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Please could someone help me with the following.
I have attempted to attach a simple diagram drawn using "Paint".
This problem is taken from Fundamental University Physics Volume 1 second edition by Alonso & Finn page 184 Q7.83:
"A uniform (rope) of mass M length L passes over a smooth peg of very small radius. When the motion starts, BC=b. Show that the accelaration and velocity when BC=(2/3) L are a=g/3 and v=√[(2g/L)((8/3)L²+2bLb²)].
Apply your results to L=12m and b=7m.
Please refer to attached diagram  the long piece which descends is BC=b initially in case you have difficulty viewing.
My first idea was to use conservation of mechanical energy. If I can calculate the centre of mass then I can equate loss of potential energy U with gain in kinetic energy K.
Calling the length of the long side x the mass of this is (M/L)x with centre of mass at (1/2)x. Similarly the short side of length (Lx) has mass (M/L)(Lx) with centre of mass (1/2)(Lx). Adding and dividing by total mass M I obtain the centre of mass of the system to be:
(1/2L)(L²2xL+2x²)
I then substitute b for x and (2/3)L for x and taking the difference between the two obtain:
U = (1/2L)Mg((4/9)L²2bL+2b²)
K = (1/2)Mv²
K+U=0
v=√[(2g/L)(bLb²(2/9)L²)]
Not the book solution but if I use the numbers given then the book answer suggests v=28.6m/s whereas my solution is a more reasonable v=√(g/2) =2.2m/s.
In order to verify my answer I tried to solve the equation of motion. This is where my problem lies.
I consider the two sides separately again with the centre of mass of the long side at y1, length 2y1, mass (M/L) 2y1 and the centre of mass of the short side at y2, length 2y2, mass (M/L) 2y2. The differential equations are then:
g(M/L) 2y1  T = (M/L) 2y1 d²y1/dt²
g(M/L) 2y2  T = (M/L) 2y2 d²y2/dt²
where T is tension in the rope. Additionally we have:
2y1 + 2y2 = L so that:
d²y2/dt² =  d²y1/dt²
I can now eliminate y2 and then T leaving the following differential equation:
d²y1/dt² (4g/L) y1 = g
A particular solution is y1 = L/4
The homogeneous equation is solved by standard techniques and my general solution is:
y1 = Ae^(t√(4g/L)) +Be^(t√(4g/L)) + L/4 with y1=(1/2)b and dy1/dt=0 at t=0
Hence A=B=(1/8)(2bL) and so:
y1 = (1/8)(2bL)(e^(t√(4g/L)) + e^(t√(4g/L))) + L/4
or y1 = (1/4)(2bL)cosh(t√(4g/L)) + L/4 position
so dy1/dt = (1/4)(2bL)√(4g/L)sinh(t√(4g/L)) speed
and d²y1/dt² = (1/4)(2bL)(4g/L)cosh(t√(4g/L)) accelaration
I can solve the last equation when accelaration is (g/3), b=7m L=12m as required by the question to get cosh(t√(4g/L)=2 hence sinh(t√(4g/L))=√3 and speed is then dy1/dt= √(g/4) m/s.I now recognise that for every 2m the rope descends then in the same time interval the centre of mass moves only 1m and therefore travels at half the speed of the rope. My answer therefore is that when the accelaration is (g/3) the speed is 2√(g/4) or √g m/s. Recall that using energy I obtained a speed of v=√(g/2) m/s.
Trying to eliminate the factor of root two I tried the following.
Abandon centre of mass. Call the end of the long piece y. Mass is y(M/L) and then the mass of the short piece is (Ly)(M/L). The equations of motion then become:
g(M/L) y  T = (M/L) y d²y/dt²
g(M/L) (Ly)  T = (M/L) (Ly) d²y/dt²
Subtracting and rearranging:
d²y/dt²  (2g/L) y = g
A particular solution is y = L/2
The homogeneous equation is solved by standard techniques and my general solution is:
y = Ae^(t√(2g/L)) +Be^(t√(2g/L)) + L/2 with y=b and dy/dt=0 at t=0
Hence A=B=(1/4)(2bL) and so:
y = (1/4)(2bL)(e^(t√(2g/L)) + e^(t√(2g/L))) + L/2
or y = (1/2)(2bL)cosh(t√(2g/L)) + L/2 position
so dy/dt = (1/2)(2bL)√(2g/L)sinh(t√(2g/L)) speed
and d²y/dt² = (1/2)(2bL)(2g/L)cosh(t√(2g/L)) accelaration
I can solve the last equation when accelaration is (g/3), b=7m L=12m as required by the question to get cosh(t√(2g/L)=2 hence sinh(t√(2g/L))=√3 and speed is then dy/dt= √(g/2) m/s in agreement with the solution using energy and confirming the book answer is incorrect.
I am uncomfortable with not being able to use my first set of equations of motion to solve this problem. I was under the impression that whenever we have an extended body, we write the equation of motion as though all forces are applied at the centre of mass and the solution tells us how the centre of mass moves. In this case it seems that concept is unhelpful and a choice of the end of the rope is better, an exception to the rule which is hard for me to understand. An immediate thought is how do I spot this next time around in a problem? Whilst this is not a rigid body problem, I thought the speed of the centre of mass and the speed of the physical rope were different by a factor of two, not root two, but I am unable to reconcile the difference between my results.
Any help would be appreciated. Thanks.
I have attempted to attach a simple diagram drawn using "Paint".
This problem is taken from Fundamental University Physics Volume 1 second edition by Alonso & Finn page 184 Q7.83:
"A uniform (rope) of mass M length L passes over a smooth peg of very small radius. When the motion starts, BC=b. Show that the accelaration and velocity when BC=(2/3) L are a=g/3 and v=√[(2g/L)((8/3)L²+2bLb²)].
Apply your results to L=12m and b=7m.
Please refer to attached diagram  the long piece which descends is BC=b initially in case you have difficulty viewing.
My first idea was to use conservation of mechanical energy. If I can calculate the centre of mass then I can equate loss of potential energy U with gain in kinetic energy K.
Calling the length of the long side x the mass of this is (M/L)x with centre of mass at (1/2)x. Similarly the short side of length (Lx) has mass (M/L)(Lx) with centre of mass (1/2)(Lx). Adding and dividing by total mass M I obtain the centre of mass of the system to be:
(1/2L)(L²2xL+2x²)
I then substitute b for x and (2/3)L for x and taking the difference between the two obtain:
U = (1/2L)Mg((4/9)L²2bL+2b²)
K = (1/2)Mv²
K+U=0
v=√[(2g/L)(bLb²(2/9)L²)]
Not the book solution but if I use the numbers given then the book answer suggests v=28.6m/s whereas my solution is a more reasonable v=√(g/2) =2.2m/s.
In order to verify my answer I tried to solve the equation of motion. This is where my problem lies.
I consider the two sides separately again with the centre of mass of the long side at y1, length 2y1, mass (M/L) 2y1 and the centre of mass of the short side at y2, length 2y2, mass (M/L) 2y2. The differential equations are then:
g(M/L) 2y1  T = (M/L) 2y1 d²y1/dt²
g(M/L) 2y2  T = (M/L) 2y2 d²y2/dt²
where T is tension in the rope. Additionally we have:
2y1 + 2y2 = L so that:
d²y2/dt² =  d²y1/dt²
I can now eliminate y2 and then T leaving the following differential equation:
d²y1/dt² (4g/L) y1 = g
A particular solution is y1 = L/4
The homogeneous equation is solved by standard techniques and my general solution is:
y1 = Ae^(t√(4g/L)) +Be^(t√(4g/L)) + L/4 with y1=(1/2)b and dy1/dt=0 at t=0
Hence A=B=(1/8)(2bL) and so:
y1 = (1/8)(2bL)(e^(t√(4g/L)) + e^(t√(4g/L))) + L/4
or y1 = (1/4)(2bL)cosh(t√(4g/L)) + L/4 position
so dy1/dt = (1/4)(2bL)√(4g/L)sinh(t√(4g/L)) speed
and d²y1/dt² = (1/4)(2bL)(4g/L)cosh(t√(4g/L)) accelaration
I can solve the last equation when accelaration is (g/3), b=7m L=12m as required by the question to get cosh(t√(4g/L)=2 hence sinh(t√(4g/L))=√3 and speed is then dy1/dt= √(g/4) m/s.I now recognise that for every 2m the rope descends then in the same time interval the centre of mass moves only 1m and therefore travels at half the speed of the rope. My answer therefore is that when the accelaration is (g/3) the speed is 2√(g/4) or √g m/s. Recall that using energy I obtained a speed of v=√(g/2) m/s.
Trying to eliminate the factor of root two I tried the following.
Abandon centre of mass. Call the end of the long piece y. Mass is y(M/L) and then the mass of the short piece is (Ly)(M/L). The equations of motion then become:
g(M/L) y  T = (M/L) y d²y/dt²
g(M/L) (Ly)  T = (M/L) (Ly) d²y/dt²
Subtracting and rearranging:
d²y/dt²  (2g/L) y = g
A particular solution is y = L/2
The homogeneous equation is solved by standard techniques and my general solution is:
y = Ae^(t√(2g/L)) +Be^(t√(2g/L)) + L/2 with y=b and dy/dt=0 at t=0
Hence A=B=(1/4)(2bL) and so:
y = (1/4)(2bL)(e^(t√(2g/L)) + e^(t√(2g/L))) + L/2
or y = (1/2)(2bL)cosh(t√(2g/L)) + L/2 position
so dy/dt = (1/2)(2bL)√(2g/L)sinh(t√(2g/L)) speed
and d²y/dt² = (1/2)(2bL)(2g/L)cosh(t√(2g/L)) accelaration
I can solve the last equation when accelaration is (g/3), b=7m L=12m as required by the question to get cosh(t√(2g/L)=2 hence sinh(t√(2g/L))=√3 and speed is then dy/dt= √(g/2) m/s in agreement with the solution using energy and confirming the book answer is incorrect.
I am uncomfortable with not being able to use my first set of equations of motion to solve this problem. I was under the impression that whenever we have an extended body, we write the equation of motion as though all forces are applied at the centre of mass and the solution tells us how the centre of mass moves. In this case it seems that concept is unhelpful and a choice of the end of the rope is better, an exception to the rule which is hard for me to understand. An immediate thought is how do I spot this next time around in a problem? Whilst this is not a rigid body problem, I thought the speed of the centre of mass and the speed of the physical rope were different by a factor of two, not root two, but I am unable to reconcile the difference between my results.
Any help would be appreciated. Thanks.
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