- **Introductory Physics Homework**
(*http://www.physicsforums.com/forumdisplay.php?f=153*)

- - **Falling string: Force and momentum changing over time**
(*http://www.physicsforums.com/showthread.php?t=307422*)

Falling string: Force and momentum changing over time1. The problem statement, all variables and given/known dataConsider a flexible string of length L and mass M that is held vertically so that its bottom just touches the floor. The string is then dropped. Let the position of the top of the string be y and the position of the floor be y = 0. 1.1 Is every piece (for y > 0) of the string moving at the same speed when it falls (you might want to consider whether the string remains straight)? 1.2 Write the mass of the string above the floor in terms of y, L, M and g. 1.3 Write the momentum of the string at time t in terms of y, dy/dt, L, M and g. 1.4 Write the momentum of the string at a time t + dt using the variables above. Note that both the length of the string and its velocity are changed from what they are at time t. 1.5 Write external force on the string at time t in terms of y, L, M and g. 1.6 Using the time rate of change of momentum determined above in Newton's 2nd law, write the equation for the second derivative of y with respect to time, d2y/dt2. 1.7 What is the normal force exerted by the floor on the rope (note that it depends on time)? 1.8 What is the acceleration of the string when it is first released? What is the acceleration of the string (portion above the floor) just prior to the last part hitting the floor? What is the velocity of the string at this point? 2. Relevant equationsp=mv F=ma v=dy/dt F=dp/dt 3. The attempt at a solution1.1 Yes, every piece is moving at the same speed as it falls. The string stays straight. 1.2 This seems easy enough: m=M*(y/L) where M is the total mass of the string, and y/L is the ratio of the string's height above the floor to its total length. 1.3 I'm not so sure about this one, but it seems like since p=mv, we would have: p=M*(y/L)*(dy/dt) Taking this a step farther, we can deduce that since dy/dt=v, and v=v_0+a*t, then dy/dt=g*t, where g is acceleration due to gravity. I'm not sure if that's right, but it seemed to make sense, and it gave me this: p=g*t*M*(y/L) 1.4 I'm confused about this question. Do they just want me to substitute t+dt in for t and show that y changes by some dy as well? I don't understand... 1.5 Well I think F=dp/dt is involved here. Would I just differentiate my solution for 1.3 with respect to time? Also, would I have to do anything special with y, since it is changing also, or can I leave it alone? Then at this point I don't think I can even speculate on parts 1.6 through 1.8 until I get the first five parts taken care of. Thanks for the help. |

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