Saturday, August 11, 2007

Irodov Problem 1.76





















I will solve this problem using two methods i) using conservation of energy and ii) newton's laws just because I love it when two different elegant approaches of explaining the world yield the same results.

Solution Using Energy Conservation
------------------------------------------- Let us first understand how the two masses move in relation to each other. The different sections of the string are marked as AB, CD, EF are depicted in the figure. Suppose, that the mass M moves down by x units, the moving pulley moves down by x units too. Thus, AB increases to AB' = AB + x. Also CD' = CD + x. Now the entire length of the string must be conserved thus AB + CD + EF = AB' + CD' + EF'. Thus, EF - EF' = 2x. In other words, if M
moves up by x units, then m moves by down 2x units. Thus, at any moment m moves twice as fast upwards as M moves downwards.

Initially the potential energy of the system is Mgh since the and the kinetic energy is 0 since none of the bodies are moving. This is total energy of the system and will be conserved until all the bodies are moving freely. When the mass M reaches the ground let the velocity of the mass be v. The velocity of m at this point will be 2v moving upwards. Also by this time the mass m moves up a distance of 2h. At this time, the mass M has completely lost all its potential energy while the mass m gains a potential energy of mg(2h). Thus we have,







The kinetic energy of m when M hits the floor is given by,





As soon as M hits the floor it stops abruptly and looses all its kinetic energy immediately - this energy is completely absorbed by the floor. From this point onwards, the mass m starts to move freely under the influence of only gravity and its total energy is conserved. Let H be the height m travels up from this point onwards before it starts to fall down under the influence of gravity. At the highest point all the kinetic energy has been converted into potential energy. Thus,







The total height is given by 2h + H given by,







Solution Using Newton's Laws
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Let the tension in the string holding M be T' and let the tension in the string holding m be T. Let the acceleration of M be w downwards then the acceleration of m will be 2w upwards.

Forces on M:
There are two forces acting on M, i) the tension in the string holding it T' and the ii) the force of gravity Mg. The dynamics is given by,



Forces on m: There are two forces acting on m, i) the tension in the string T and ii) the force of gravity mg. The dynamics are thus given by,



Forces acting on the moving pulley: There are three forces acting on the moving pulley, i) the tension T in the two parts of the string holding the mass m and ii) the tension T' in the string holding the mass M. The Pulley accelerates same as M at a rate w downwards and its mass is 0. Thus we have,




Now we have all the information need to solve for the system. From (5) and (3) we have,



From (4) and (6) we obtain,





The acceleration of the mass m is given by 2w. At the time mass M hits the floor, m has moved
2h units upwards. Its velocity at that time can be obtained as






After this point there is not tension in any of the strings and the mass m moves freely under the influence of gravity thus it decelerates at a rate g. The extra height traveled by m under this influence of gravity is given by,





Thus, the total height reached by m when it starts to fall down is given by,







Isn't it nice that law of conservation of energy and newton's laws are in complete agreement!

Irodov Problem 1.75






















Let M and m be the masses of the two rod and the bead respectively, as shown in the figure. Let be the tension in the string holding the ball and be the tension in the string holding the rod. Further let, be the upward acceleration of the ball and the downward acceleration of the rod.
Now let us consider the forces acting on the bodies.


Forces on the rod: There are two forces on the rod i) the force of gravity pulling it down Mg and the tension in the string pulling it up. The acceleration of the rod is downwards. Hence we have,




Forces on the bead: There are two forces acting on the bead, i) the gravity mg pulling it down and ii) the tension in the string pulling it up. The acceleration of the bead is in the upward direction. Hence we have,




Forces on the moving pulley: There are three forces acting on the pulley, i) the tensions in the two parts of the string pulling it down and ii) the tension in the string connecting the bead pulling it up. The Pulley has no mass and accelerates at a rate downwards (since the bead moves with an acceleration upwards). Thus, we have,



























Relating the accelerations
and : The acceleration of the bead and the rod are not independent of each other. The Figure shown snapshots of the system at two instants of time. AB, CD, EF and GH are the length of various sections of the strings initially and A'B, CD', E'F' and G'H are the corresponding sections of strings after some time. Suppose that the bead moves x units upwards. Then A'B = AB-x. Since the total length of the string AB + CD must be conserved, C'D' = CD +x in other wards the pulley moves x units down. Since the pulley moves x units down
the length GH reduces by x units or G'H = GH-x. Since the total length of this string GH+EF must be conserved, E'F' = EF + x. The net distance that the rod moves is the sum of distance moved by the pulley holding it and the increase in length EF. This is equal to 2x. In other words if the ball moves a distance x upwards, the rod moves 2x downwards. Thus, we have,


Now we have everything need to solve the problem. From (4) and (1) we have,



From (3) and (2) we have,



From (5) and (6) we have,





The relative acceleration of the two bodies is thus, given by,




The time takes to cross a distance of l (the length of the rod) is given by,