course Phy 201 »\Ü{úÛw°û`›ò¼bˤb¤T„ùæàÊÀassignment #024
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20:04:58 `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 --> If we use v^2/r we get the magnitude of (3m/s)^2/5m=1.8m/s^2. We now find he F=m*a which gives us, Fcentripetal=12kg*1.8m/s^2,Fcent=21.6newtons confidence assessment: 2
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20:05:20 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 self critique assessment: 3
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20:14:27 `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 --> Since F=ma=m(v^2/r) we can use this equation to solve for v.We mulitply both sides by we get mv^2=F*r then divide by m we get v^2=F*r/m then the sqrt we get v=sqrt(F*r/m. We fill in the equation and get, v=sqrt(25N*.7m/.05kg),v=18.7m/s confidence assessment: 3
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20:15:30 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 self critique assessment: 3
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20:24:04 `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 we know that the circumference=2*pi*r and we know the velocity we can solve. 2*3.14*.7m=4.4m. The velocity is 18.7m/s and if we divide this by the distance around the circle we get 18.7m/s/4.4m=4.3 times confidence assessment: 1
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20:26:10 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 self critique assessment: 3
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20:29:47 `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 --> Lets say if you have a motorcycle riding in a circular motion around the inside diameter of a ball. The motorcycle must keep a certian speed to resist the earths gravity. If he goes to slow he will fall to the botttom of the ball confidence assessment: 3
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20:30:26 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 self critique assessment: 3
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