course phy 201 ?????????????????assignment #024
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21:34:51 `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 --> m=12kg v = 3m/s r = 5m a = v^2/r a = 9/5 = 1.8m/s/s F=ma F = 12kg * 1.8 F = 21.6J confidence assessment: 1
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21:36:01 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 --> I am a little hazy on the equations for gravitational field and when that equation will be used. self critique assessment: 2
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21:39:25 `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 = 50g r =.07m F = 25N F = ma a = v^2/r F = m (v^2/r) v^2 = Fr/m v = sqrt 25N .07m / .5kg v = 1.8m/s confidence assessment: 2
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21:44:00 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 --> I understand the concepts much more through the equations than the notes it is easier to follow the steps by remembering former equations and relations. I still revert back to my bad habit of not checking my conversions. self critique assessment: 2
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21:49:20 `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 --> v = 18.1m/s Fnet * dt = mv 25 * dt = .05 * 18.1 dt = .03 sec 18.1 * .03 = .5 turn no clue confidence assessment: 1
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21:56:03 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 --> sphere = 2pir r = .7m v = 18.1 area would be 2*3.14 *.7 = 4.4 18.7/4.4 = 4.25 still unsure about this problem why would the diameter not be used instead of the area? self critique assessment: 1
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21:58:04 `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 --> force is made up of vectors. As the object moves it has forces keeping it from flying off into space and another is required to keep it in orbit. These parallel and perpindicular forces are both required to keep the object in motion. confidence assessment: 2
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22:01:02 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 --> force is used to hold the object in orbit such as a satellite but keep a certain velocity so that it doesnt accelerate too much and fly out of orbit or too slow so gravity takes over and the obj. falls back towards earth self critique assessment: 3
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