System with varying mass - String

Click For Summary
SUMMARY

The discussion focuses on a physics problem involving a string of mass M and length L that is partially hanging off a table. The key equations derived include the velocity of the string as a function of the length under the table, v = gt + C - (g/L)yt, and the length of the string as a function of time, y = (1/L)(C/g - v/g)t. The velocity when the entire string leaves the table is expressed as v = gt + C, and the time for the complete departure of the string is t = (C/g)L. These equations are derived using principles from classical mechanics and differential equations.

PREREQUISITES
  • Understanding of classical mechanics principles, particularly momentum and forces.
  • Familiarity with differential equations, specifically the form dy/dx + y P(x) = Q(x).
  • Knowledge of integration techniques and initial condition applications.
  • Basic concepts of kinematics, including velocity and acceleration.
NEXT STEPS
  • Study the application of differential equations in physics problems.
  • Explore the concepts of momentum conservation in variable mass systems.
  • Learn about the integration of functions in the context of motion equations.
  • Investigate similar problems involving strings and pulleys to reinforce understanding.
USEFUL FOR

Students and educators in physics, particularly those focusing on mechanics and differential equations, as well as anyone interested in solving dynamic systems involving variable mass.

Microzero
Messages
21
Reaction score
0
System with varying mass --- String

A string of mass M and length L is placed near a hole on the top of a horizontal and smooth table. A slight disturbance is given to one end of the string at time t = 0 so that it leaves the table through the hole. Assume the string on the table remains at rest while the remaining part is moving down. Denote the length of the string under the table as y and the speed of the string that is moving down by v.

Find:
1. v as a function of y
2. y as a function of t
3. the velocity of string when the whole string leaves the table
4. the time when the whole string leaves the table

[ Use the D.E. dy/dx + y P(x) = Q(x) to solve the problem. ]

I don't know how to do the first 2 questions. Please give me some ideas.
~ Thank you ~
----------------------------------------------
Here is my idea:
By considering the momentum
Fext= M dv/dt + (v-u) dM/dt -----★
(Resnick, Halliday--Physics 4th edt., Ch.9)
Fext= Mg , u=0 (remaining part rest at table)
dM/dt = ρdy/dt
∴ ★ becomes :
dv/dt + v(1/L)dy/dt = g
 
Last edited:
Physics news on Phys.org
integrating this equation, we get:v = gt + C where C is an integration constant. From the initial condition, we can obtain v = gt when y=0 Therefore, we get v = gt + C - (g/L)yt 1. v as a function of y : v = gt + C - (g/L)yt 2. y as a function of t : y = (1/L)(C/g - v/g)t 3. The velocity of string when the whole string leaves the table : v = gt + C 4. Time when the whole string leaves the table : t = (C/g)L
 

Similar threads

  • · Replies 27 ·
Replies
27
Views
2K
Replies
13
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
Replies
7
Views
3K
Replies
23
Views
3K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 7 ·
Replies
7
Views
4K
Replies
17
Views
2K