course Phy 231 7/9 9:30 024. Centripetal Acceleration Question: `q001. Note that this assignment contains 4 questions.
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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): OK Self-critique Assessment: OK ********************************************* 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? Your solution: 25 N = .05kg * a A = 500 m/s^2 500 m/s^2 = v^2 / .7m V^2 = 350m^2/s^2 V = 18.7m/s confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The centripetal acceleration as speed v will be v^2 / r, where r = 70 cm = .7 meters. The centripetal force will therefore be m v^2 / r, where m is the 50 g = .05 kg mass. If F stands for the 25 Newton breaking force, then we have m v^2 / r = F, which we solve for v to obtain v = `sqrt(F * r / m). Substituting the given values we obtain v = `sqrt( 25 N * .7 meters / (.05 kg) ) = `sqrt( 25 kg m/s^2 * .7 m / (.05 kg) ) = `sqrt(350 m^2 / s^2) = 18.7 m/s. Self-critique (if necessary): OK Self-critique Assessment: OK ********************************************* Question: `q003. What is the maximum number of times per second the mass in the preceding problem can travel around its circular path before the string breaks? Your solution: Velocity = 18.7m/s Radius = .7m Circumference = 2pi*.7 = 4.398 m 18.7m/s = 4.398m (x) X = 4.25 rotations / second confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK Self-critique Assessment: OK ********************************************* Question: `q004. Explain in terms of basic intuition why a force is required to keep a mass traveling any circular path. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: You always have to have a force to make an object’s path curved. A straight line path does not require any force; Newton’s first law is that an object will keep moving in the direction it was going before until a force acts upon it. In a circular path, the direction of the position is always changing and so is the velocity and acceleration, so there has to be a force on it. confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK Self-critique Assessment: OK "