Question
Consider an object with mass m=5kg dropped from a high height.Draw 2 free body diagrams one in which air resistance is neglectedand the other where aerodynamic drag is not neglected. For thefirst case show that accleration is a constant and integrate to getvelocity as a function of time. For the second case assume thatdrag force Fdrag=0.02v2 (in N when v is inm/s). Show that accelration is neither a constant nor a functionthat can be easily integrated with respect to time. part 2 is to use MATLAB to solve accleration equation in thepresence of drag from part 1 to get the velocity as function oftime. I think I can do this once I havepart 1 so ignore part 3 modifyprogram to compute the position as a function of timeCould use help in setting thisup Consider an object with mass m=5kg dropped from a high height.Draw 2 free body diagrams one in which air resistance is neglectedand the other where aerodynamic drag is not neglected. For thefirst case show that accleration is a constant and integrate to getvelocity as a function of time. For the second case assume thatdrag force Fdrag=0.02v2 (in N when v is inm/s). Show that accelration is neither a constant nor a functionthat can be easily integrated with respect to time. part 2 is to use MATLAB to solve accleration equation in thepresence of drag from part 1 to get the velocity as function oftime. I think I can do this once I havepart 1 so ignore part 3 modifyprogram to compute the position as a function of timeCould use help in setting thisup
Explanation / Answer
First FBD is just oneforce: mg down. . Then using Newtons secondlaw: totalforce =ma mg = ma g = a . (acceleration is constant... a = g). . To get velocity as a function of time, justuse v = a dt = g dt = g t . So: v = g t . In other words, part 1 is very simple. . For part 2, there are two forces on theFBD: drag up, mgdown. Use down as positive direction,and... . total of forces = ma . mg - 0.2 v2 = ma . 5 * 9.8 - 0.2 v2 = 5 * a . a = 9.8 - 0.04v2 (not constant) . Thenstill... v = a dt = ( 9.8 - 0.04 v2 )dt . which is difficult to integrate in this form because vis a function of t... and you need it in the integral, but you aretrying to solve for it. . But this is weird... because I dont understand whythey're doing this to you. The velocity can easily be determined byusing: . a = dv / dt . 9.8 - 0.04 v2 = dv/dt a separable differential equation... . dt = dv / (9.8 -0.04v2) . dt = dv / (9.8 -0.04v2 ) . Anyway... t = dv / g ( 1 - b2 v2) where g = 9.8 and b2 = 0.004082 . t = (1/g) (1 -b2v2 )-1 dv . You can plug the numbers back infor g and b if you need to enter it in matlab. Once itsolves the integral on the right side, you will have t as a function of v. You can then solvefor v and you'll have v as a function of t. . To get part 3... position as a function oftime is given by: . x = v dt where v is the function from part 2. Note that theposition will be in meters below the starting point. . Thenstill... v = a dt = ( 9.8 - 0.04 v2 )dt . which is difficult to integrate in this form because vis a function of t... and you need it in the integral, but you aretrying to solve for it. . But this is weird... because I dont understand whythey're doing this to you. The velocity can easily be determined byusing: . a = dv / dt . 9.8 - 0.04 v2 = dv/dt a separable differential equation... . dt = dv / (9.8 -0.04v2) . dt = dv / (9.8 -0.04v2 ) . Anyway... t = dv / g ( 1 - b2 v2) where g = 9.8 and b2 = 0.004082 . t = (1/g) (1 -b2v2 )-1 dv . You can plug the numbers back infor g and b if you need to enter it in matlab. Once itsolves the integral on the right side, you will have t as a function of v. You can then solvefor v and you'll have v as a function of t. . To get part 3... position as a function oftime is given by: . x = v dt where v is the function from part 2. Note that theposition will be in meters below the starting point.