 T T 5.5s v 1 T 1 5.5 0.18r / s 60 s / min f 11r / min 10. A person is swinging a 0.150 kg ball on a string in a horizontal circle of radius 0.600 m. The ball makes 2.00 revolutions per second. a. Although the circle is horizontal, the string is not. Explain why not. If we draw a free body diagram, we see that the string cannot be perfectly horizontal since there is no force other than the string to balance the force of gravity: A horizontal string cannot exert a vertical force to balance the downward force of gravity; therefore, the string must have a vertical component. b. Ignoring the fact that the string is not perfectly horizontal (due to the weight of the ball), calculate the tension in the string. m 0.150kg r 0.600m f 2.00 Hz FT ? T 1 f 1 2.00 0.500 s UNIT 2 2D Motion 2 r v T 2 (0.600) 0.500 7.54m / s RRHS PHYSICS v2 ac r (7.54) 2 0.600 94.8m / s 2 Page 69 of 136 Since we are assuming that the string is horizontal, the tension in the string will be providing the centripetal force: Fc mac FT mac (0.150)(94.8) 14.2 N c. Find the tension in the string (and the angle it makes with the horizontal) if we do not ignore the effect of the weight of the ball. 5.91, 14.3N Vertical ma F 0 Fg FTy 0 Fg FTy FTy mg (0.150)(9.80) 1.47 N Horizontal mac F mac FTx (0.150)(94.8) FTx FTx 14.2 N tan FT FTx2 FTy2 FTy FTx 1.47 14.2 5.91 (14.2)2 (1.47) 2 14.3N As can be seen the tension is less than 1% different when accounting for the angle of the string. UNIT 2 2D Motion RRHS PHYSICS Page 70 of 136 11. A projected space station consists of a circular tube which is set rotating about its center (like a tubular bicycle tire). The circle formed by the tube has a diameter of 1.600 km. a. On which part of the inside of the tube will people be able to walk? The people will be able to walk on the inside outer edge of the tube. It is similar to placing a marble on a turntable – as the turntable spins, the marble will roll and will continue in a straight line toward the outer edge of the turntable. In the space station, the people will move in a straight line toward the inside outer edge. From their point of view, it will be as if there is a force pushing them toward this part of the space station. b. What must be the rotation speed (revolutions per day) if an effect equal to gravity at the surface of the earth (1 g ) is to be felt? r 800.m ac 9.80m / s 2 f ? 2 r T 2 (800.) 88.5 T T 56.8s v2 r v2 9.80 800. v 88.5m / s ac f v 1 T 1 56.8 0.0176r / s 86400 s / day 1520r / day UNIT 2 2D Motion RRHS PHYSICS Page 71 of 136 12. When you drive rapidly on a hilly road or ride in a roller coaster, you feel lighter as you go over the top of a hill and heavier when you go through a valley. a. Sketch each situation, including the relevant forces, and explain these sensations. When the car is going over the top of the hill, the center of curvature for the hill is below the car, so there is a downward centripetal acceleration. This means that the downward force of gravity is greater than the upward normal force on the person. Since the normal force is smaller than if the car were at rest, the person will feel lighter. When the car is going through a valley, the center of curvature for the valley is above the car, so there is an upward centripetal acceleration. This means that the upward normal force is greater than the downward force of gravity on the person. Since the normal force is larger than if the car were at rest, the person will feel heavier. b. Consider a car going over the top of a hill (radius = 95 m). How fast would the car have to go so that the normal force acting on the driver is zero? Since the normal force is zero, the only force acting on the person will be the force of gravity so the acceleration must be 9.80 m/s 2. The person will feel weightless in this situation. UNIT 2 2D Motion RRHS PHYSICS Page 72 of 136 r 95m ac 9.80m / s 2 v? v2 r v2 9.80 95 v 30.5m / s ac 110km / h 13. A bucket of water can be whirled in a vertical circle without the water spilling out, even at the top of the circle when the bucket is upside down. Explain. This is similar to the upside down roller coaster problem. If the water were to fall out of the bucket, it will fall with an acceleration of 9.80 m/s 2. As long as the bucket is being swung fast enough so that its downward centripetal acceleration at the top of the path is at least 9.80 m/s2, it will be pushing down on the water and the water will remain with the bucket. If the bucket slows down so that its centripetal acceleration is less than 9.80 m/s 2, the water will now have a greater downward acceleration and will leave the bucket. 14. For a car travelling with speed v around a curve of radius r , determine a formula for the angle at which a road should be banked so that no friction is required. In this case, the centripetal acceleration is horizontal toward the inside of the curve. We will use a horizontal coordinate system and break the normal force up into horizontal and vertical components. The angle in the free body diagram will be the same angle as the angle that the road makes with the horizontal. UNIT 2 2D Motion RRHS PHYSICS Page 73 of 136 Since there is no friction, the only force with any horizontal components is the normal force exerted by the road. The horizontal component of this normal force must provide the required centripetal force. Fc mac FNx mac v2 r v2 mg tan m r 2 v tan gr FNy tan m v2 tan gr 1 UNIT 2 2D Motion RRHS PHYSICS Page 74 of 136