#$&* course PHY 121 If your solution to stated problem does not match the given solution, you should self-critique per instructions at
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Given Solution: The centripetal acceleration of the object is v^2 / r = (3 meters/second) ^ 2/(5 meters) = 1.8 meters/second ^ 2. The centripetal force, by Newton's Second Law, must therefore be Fcent = 12 kg * 1.8 meters/second ^ 2. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `q002. How fast must a 50 g mass at the end of a string of length 70 cm be spun in a circular path in order to break the string, which has a breaking strength of 25 Newtons? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Velocity= sqrt( 25 N * .7 meters / (.05 kg) ) =sqrt( 25 kg m/s^2 * .7 m / (.05 kg) ) =sqrt(350 m^2 / s^2) = 18.7 m/s confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 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. STUDENT COMMENT: I read through the solution but still wouldn't be able to solve this. INSTRUCTOR RESPONSE The question comes down to this: At 18.7 m/s (the result found in the preceding), how many times will the mass travel around a circle of radius .7 meters in 1 second? The circumference of the circle is about 4.4 meters, so at 18.7 m/s the object will go around the circle a little over 4 times in 1 second. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I definitely am not understanding this at all. Even with the breakdown in the instructor response. I understand the max speed. But where does the 2 pi r equation come in ? and how do I know to use this then? ------------------------------------------------ Self-critique rating:3
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Given Solution: 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. STUDENT RESPONSE (good intuition but statement isn't quite right) Something has to keep the momentum going for anything in a circular path to continue. Otherwise, it will fly off in a vector. INSTRUCTOR CRITIQUE Nothing is required to keep something moving in a straight line; in the absence of a force it will maintain its momentum, in the same direction as the original. The force is required to cause the object to deviation from its 'natural' straight-line motion. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!