System with varying mass - String

AI Thread 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 points include deriving the velocity of the string as it moves downward, expressed as a function of the length under the table (y) and time (t). The momentum approach leads to the equations v = gt + C - (g/L)yt for velocity and y = (1/L)(C/g - v/g)t for length over time. The final velocity when the entire string leaves the table is also addressed, along with the time it takes for this to occur. The problem emphasizes the application of differential equations to analyze the system's dynamics.
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
 
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
Back
Top