course Phy 121 I could not get Query 24 to run properly, and that is why it is not included with this part of the assignment. I have discussed this with you via e-mail, and as soon as we can get it to work I will send the remainder of the assignment. 琤の涗咐悲樊隇咱皯j阛ssignment #024
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19:41:40 `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 --> P. 109 of textbook Fr = ma(sub r) = m (v^2/r) Therefore: 12 kg (3m/s^2/ 5m) = 22 N confidence assessment: 1
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19:43:08 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 think I got the correct answer. self critique assessment: 2
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19:49:57 `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 --> F (sub r) = m (v^2 / r) Since the length of the string is 70 cm, the radius is half the diameter which gives a radius of 35 cm. Plug what we know into the formula. 25 N = 50 g (v^2 / 35cm) Solve for v^2. 25 N = 50v^2 / 1750 Multiply both sides by 1750 to get it to cancel on the right side of the equation. 43750 = 50v^2 Divide both sides by 50. 875 = v^2 Take the square root of both sides. 30 m/s = v confidence assessment: 1
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19:51:01 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 --> Well I was at least on the right track. I didn't convert to meters, which was part of the problem but I think I did everything else correctly. self critique assessment: 2
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20:00:34 `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 --> Since I didn't get the correct solution for the previous problem it makes it difficult to answer this one. However, I suspect it will take a combination of the following formulas: T = 1 / f v = 2pi r / T confidence assessment: 0
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20:01:06 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 --> I am pretty sure I had the correct approach. self critique assessment: 2
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20:09:22 `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 --> Concerning a very basic explanation for an object traveling in a circular path, consider Newton's first law. This object will continue to function as it is currently functioning as long as no other forces act upon the object. The subtraction or addition of any forces from the current situation will cause the object to deviate from its current path. confidence assessment: 1
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20:09:49 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 --> I think my answer is at least partially correct. self critique assessment: 2
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