- #1
manosairfoil
- 5
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Hi! I am struggling with this problem the last two days and I cannot decide which solution is correct.
Homework Statement
A cylinder of radius R is fixed horizontally on the floor. A uniform chain of mass M and length L (L<πR/2) is placed on the cylinder in such a way that one end of the chain is on a highest point of the cylinder. We release the chain. All drag and friction shall be negleted. Gravitational acceleration is g.
(This is a 2D problem, the cylinder could also be a circular disk.)
I am asked to find:
1)The accelaration of the chain once released.
2) An expression for the angular velocity and acceleration as a function of the total angle the chain has fallen.
The attempt at a solution
1)
First Approach :
We find the CM of the chain using the formula r=[Rsin(L/(2R))]/(L/((2R))
We then decompose the weight to the tangential component. ( Mgsin(L/2R) )
Acceleration of cm is gsin(L/2R). Since there is no velocity at this moment centripital acceleration is 0, so the magnitude of the accelerarion of each point of the chain is gsin(L/2R) * R/r .
I don't know if this approach is correct. I believe that if we sum all the normal forces from the cylinder the net normal force will have a tangential component. ( for every point on the chain the normal force has different magnitude as the radial component of the weight gets smaller and smaller as we move down).
Approach B:
For each dm we write the N2L equation:
F1 + dm1gsin(θο + θ1) - F2 = dm1a
Where F1 and F2 are the contact forces from the "next" and "previous" dm.
If we add all those equations for all dm's the F1,F2 forces cancell up. (By Newton's 3rd law)
We get:
g * Integral(from 0 to M) of (sin[θο + θ(m)]dm =Ma
Assuming a linear density λ = M/L
We get θ(m) =m/(λR)
Substituting back to the integral (with θο = 0 )we have
a = gR/L (1-cos(L/R)
I can't find a logical error in the second approach so i think it might be correct. But the first approach seems simple and nice to turn down immediately :/For the question 2) for the acceleration we use exacly the same apporaches (but with θο != 0 )
To find the velocity can we take the difference of the potential energies? (With respect to the postition of the cm). Do the normal forces lroduce any work? (If the is a tangential component of the total N force, shouldn't they?)
This problem has made me very confused. I would be grateful if someone could help me understand the situation better.. :D
Homework Statement
A cylinder of radius R is fixed horizontally on the floor. A uniform chain of mass M and length L (L<πR/2) is placed on the cylinder in such a way that one end of the chain is on a highest point of the cylinder. We release the chain. All drag and friction shall be negleted. Gravitational acceleration is g.
(This is a 2D problem, the cylinder could also be a circular disk.)
I am asked to find:
1)The accelaration of the chain once released.
2) An expression for the angular velocity and acceleration as a function of the total angle the chain has fallen.
The attempt at a solution
1)
First Approach :
We find the CM of the chain using the formula r=[Rsin(L/(2R))]/(L/((2R))
We then decompose the weight to the tangential component. ( Mgsin(L/2R) )
Acceleration of cm is gsin(L/2R). Since there is no velocity at this moment centripital acceleration is 0, so the magnitude of the accelerarion of each point of the chain is gsin(L/2R) * R/r .
I don't know if this approach is correct. I believe that if we sum all the normal forces from the cylinder the net normal force will have a tangential component. ( for every point on the chain the normal force has different magnitude as the radial component of the weight gets smaller and smaller as we move down).
Approach B:
For each dm we write the N2L equation:
F1 + dm1gsin(θο + θ1) - F2 = dm1a
Where F1 and F2 are the contact forces from the "next" and "previous" dm.
If we add all those equations for all dm's the F1,F2 forces cancell up. (By Newton's 3rd law)
We get:
g * Integral(from 0 to M) of (sin[θο + θ(m)]dm =Ma
Assuming a linear density λ = M/L
We get θ(m) =m/(λR)
Substituting back to the integral (with θο = 0 )we have
a = gR/L (1-cos(L/R)
I can't find a logical error in the second approach so i think it might be correct. But the first approach seems simple and nice to turn down immediately :/For the question 2) for the acceleration we use exacly the same apporaches (but with θο != 0 )
To find the velocity can we take the difference of the potential energies? (With respect to the postition of the cm). Do the normal forces lroduce any work? (If the is a tangential component of the total N force, shouldn't they?)
This problem has made me very confused. I would be grateful if someone could help me understand the situation better.. :D