course Phy 121 œõ×ò¸Šý—쀀·N¬þúgûɾ´¯Å—Student Name:
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17:56:21 `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 --> a = v^2 / r a = (3 m/s)^2 / 5 m a = 9 m^2/s^2 / 5 m a = 1.8 m/s^2 F = m(a) F = 12 kg (1.8 m/s^2) F = 21.6 N
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17:56:55 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, this question was pretty easy.
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18:02:24 `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 --> r = 70 cm / 2 r = 35 cm r = .35 m m = 50 g m = .05 kg F = m(a) 25 N = .05kg (a) 500 m/s^2 = a a = v^2 / r 500 m/s^2 = v^2 / .35 m 175 m^2/s^2 = v^2 13.229 m/s = v
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18:04:34 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 --> On this problem, I thought that you had to divide 70 by 2 to get the radius but I now see that 70 is the radius since you are spinning the string in a circle. Also, I learned how to put the two equations that I used earlier into one to save some steps in my work.
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18:09:20 `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 --> 25 kg m/s^2 / 18.7 m/s 1.337 kg 1.337 kg / .7 kg = 1.91 times
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18:10:59 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 --> This makes sense now but before I was confused on which formula to use.
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18:13:07 `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 --> A force is required to keep a mass traveling any circular path because it is used to keep changing the direction of motion. When the direction of motion is changed, a change in velocity also happens. The force is directed toward the center of the circle.
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18:14:12 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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