course Phy 121
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10:23:50 `q001. Note that this assignment contains 4 questions. . Note that this assignment contains 4 questions. When an object moves a constant speed around a circle a force is necessary to keep changing its direction of motion. This is because any change in the direction of motion entails a change in the velocity of the object. This is because velocity is a vector quantity, and if the direction of a vector changes, then the vector and hence the velocity has changed. The acceleration of an object moving with constant speed v around a circle of radius r has magnitude v^2 / r, and the acceleration is directed toward the center of the circle. This results from a force directed toward the center of the circle. Such a force is called a centripetal (meaning toward the center) force, and the acceleration is called a centripetal acceleration. If a 12 kg mass travels at three meters/second around a circle of radius five meters, what centripetal force is required?
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RESPONSE --> centripetal acceleration = v^2 / r = (3m/s) ^ 2/(5 m) = 1.8 m/s ^ 2 centripetal force = Fcent = 12 kg * 1.8 m/s ^ 2 = 21.6 kg * m/s ^ 2
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10:23:54 The centripetal acceleration of the object is v^2 / r = (3 meters/second) ^ 2/(5 meters) = 1.8 meters/second ^ 2. The centripetal force, by Newton's Second Law, must therefore be Fcent = 12 kg * 1.8 meters/second ^ 2.
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RESPONSE --> ok
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10:27:13 `q002. How fast must a 50 g mass at the end of a string of length 70 cm be spun in a circular path in order to break the string, which has a breaking strength of 25 Newtons?
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RESPONSE --> m v^2 / r = F solve for v v = `sqrt(F * r / m) v = `sqrt( 25 N * .7 m / (.05 kg) ) = `sqrt( 25 kg m/s^2 * .7 m / (.05 kg) ) = `sqrt(350 m^2 / s^2) = 18.7 m/s
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10:27:18 The centripetal acceleration as speed v will be v^2 / r, where r = 70 cm = .7 meters. The centripetal force will therefore be m v^2 / r, where m is the 50 g = .05 kg mass. If F stands for the 25 Newton breaking force, then we have m v^2 / r = F, which we solve for v to obtain v = `sqrt(F * r / m). Substituting the given values we obtain v = `sqrt( 25 N * .7 meters / (.05 kg) ) = `sqrt( 25 kg m/s^2 * .7 m / (.05 kg) ) = `sqrt(350 m^2 / s^2) = 18.7 m/s.
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RESPONSE --> ok
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10:30:34 `q003. What is the maximum number of times per second the mass in the preceding problem can travel around its circular path before the string breaks?
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RESPONSE --> The distance traveled along this path in a single revolution is 2 `pi r = 2 `pi * .7 meters = 4.4m. 18.7/4.4 = 4.25 times every second
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10:30:47 The maximum possible speed of the mass was found in the preceding problem to be 18.7 meters/second. The path of the mass is a circle of radius 70 cm = .7 meters. The distance traveled along this path in a single revolution is 2 `pi r = 2 `pi * .7 meters = 4.4 meters, approximately. At 18.7 meters/second, the mass will travel around the circle 18.7/4.4 = 4.25 times every second.
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RESPONSE --> ok
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10:37:05 `q004. Explain in terms of basic intuition why a force is required to keep a mass traveling any circular path.
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RESPONSE --> You can't change the direction of motion of a object without giving it a push. Without the push an object in motion will remain in motion along a straight line and with no change in speed. The force of the object keeps it going pulling it toward the center of the circle just like spinning a bolt tied to the end of a string.
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10:37:08 We simply can't change the direction of motion of a massive object without giving it some sort of a push. Without such a force an object in motion will remain in motion along a straight line and with no change in speed. If your car coasts in a circular path, friction between the tires and the road surface pushes the car toward the center of the circle, allowing it to maintain its circular path. If you try to go too fast, friction won't be strong enough to keep you in the circular path and you will skid out of the circle. In order to maintain a circular orbit around the Earth, a satellite requires the force of gravity to keep pulling it toward the center of the circle. The satellite must travel at a speed v such that v^2 / r is equal to the acceleration provided by Earth's gravitational field.
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RESPONSE --> ok
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ìVܥjWq assignment #024 [뺗܃ܝ Physics I 03-24-2006
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10:43:44 Why was it necessary to let the string go slack at the top of the circle in order to get the desired results?
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RESPONSE --> If it is not slack the weight is being pulled toward the center by a force that exceeds its weight. If the string goes slack before the weight reaches the top of its arc then the path isn't circular.
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10:43:48 ** If the string goes slack just at the instant the weight reaches the 'top' of its circular path then we are assured that the centripetal acceleration is equal to the acceleration of gravity. If there is tension in the string then the weight is being pulled downward and therefore toward the center by a force that exceeds its weight. If the string goes slack before the weight reaches the top of its arc then the path isn't circular and our results won't apply to an object moving in a circular arc. **
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RESPONSE --> ok
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10:46:51 Why do you expect that, if the string is released exactly at the top of the circle, the initial velocity of the washer will be horizontal?
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RESPONSE --> When the washer is at the top of its arc it is directly above the center so the radial line is vertical. Its velocity is perpendicular to this vertical making it horizontal.
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10:46:55 ** The direction of an object moving in a circular arc is perpendicular to a radial line (i.e., a line from the center to a point on the circle). When the object is at the 'top' of its arc it is directly above the center so the radial line is vertical. Its velocity being perpendicular to this vertical must be wholly in the horizontal direction. **
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RESPONSE --> ok
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10:47:35 What is the centripetal acceleration of the washer at the instant of release, assuming that it is released at the top of its arc and that it goes slack exactly at this point, and what was the source of this force?
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RESPONSE --> The centripetal acceleration will be equal to the acceleration of gravity.
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10:47:38 ** Under these conditions, with the string slack and not exerting any force on the object, the centripetal acceleration will be equal to the acceleration of gravity. **
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RESPONSE --> ok
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]{{wMx˯꜊ Student Name: assignment #025
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10:54:18 `q001. Note that this assignment contains 5 questions. . A pendulum consists of a 150 g mass suspended from a light string. Another light string is attached to the mass, which is then pulled back from its equilibrium position by that string until the first string makes an angle of 15 degrees with vertical. The second string remains horizontal. Let the x axis be horizontal and the y axis vertical. Assume that the mass is pulled in the positive x direction. If T stands for the tension in the pendulum string, then in terms of the variable T what are the x and y components of the tension?
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RESPONSE --> The tension force will be directed 90 degrees + 15 degrees = 105 degrees x component T cos(105 degrees) = -.2588 y component T sin(105 degrees) = .9659
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10:54:21 The pendulum string makes an angle of 15 degrees with vertical. Since we have assumed that the pendulum is pulled in the positive x direction, the direction of the tension in the string will be upward and to the left at an angle of 15 degrees with vertical. The tension force will therefore be directed at 90 degrees + 15 degrees = 105 degrees as measured counterclockwise from the positive x axis. The tension will therefore have x component T cos(105 degrees) and y component T sin(105 degrees).
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RESPONSE --> ok
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10:56:17 `q002. Continuing the preceding problem, we see that we have a vertical force of T sin(105 deg) from the tension. What other vertical force(s) act on the mass? What is the magnitude and direction of each of these forces? What therefore must be the magnitude of T sin(105 deg).
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RESPONSE --> gravitational force = .150 kg * 9.8 m/s^ 2 = 1.47 Newtons the mass is not accelerating so the net force in the y direction is zero. T sin(105 deg) - 1.47 Newtons = 0 T sin(105 deg) = 1.47 Newtons
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10:56:20 The only other vertical force acting on the mass will be the gravitational force, which is .150 kg * 9.8 meters/second ^ 2 = 1.47 Newtons. The direction of this force is vertically downward. Since the mass is in equilibrium, i.e., not accelerating, the net force in the y direction must be zero. Thus T sin(105 deg) - 1.47 Newtons = 0 and T sin(105 deg) = 1.47 Newtons.
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RESPONSE --> ok
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11:00:35 `q003. Continuing the preceding two problems, what therefore must be the tension T, and how much tension is there in the horizontal string which is holding the pendulum back?
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RESPONSE --> T sin(105 deg) = 1.47 Newtons T = 1.47 Newtons / (sin(15 deg)) = 1.47 Newtons/.97 = 1.52 Newtons horizontal component of tension will be T cos(105 deg) = 1.52 Newtons * cos(105 deg) = 1.52 Newtons * (-.26) = -.39 Newtons T2 + (-.39 N) = 0 and T2 = .39 N tension in the second string is .39 Newtons.
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11:00:42 If T sin(105 deg) = 1.47 Newtons then T = 1.47 Newtons / (sin(15 deg)) = 1.47 Newtons/.97 = 1.52 Newtons. Thus the horizontal component of the tension will be T cos(105 deg) = 1.52 Newtons * cos(105 deg) = 1.52 Newtons * (-.26) = -.39 Newtons, approximately. Since the mass is in equilibrium, the net force in the x direction must be zero. The only forces acting in the x direction are the x component of the tension, to which we just found to be -.39 Newtons, and the tension in the second string, which for the moment will call T2. Thus T2 + (-.39 N) = 0 and T2 = .39 N. That is, the tension in the second string is .39 Newtons. STUDENT COMMENT: I'm really confused now. If we started out with a .15 kg mass that is equal to 1.47 Newtons. How did we create more weight to get 1.52 Newtons? Is the horizontal string not helping support the weight or is it puling on the weight adding more force? INSTRUCTOR RESPONSE: A horizontal force has no vertical component and cannot help to support an object against a vertical force. The vertical component of the tension is what supports the weight, so the tension has to be a bit greater than the weight. The tension in the string is resisting the downward weight vector as well as the horizontal pull, so by the Pythagorean Theorem it must be greater than either.
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RESPONSE --> ok
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11:04:32 `q004. If a 2 kg pendulum is held back at an angle of 20 degrees from vertical by a horizontal force, what is the magnitude of that horizontal force?
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RESPONSE --> direction of string = 90 deg + 20 deg = 110 deg x component of the tension will be T cos(110 deg) y component of the tension will be T sin(110 deg) The weight of the 2 kg pendulum is 2 kg * 9.8 meters/second ^ 2 = 19.6 Newtons in the negative direction. Because the pendulum are in equilibrium the net vertical force is zero. This equation is easily solved for the tension: T = 19.6 N / (sin(110 deg) ) = 19.6 N / (.94) = 20.8 Newtons The horizontal component of the tension = T cos(110 deg) = 20.8 N * cos(110 deg) = 20.8 N * (-.34) = -7 N To get equilibrium, the other horizontal force will be + 7 Newtons.
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11:04:36 At the 20 degree angle the tension in the pendulum string will have a vertical component equal and opposite to the force exerted by gravity. The tension with therefore have a horizontal component. To achieve equilibrium by exerting the horizontal force, this horizontal force must balance the horizontal component of the tension. We therefore begin by letting T stand for the tension in the pendulum string. We also assumed that the pendulum is displaced in the positive x, so that the direction of the string as measured counterclockwise from the positive x axis will be 90 degrees + 20 degrees = 110 degrees. Thus the x component of the tension will be T cos(110 deg) and the y component of the tension will be T sin(110 deg). The weight of the 2 kg pendulum is 2 kg * 9.8 meters/second ^ 2 = 19.6 Newtons, directed in the negative vertical direction. Since the pendulum are in equilibrium, the net vertical force is zero: T sin(110 deg) + (-19.6 N) = 0 This equation is easily solved for the tension: T = 19.6 N / (sin(110 deg) ) = 19.6 N / (.94) = 20.8 Newtons, approximately. The horizontal component of the tension is therefore T cos(110 deg) = 20.8 N * cos(110 deg) = 20.8 N * (-.34) = -7 N, approx.. To achieve equilibrium, the additional horizontal force needed will be + 7 Newtons.
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RESPONSE --> ok
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11:06:30 `q005. The 2 kg pendulum in the previous exercise is again pulled back to an angle of 20 degrees with vertical. This time it is held in that position by a chain of negligible mass which makes an angle of 40 degrees above horizontal. Describe your sketch of the forces acting on the mass of the pendulum. What must be the tension in the chain?
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RESPONSE --> I am not sure how to do this so I am going to check out how you solved and see what I can figure out.
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11:06:44 The weight of the pendulum is partially supported by the tension in the chain. Thus the tension in the pendulum string is not the same as before. The horizontal component of the tension in the chain will be equal and opposite to the horizontal component of the tension in the pendulum string. Your picture should show the weight vector acting straight downward, the tension in the pendulum string acting upward and to the left at an angle of 20 degrees to vertical and the tension in the chain should act upward into the right at an angle of 40 degrees above horizontal. The lengths of the vectors should be adjusted so that the horizontal components of the two tensions are equal and opposite, and so that the sum of the vertical components of the two tensions is equal of opposite to the weight vector. Since both tensions are unknown we will let T1 stand for the tension in the pendulum and T2 for the tension in the chain. Then T1, as in the preceding problem, acts at an angle of 110 degrees as measured counterclockwise from the positive x axis, and T2 acts at an angle of 40 degrees. At this point whether or not we know where we are going, we should realize that we need to break everything into x and y components. It is advisable to put this information into a table something like the following: x comp y comp T1 T1 * cos(110 deg) T1 * sin(110 deg) in T2 T2 * cos(40 deg) T2 * sin(40 deg) Weight 0 -19.6 N The pendulum is held in equilibrium, so the sum of all the x components must be 0, as must the sum of all y components. We thus obtain the two equations T1 * cos(110 deg) + T2 * cos(40 deg) = 0 and T1 * sin(110 deg) + T2 * sin(40 deg) - 19.6 N = 0. The values of the sines and cosines can be substituted into the equations obtain the equations -.33 T1 + .77 T2 = 0 .95 T1 + .64 T2 - 19.6 N = 0. We solve these two simultaneous equations for T1 and T2 using one of the usual methods. Here we will solve using the method of substitution. If we solve the first equation for T1 in terms of T2 we obtain T1 = .77 T2 / .33 = 2.3 T2. Substituting 2.3 T2 for T1 in the second equation we obtain .95 * 2.3 T2 + .64 T2 - 19.6 N = 0, which we easily rearrange to obtain 2.18 T2 + .64 T2 = 19.6 Newtons, or 2.82 T2 = 19.6 N, which has solution T2 = 19.6 Newtons/2.82 = 6.9 N, approximately. Since T1 = 2.3 T2, we have T1 = 2.3 * 6.9 N = 15.9 N, approximately. Thus the pendulum string has tension approximately 15.9 Newtons and the chain the tension of approximately 6.9 Newtons.
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RESPONSE --> It makes since now.
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fwxI[Z{ Student Name: assignment #026
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11:11:15 `q001. Note that this assignment contains 3 questions. . Water has a density of 1 g per cm^3. If an object is immersed in water, it experiences a buoyant force which is equal to the weight of the water it displaces. Suppose that an object of mass 400 grams and volume 300 cm^3 is suspended from a string of negligible mass and volume, and is submerged in water. If the mass is suspended in equilibrium, what will be the tension in the string?
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RESPONSE --> Force of gravity on 400 g mass = .4 kg * 9.8 m/s^2 = 3.92 Newtons. It will have the upward buoyant force equal to the weight of the 300 cm^3 of water it displaces. weight of .3 kg * 9.8 m/s^2 = 2.94 Newtons. -3.92 Newtons + 2.94 Newtons + T = 0 T = .98 Newtons
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11:11:19 The 400 g mass will experience a downward gravitational force of .4 kg * 9.8 meters/second^2 = 3.92 Newtons. It will also experience in upward buoyant force equal to the weight of the 300 cm^3 of water it displaces. This volume of water, at 1 g per cm^3, will have a mass of 300 grams and therefore a weight of .3 kg * 9.8 meters/second^2 = 2.94 Newtons. The forces acting on the mass are therefore the downward 3.92 Newtons of gravity, the upward 2.94 Newtons of the buoyant force and the tension, which we will call T, in the string. Since the system is in equilibrium these forces must add up to 0. We thus have -3.92 Newtons + 2.94 Newtons + T = 0, which has solution T = .98 Newtons. In common sense terms, gravity pulls down with 3.92 Newtons of force and the buoyant force pushes of with 2.94 Newtons of force so to keep all forces balanced the string must pull up with a force equal to the .98 Newton difference.
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RESPONSE --> ok
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11:13:01 `q002. A solid cylinder has a cross-sectional area of 8 cm^2. If this cylinder is held with its axis vertical and is immersed in water to a depth of 12 cm, what will be the buoyant force on the cylinder?
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RESPONSE --> volume of the immersed portion = 12 cm * 8 cm^2 = 96 cm^3 weight of displace water = .096 kg * 9.8 m/s^2 = .94 Newtons which is the bouyant force
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11:14:58 At a depth of 12 cm, the volume of the immersed portion will be 12 cm * 8 cm^2 = 96 cm^3. This portion will therefore displace 96 grams of water. The weight of this displace water will be .096 kg * 9.8 meters/second^2 = .94 Newtons. This will be the buoyant force on the cylinder.
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RESPONSE --> ok
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11:18:01 `q003. The solid cylinder in the preceding problem has a total length of 18 cm and a mass of 80 grams. If the cylinder is immersed as before to a depth of 12 cm then released, what will be the net force acting on it at the instant of release?
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RESPONSE --> Gravity exerts a downward force of .080 kg * 9.8 m/s^2 = .79 Newtons The net force on the cylinder = .94 N - .79 N = .15 N upward a = F / m = .15 N / .080 kg = 1.875 m/s^2
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11:18:05 The buoyant force on the cylinder is still .94 Newtons, directed upward. Gravity exerts a downward force of .080 kg * 9.8 meters/second^2 = .79 Newtons, approximately. The net force on the cylinder is therefore .94 N - .79 N = .15 N, directed upward. This will give its 80 gram mass and acceleration a = F / m = .15 N / .080 kg = 1.875 m/s^2. Note that as the cylinder rises less and less of its volume is submerged, so the buoyant force will decrease while the weight remains the same. Until the buoyant force has decreased to become equal and opposite to the weight, the net force will continue to be upward and the cylinder will continue to gain velocity. After this instant the cylinder will continue to rise, but the net force will be downward so that the cylinder will begin slowing down. Eventually the cylinder will come to rest and the net downward force will cause it to start descending once more. It will continue descending until the net force is again 0, at which the time it will have a downward velocity that will carry it beyond this point until it again comes to rest and the cycle will start over again.
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RESPONSE --> ok
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