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The problem is attached in the photo. The correct answer, according to the teacher's solution, was obtained using conservation of energy. Initially I tried using Newton's law/kinematics and got the wrong answer. Why didn't this work? Can you ever use Newton's law/kinematics to solve pulley problems (I feel like we did in class for some of them?)
Here are the two approaches with my two different answers:
Given:
m_{1} = 26 kg (the left block)
m_{2} = 17 kg (The right block)
M = 9 kg (the pulley
R = 0.084 m
d = 1.8 m
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NEWTON’S LAW/KINEMATIC APPROACH:
Definition of torque:
(1) т = r × F
Newton’s Second law for rotation:
(2) т_{net} = т_{m1} + т_{m2} = I α
Combine (1) and (2):
(3) r_{1} × F_{1} + r_{2} × F_{2} = I α
Clearly r_{1}×F_{1} points in the +z direction and r_{2}×F_{2} points in the –z direction, so we can say that:
(4) r_{1} F_{1} sinϕ_{1} – r_{2}F_{2} sinϕ_{2} = I α
I am assuming that r and F are perpendicular to each other in both cases (maybe this is the error in my logic?). In my free body diagram I drew r_{1} pointing to the left and F_{1} pointing down, and r_{2} pointing to the right and F_{2} pointing down. Anyways, if I can say that ϕ_{1} and ϕ_{2} are both equal to 90°, the above equation becomes:
(5) r_{1} F_{1} – r_{2} F_{2} = I α
Now given that:
(6) α = a / R
And, for a rotating disk:
(7) I = ½ M R^{2}
And, force due to gravity is given by:
(8) F = m g
And, the magnitudes of r_{1} and r_{2} are simply the radius of the disk R:
(9) r_{1} = r_{2} = R
If we plug (6), (7), (8), and (9) into (5), cancel out the R’s, and rearrange to isolate a on one side of the equation, we get:
(10) a = 2 g (m_{1} – m_{2}) / M
Because the net force is constant throughout, that means that the net torque, angular acceleration, and linear acceleration are constant throughout. Thus we can use the kinematic equation:
(11) v^{2} = v_{0}^{2} + 2 a ∆r, describing the motion of the m_{1}m_{2} system
Since we start at rest, v_{0}^{2} is zero. Deleting this term, plugging (10) into (11), and finding the square root of both sides, gives:
(12) v = sqrt{ 4 g (m_{1} – m_{2}) ∆r / M }
∆r will be 0.9 m – they will each travel half of the initial vertical distance between them. Thus, I’ll replace ∆r with d/2. Simplifying this gives the final equation:
(13) v = sqrt{ 2 g (m_{1} – m_{2}) d / M }
v = 5.94 m/s
_______________________________________________________________________________________________________________
CONSERVATION OF ENERGY APPROACH:
The law of conservation of energy, where the subscript _{M} indicates variables pertaining to the pulley, the subscript _{m1} indicates variables pertaining to mass 1, etc.:
PE_{m1,i} + PE_{m2,i} + PE_{M,i} + KE_{m1,i} + KE_{m2,i} + KE_{M,i} = PE_{m1,f} + PE_{m2,f} + PE_{M,f} + KE_{m1,f} + KE_{m2,f} + KE_{M,f}
All the KE terms at the beginning are zero so we can delete them. PE_{M} does not change so we delete that term. Thus, the above simplifies to:
PE_{m1,i} + PE_{m2,i} = KE_{m1,f } + KE_{m2,f} + KE_{M,f} + PE_{m1,f} + PE_{m2,f}
Which can be rewritten as:
m_{1} g h_{1,i} + m_{2} g h_{2,i} = ½ m_{1} v^{2} + ½ m_{2} v^{2} + ½ I ω^{2}+ m_{1} g h_{1,f} + m_{2} g h_{2,f}
Noting that I = ½ M R^{2} and ω = v / R, we have:
m_{1} g h_{1} + m_{2} g h_{2} = ½ m_{1} v^{2} + ½ m_{2} v^{2} + ¼ M v^{2}+ m_{1} g h_{1,f} + m_{2} g h_{2,f}
Let h_{1,i} = 1.8 m (the initial vertical distance (d) between the blocks) and h_{2,i} = 0 m. Thus, the second term on the left can be deleted, h_{1,i} can be replaced with d. Also, since the blocks end up at the same height in the middle, h_{1,f} = h_{2,f} = 0.9 m = ½ d. Thus:
m_{1}gd = ½ m_{1} v^{2} + ½ m_{2} v^{2} + ¼ M v_{2} + ½ m_{1} g d + ½ m_{2} g d
Rearranging and solving for v gives:
v = sqrt{ [ (m_{1} – m_{2}) g d) / (m_{1} + m_{2} + ½ M) }
v= 1.83 m/s
Here are the two approaches with my two different answers:
Given:
m_{1} = 26 kg (the left block)
m_{2} = 17 kg (The right block)
M = 9 kg (the pulley
R = 0.084 m
d = 1.8 m
___________________________________________________________________________________________________________________
NEWTON’S LAW/KINEMATIC APPROACH:
Definition of torque:
(1) т = r × F
Newton’s Second law for rotation:
(2) т_{net} = т_{m1} + т_{m2} = I α
Combine (1) and (2):
(3) r_{1} × F_{1} + r_{2} × F_{2} = I α
Clearly r_{1}×F_{1} points in the +z direction and r_{2}×F_{2} points in the –z direction, so we can say that:
(4) r_{1} F_{1} sinϕ_{1} – r_{2}F_{2} sinϕ_{2} = I α
I am assuming that r and F are perpendicular to each other in both cases (maybe this is the error in my logic?). In my free body diagram I drew r_{1} pointing to the left and F_{1} pointing down, and r_{2} pointing to the right and F_{2} pointing down. Anyways, if I can say that ϕ_{1} and ϕ_{2} are both equal to 90°, the above equation becomes:
(5) r_{1} F_{1} – r_{2} F_{2} = I α
Now given that:
(6) α = a / R
And, for a rotating disk:
(7) I = ½ M R^{2}
And, force due to gravity is given by:
(8) F = m g
And, the magnitudes of r_{1} and r_{2} are simply the radius of the disk R:
(9) r_{1} = r_{2} = R
If we plug (6), (7), (8), and (9) into (5), cancel out the R’s, and rearrange to isolate a on one side of the equation, we get:
(10) a = 2 g (m_{1} – m_{2}) / M
Because the net force is constant throughout, that means that the net torque, angular acceleration, and linear acceleration are constant throughout. Thus we can use the kinematic equation:
(11) v^{2} = v_{0}^{2} + 2 a ∆r, describing the motion of the m_{1}m_{2} system
Since we start at rest, v_{0}^{2} is zero. Deleting this term, plugging (10) into (11), and finding the square root of both sides, gives:
(12) v = sqrt{ 4 g (m_{1} – m_{2}) ∆r / M }
∆r will be 0.9 m – they will each travel half of the initial vertical distance between them. Thus, I’ll replace ∆r with d/2. Simplifying this gives the final equation:
(13) v = sqrt{ 2 g (m_{1} – m_{2}) d / M }
v = 5.94 m/s
_______________________________________________________________________________________________________________
CONSERVATION OF ENERGY APPROACH:
The law of conservation of energy, where the subscript _{M} indicates variables pertaining to the pulley, the subscript _{m1} indicates variables pertaining to mass 1, etc.:
PE_{m1,i} + PE_{m2,i} + PE_{M,i} + KE_{m1,i} + KE_{m2,i} + KE_{M,i} = PE_{m1,f} + PE_{m2,f} + PE_{M,f} + KE_{m1,f} + KE_{m2,f} + KE_{M,f}
All the KE terms at the beginning are zero so we can delete them. PE_{M} does not change so we delete that term. Thus, the above simplifies to:
PE_{m1,i} + PE_{m2,i} = KE_{m1,f } + KE_{m2,f} + KE_{M,f} + PE_{m1,f} + PE_{m2,f}
Which can be rewritten as:
m_{1} g h_{1,i} + m_{2} g h_{2,i} = ½ m_{1} v^{2} + ½ m_{2} v^{2} + ½ I ω^{2}+ m_{1} g h_{1,f} + m_{2} g h_{2,f}
Noting that I = ½ M R^{2} and ω = v / R, we have:
m_{1} g h_{1} + m_{2} g h_{2} = ½ m_{1} v^{2} + ½ m_{2} v^{2} + ¼ M v^{2}+ m_{1} g h_{1,f} + m_{2} g h_{2,f}
Let h_{1,i} = 1.8 m (the initial vertical distance (d) between the blocks) and h_{2,i} = 0 m. Thus, the second term on the left can be deleted, h_{1,i} can be replaced with d. Also, since the blocks end up at the same height in the middle, h_{1,f} = h_{2,f} = 0.9 m = ½ d. Thus:
m_{1}gd = ½ m_{1} v^{2} + ½ m_{2} v^{2} + ¼ M v_{2} + ½ m_{1} g d + ½ m_{2} g d
Rearranging and solving for v gives:
v = sqrt{ [ (m_{1} – m_{2}) g d) / (m_{1} + m_{2} + ½ M) }
v= 1.83 m/s
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