#$&* course PHY 201 11:25 AM on 04/4/13 024. Centripetal Acceleration
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Given Solution: The centripetal acceleration of the object is v^2 / r = (3 meters/second) ^ 2/(5 meters) = 1.8 meters/second ^ 2. The centripetal force, by Newton's Second Law, must therefore be Fcent = 12 kg * 1.8 meters/second ^ 2. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `q002. How fast must a 50 g mass at the end of a string of length 70 cm be spun in a circular path in order to break the string, which has a breaking strength of 25 Newtons? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: m = 50 g * .001 kg / 1 g = 0.05 kg F = 25 N radius = 70 cm = .7 m F = ma = m (v)^2 / r 25 kg*m/s^2 = 0.05 kg * v^2 / 0.7 m 0.7 kg * 25 kg*m/s^2 = 0.05 kg * v^2 => 17.5 kg^2*m/s^2 = 0.05 kg * v^2 v = sqrt((17.5 kg^2*m/s^2) / 0.05 kg) = 18.7 m/s confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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: The formula for a revolution in circle is 2*pi*R/T where R is the radius, and T is the period. T = 2*pi*R = 2*3.142*0.7m = 4.4 m I know I must divide the speed of 18.7 m/s by this revolution of 4.4 m to cancel the meters out and be left with a time of seconds only. (18.7 m/s) / 4.4 m = 4.25 s confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The maximum possible speed of the mass was found in the preceding problem to be 18.7 meters/second. The path of the mass is a circle of radius 70 cm = .7 meters. The distance traveled along this path in a single revolution is 2 `pi r = 2 `pi * .7 meters = 4.4 meters, approximately. At 18.7 meters/second, the mass will travel around the circle 18.7/4.4 = 4.25 times every second. STUDENT COMMENT: I read through the solution but still wouldn't be able to solve this. INSTRUCTOR RESPONSE The question comes down to this: At 18.7 m/s (the result found in the preceding), how many times will the mass travel around a circle of radius .7 meters in 1 second? The circumference of the circle is about 4.4 meters, so at 18.7 m/s the object will go around the circle a little over 4 times in 1 second. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `q004. Explain in terms of basic intuition why a force is required to keep a mass traveling any circular path. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: A ""force"" is required to keep a mass traveling in a circular path because an object/body wants to remain traveling in a straight line. The centrifugal force will try to move the body in the direction of the radius of the path. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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: If you understand the assignment and were able to solve the previously given problems from your worksheets, you should be able to complete most of the following problems quickly and easily. If you experience difficulty with some of these problems, you will be given notes and we will work to resolve difficulties. ********************************************* Question: `q004. The centripetal force on a 0.2 kg mass, as it travels around a circle of radius 80 cm, is 12 Newtons. How fast is it moving? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: m = 0.2 kg, r = 80 cm = 0.8 m, F = 12 N 12 kg*m/s^2 = 0.2 kg * (v^2) / 0.8 m 0.8 m * 12 kg*m/s^2 = 0.2 kg(v^2) 9.6 kg*m^2/s^2 = 0.2 kg(v^2) v = sqrt((9.6 kg*m^2/s^2 / 0.2 kg)) = 6.9 m/s confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: