A problem from giamcoli ( it is )

[oahsen]

a problem from giamcoli ( it is urgent )

i am trying to solve this problem for several days. pls help me. it is from giancoli 3. edith.
a heavy steel cable of length L and mass M passes over a small massless,frictionless pulley. a)if a length y ahngs on one side of the pulley ( so L-y hangs on the other side) calculate the acc of the cable as a fuction of y b)calculate the vf of the cable as a function of y( y is changing)... (the book gives a hint that "use the chain rule and integrate)

i ve found the acc= g*(2*m*y-m*L)/m*L ( i hope it is so )
but how can i find the V... i know a=dv/dt but y,v,t is not constant there and i don't know what i do with 3 non-onstant terms?

 

[BerryBoy]

Steel Cable Question

a = g \frac{2y - L}{L} (your masses cancel)

I made the substition off 'acc'. Differentiate it with respect to time and then substitute your 'dt'. You'll should get a \delta t / \delta y term in this; but this is \frac{1}{v}, so now your integral will calculate v^2.

 

[oahsen]

thank you very much but;

thanks but i haven't unsdetstood well what you wrote. could u give a bit more explanation?

 

[OlderDan]

I'm not sure if this is what BerryBoy was saying, but maybe it is what he meant

a = dv/dt = (dv/dy)(dy/dt) = v(dv/dy)

You can use this to separate variables and integrate.

 

[oahsen]

i understood

i understood what you say. and i found this answer( i hope it is true);
v=(((2*y*y*g)/L)-2*g*y))^1/2...
thanks berrboy and olderdan.

 

[BerryBoy]

Maybe I made a mistake, but I got the y^2 with the minus sign.

v_{f}^2 = 2gy ( 1 - \frac{y}{L} )

Regards,
Sam

 

[OlderDan]

Maybe I made a mistake, but I got the y^2 with the minus sign.

v_{f}^2 = 2gy ( 1 - \frac{y}{L} )

Regards,
Sam

The indefinite integral should have the opposite sign from what you have. Unfortunately, the indefinite integral creates the impression that v^2 is negative when the sign is corrected. That of course cannot be. A constant needs to be added to the indefinite integral to ensure that v^2 is zero at some initial y value (of not less than L/2), where the initial velocity is known (probably zero, but it could start with some velocity.)

 

[BerryBoy]

The indefinite integral should have the opposite sign from what you have. Unfortunately, the indefinite integral creates the impression that v^2 is negative when the sign is corrected. That of course cannot be. A constant needs to be added to the indefinite integral to ensure that v^2 is zero at some initial y value (of not less than L/2), where the initial velocity is known (probably zero, but it could start with some velocity.)

But I used a definite integral (from t_0 to t_f). I did this because y(t) at these points becomes y and L respectively.

 

[BerryBoy]

OK, realized that in my equation, I've made the assumption:
y(t_f) = L
So it is only valid if the steel cable falls completely to the side of y.

OlderDan: But I still disagree with with your statement on my equation. y/L is always less than 1, if it were the other way around then you would get a complex answer.

I also did this problem again and now decide there is no factor of '2' in the answer.

Best Regards,
Sam

 

[OlderDan]

OK, realized that in my equation, I've made the assumption:
y(t_f) = L
So it is only valid if the steel cable falls completely to the side of y.

OlderDan: But I still disagree with with your statement on my equation. y/L is always less than 1, if it were the other way around then you would get a complex answer.

I also did this problem again and now decide there is no factor of '2' in the answer.

Best Regards,
Sam

OK.. It's not that we are getting different results. We are just using different letters to represent things. The way the problem was stated, and the way the acceleration was expressed in your post #2, y is a variable. In your last result, y is a constant (the initial value of y) and you are finding the velocity only at "the end" when the variable has the value L.

ma = \frac{m}{L}g\left[ {y - \left( {L - y} \right)} \right] = mg\left( {\frac{{2y - L}}{L}} \right)

a = g\left( {\frac{{2y - L}}{L}} \right)

v\frac{{dv}}{{dy}} = g\left( {\frac{{2y - L}}{L}} \right)

vdv = g\left( {\frac{{2y}}{L} - 1} \right)dy

\frac{{v^2 }}{2} = g\left( {\frac{{y^2 }}{L} - y} \right) + C

C = g\left( {y_o - \frac{{y_o ^2 }}{L}} \right) where v is zero at y_o

v^2 = 2g\left( {\frac{{y^2 }}{L} - y + y_o - \frac{{y_o ^2 }}{L}} \right) for any y starting from rest at y_o

v^2 = 2g\left( {\frac{{y^2 }}{L} - y + \frac{L}{4}} \right) if y_o = \frac{L}{2}

v_L ^2 = 2g\left( {y_o - \frac{{y_o ^2 }}{L}} \right) = 2gy_o \left( {1 - \frac{{y_o }}{L}} \right) when y = L

v_{\max } ^2 = g\frac{L}{2} when y_o = \frac{L}{2} and y = L

 

[BerryBoy]

OK, I'm happy now. I get the same answer (a different way mind - I never like using indefinite integrals). Sorry to have caused confusion; I should've written y_0 when I was referring to the initial overhang. Also sorry that OlderDan had to write all his working, it would've taken me a long time to write all that.

Regards,
Sam