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Applied Maths - Mechanics 46
A pulley of mass m is suspended from a fixed point by an elastic spring of modulus 12mg and natural length a. A light inextensible string passes over the pulley and carries particles of mass m and 2m, one attached to each end. Initially the system is held at rest with the elastic spring extended a distance 0.5a.... 顯示更多 A pulley of mass m is suspended from a fixed point by an elastic spring of modulus 12mg and natural length a. A light inextensible string passes over the pulley and carries particles of mass m and 2m, one attached to each end. Initially the system is held at rest with the elastic spring extended a distance 0.5a. When the system has been released and the elastic spring is of length a+x, determine an expression in terms of g, a and x for the acceleration of the pulley. Show that its motion is simple harmonic and find the period of the oscillations. 更新: Ans : d^2 x / dt^2 = g - 36gx/11a ; T = pi/3 (11a/g)^0.5
最佳解答:
http://www.flickr.com/photos/27778998@N04/3825187262/ Let A be the acceleration of pulley. Let Ar be the relative acceleration of m and 2m respect to the pulley. Let T be the tension in the string. Consider 2m, 2mg - T = 2m(Ar - A) Ar = [(2mg - T)/(2m) + A]...............(1) Consider m, T - mg = m(A + Ar)...................(2) Sub (1) into (2), T - mg = m[(2mg - T)/(2m) + A + A] T = 4m(g+A)/3...........................(3) Consider the pulley, -kx - 2T - mg = mA, where k = 12mg/a -12mgx/a - 2T - mg = mA.................(4) Sub (3) into (4), -12mgx/a - 2[4m(g+A)/3] - mg = mA A = - 36gx/(11a) - g Changing the variable x with h, where h is the displacement from the equilibrium position. We have, A = - 36gh/(11a) So, A is proportional to h. i.e. The motion is S.H.M. Compare A= - ω2 h ω = √[36g/(11a)] Period = 2(pi)/ω = 2(pi)√[11a/(36g))] = [(pi)/3]√(11a/g) 2009-08-16 10:41:14 補充： The acceleration of the pulley is different from your ans. I don't whether I am correct 2009-08-16 11:05:07 補充： I think I was wrong before. here comes a correct version: http://www.flickr.com/photos/27778998@N04/3825265506/ 2009-08-16 11:05:17 補充： Let A be the acceleration of pulley. Let Ar be the relative acceleration of m and 2m respect to the pulley. Let T be the tension in the string. Consider 2m, 2mg - T = 2m(Ar + A)....................(1) Consider m, mg - T = m(A -Ar)...................(2) 2009-08-16 11:05:25 補充： Solving (1) and (2), we have T = 4m(g - A)/3.............(3) Consider the pulley, -kx + 2T + mg = mA, where k = 12mg/a -12mgx/a + 2T + mg = mA.................(4) 2009-08-16 11:05:34 補充： Sub (3) into (4), -12mgx/a + 2[4m(g - A)/3] + mg = mA A = g - 36gx/(11a) Changing the variable x with h, where h is the displacement from the equilibrium position. We have, A = - 36gh/(11a) So, A is proportional to h. i.e. The motion is S.H.M. 2009-08-16 11:05:42 補充： Compare A= - ω2 h ω = √[36g/(11a)] Period = 2(pi)/ω = 2(pi)√[11a/(36g))] = [(pi)/3]√(11a/g) 2009-08-16 11:32:33 補充： Ignore the old version, it is WRONG.
其他解答:
No diagram for this question .|||||I do not understand the question, is a diagram available? 2009-08-16 11:39:05 補充： Taking the downward direction as positive. Let the position of mass m below the hanging point be y Let the position of mass 2m below the hanging point be z Let the tension in the inelastic string be T 2009-08-16 11:39:45 補充： Length of the string = y – (a + x) + z (a + x) is a constant Therefore d2y/dt2 + d2z/dt2 = 2d2x/dt2 … (1) The tension in the spring is -12mgx/a 2009-08-16 11:40:00 補充： Force equations of the pulley, m and 2m respectively are: -12mgx/a + 2T + mg = md2x/dt2 … (2) mg – T = md2y/dt2 … (3) 2mg – T = 2nd2z/dt2 … (4) 2009-08-16 11:40:13 補充： (3) => d2y/dt2 = g – T/m … (5) (4) => d2z/dt2 = g – T/2m … (6) Combining (1), (5) and (6) => g – T/m + g – T/2m = 2d2x/dt2 2mg – 2T + 2mg – T = 4md2x/dt2 T = 4mg/3 – (4m/3)d2x/dt2 … (7) 2009-08-16 11:40:23 補充： Sub (7) into (2), -12mgx/a + 2[4mg/3 – (4m/3)d2x/dt2] + mg = md2x/dt2 -12gx/a + 8g/3 – (8/3)d2x/dt2 + g = d2x/dt2 -12gx/a + 11g/3 = (11/3)d2x/dt2 d2x/dt2 = g – 36gx/11a => SHM Period = 2π√(11a/36g) = (π/3)√(11a/g)
Applied Maths - Mechanics 46
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發問:A pulley of mass m is suspended from a fixed point by an elastic spring of modulus 12mg and natural length a. A light inextensible string passes over the pulley and carries particles of mass m and 2m, one attached to each end. Initially the system is held at rest with the elastic spring extended a distance 0.5a.... 顯示更多 A pulley of mass m is suspended from a fixed point by an elastic spring of modulus 12mg and natural length a. A light inextensible string passes over the pulley and carries particles of mass m and 2m, one attached to each end. Initially the system is held at rest with the elastic spring extended a distance 0.5a. When the system has been released and the elastic spring is of length a+x, determine an expression in terms of g, a and x for the acceleration of the pulley. Show that its motion is simple harmonic and find the period of the oscillations. 更新: Ans : d^2 x / dt^2 = g - 36gx/11a ; T = pi/3 (11a/g)^0.5
最佳解答:
http://www.flickr.com/photos/27778998@N04/3825187262/ Let A be the acceleration of pulley. Let Ar be the relative acceleration of m and 2m respect to the pulley. Let T be the tension in the string. Consider 2m, 2mg - T = 2m(Ar - A) Ar = [(2mg - T)/(2m) + A]...............(1) Consider m, T - mg = m(A + Ar)...................(2) Sub (1) into (2), T - mg = m[(2mg - T)/(2m) + A + A] T = 4m(g+A)/3...........................(3) Consider the pulley, -kx - 2T - mg = mA, where k = 12mg/a -12mgx/a - 2T - mg = mA.................(4) Sub (3) into (4), -12mgx/a - 2[4m(g+A)/3] - mg = mA A = - 36gx/(11a) - g Changing the variable x with h, where h is the displacement from the equilibrium position. We have, A = - 36gh/(11a) So, A is proportional to h. i.e. The motion is S.H.M. Compare A= - ω2 h ω = √[36g/(11a)] Period = 2(pi)/ω = 2(pi)√[11a/(36g))] = [(pi)/3]√(11a/g) 2009-08-16 10:41:14 補充： The acceleration of the pulley is different from your ans. I don't whether I am correct 2009-08-16 11:05:07 補充： I think I was wrong before. here comes a correct version: http://www.flickr.com/photos/27778998@N04/3825265506/ 2009-08-16 11:05:17 補充： Let A be the acceleration of pulley. Let Ar be the relative acceleration of m and 2m respect to the pulley. Let T be the tension in the string. Consider 2m, 2mg - T = 2m(Ar + A)....................(1) Consider m, mg - T = m(A -Ar)...................(2) 2009-08-16 11:05:25 補充： Solving (1) and (2), we have T = 4m(g - A)/3.............(3) Consider the pulley, -kx + 2T + mg = mA, where k = 12mg/a -12mgx/a + 2T + mg = mA.................(4) 2009-08-16 11:05:34 補充： Sub (3) into (4), -12mgx/a + 2[4m(g - A)/3] + mg = mA A = g - 36gx/(11a) Changing the variable x with h, where h is the displacement from the equilibrium position. We have, A = - 36gh/(11a) So, A is proportional to h. i.e. The motion is S.H.M. 2009-08-16 11:05:42 補充： Compare A= - ω2 h ω = √[36g/(11a)] Period = 2(pi)/ω = 2(pi)√[11a/(36g))] = [(pi)/3]√(11a/g) 2009-08-16 11:32:33 補充： Ignore the old version, it is WRONG.
其他解答:
No diagram for this question .|||||I do not understand the question, is a diagram available? 2009-08-16 11:39:05 補充： Taking the downward direction as positive. Let the position of mass m below the hanging point be y Let the position of mass 2m below the hanging point be z Let the tension in the inelastic string be T 2009-08-16 11:39:45 補充： Length of the string = y – (a + x) + z (a + x) is a constant Therefore d2y/dt2 + d2z/dt2 = 2d2x/dt2 … (1) The tension in the spring is -12mgx/a 2009-08-16 11:40:00 補充： Force equations of the pulley, m and 2m respectively are: -12mgx/a + 2T + mg = md2x/dt2 … (2) mg – T = md2y/dt2 … (3) 2mg – T = 2nd2z/dt2 … (4) 2009-08-16 11:40:13 補充： (3) => d2y/dt2 = g – T/m … (5) (4) => d2z/dt2 = g – T/2m … (6) Combining (1), (5) and (6) => g – T/m + g – T/2m = 2d2x/dt2 2mg – 2T + 2mg – T = 4md2x/dt2 T = 4mg/3 – (4m/3)d2x/dt2 … (7) 2009-08-16 11:40:23 補充： Sub (7) into (2), -12mgx/a + 2[4mg/3 – (4m/3)d2x/dt2] + mg = md2x/dt2 -12gx/a + 8g/3 – (8/3)d2x/dt2 + g = d2x/dt2 -12gx/a + 11g/3 = (11/3)d2x/dt2 d2x/dt2 = g – 36gx/11a => SHM Period = 2π√(11a/36g) = (π/3)√(11a/g)
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