Using conservation of energy vs. Newton's laws in a pulley problem

• JJ__
In summary, 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?f

JJ__

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:

m1 = 26 kg (the left block)

m2 = 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) тnet = тm1 + тm2 = I α

Combine (1) and (2):

(3) r1 × F1 + r2 × F2 = I α

Clearly r1×F1 points in the +z direction and r2×F2 points in the –z direction, so we can say that:

(4) r1 F1 sinϕ1r2F2 sinϕ2 = 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 r1 pointing to the left and F1 pointing down, and r2 pointing to the right and F2 pointing down. Anyways, if I can say that ϕ1 and ϕ2 are both equal to 90°, the above equation becomes:

(5) r1 F1 r2 F2 = I α

Now given that:

(6) α = a / R

And, for a rotating disk:

(7) I = ½ M R2

And, force due to gravity is given by:

(8) F = m g

And, the magnitudes of r1 and r2 are simply the radius of the disk R:

(9) r1 = r2 = 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 (m1m2) / 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) v2 = v02 + 2 a r, describing the motion of the m1-m2 system

Since we start at rest, v02 is zero. Deleting this term, plugging (10) into (11), and finding the square root of both sides, gives:

(12) v = sqrt{ 4 g (m1m2) ∆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 (m1m2) d / M }

v = 5.94 m/s

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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.:

PEm1,i + PEm2,i + PEM,i + KEm1,i + KEm2,i + KEM,i = PEm1,f + PEm2,f + PEM,f + KEm1,f + KEm2,f + KEM,f

All the KE terms at the beginning are zero so we can delete them. PEM does not change so we delete that term. Thus, the above simplifies to:

PEm1,i + PEm2,i = KEm1,f + KEm2,f + KEM,f + PEm1,f + PEm2,f

Which can be rewritten as:

m1 g h1,i + m2 g h2,i = ½ m1 v2 + ½ m2 v2 + ½ I ω2+ m1 g h1,f + m2 g h2,f

Noting that I = ½ M R2 and ω = v / R, we have:

m1 g h1 + m2 g h2 = ½ m1 v2 + ½ m2 v2 + ¼ M v2+ m1 g h1,f + m2 g h2,f

Let h1,i = 1.8 m (the initial vertical distance (d) between the blocks) and h2,i = 0 m. Thus, the second term on the left can be deleted, h1,i can be replaced with d. Also, since the blocks end up at the same height in the middle, h1,f = h2,f = 0.9 m = ½ d. Thus:

m1gd = ½ m1 v2 + ½ m2 v2 + ¼ M v2 + ½ m1 g d + ½ m2 g d

Rearranging and solving for v gives:

v = sqrt{ [ (m1m2) g d) / (m1 + m2 + ½ M) }

v= 1.83 m/s

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Are you assuming that F1 and F2 are the weights of the hanging masses?

(To properly use Newton's law & kinematics, you must analyze all three bodies, not just the pulley.)

And, force due to gravity is given by:

(8) F = m g
While this is true, the tension in the strings are not given by this as that would ignore the acceleration of the masses themselves.

While this is true, the tension in the strings are not given by this as that would ignore the acceleration of the masses themselves.
Why would this be the case? What are the forces exerted on the pulley?

Are you assuming that F1 and F2 are the weights of the hanging masses?

(To properly use Newton's law & kinematics, you must analyze all three bodies, not just the pulley.)
Yes that's what I was assuming. Is that not correct? What exactly are the forces that are acting on the pulley and causing it to rotate?

What are the forces exerted on the pulley?

What exactly are the forces that are acting on the pulley and causing it to rotate?

It is the ropes that exert forces on the pulley--via the tensions in the rope. That tension does not equal the weight of the masses. (If it did, the masses would be in equilibrium.) You must solve for the tensions in the rope. (That's why you need additional equations besides that for the pulley.)

The weight of the masses acts on the masses, not directly on the pulley.

It is the ropes that exert forces on the pulley--via the tensions in the rope. That tension does not equal the weight of the masses. (If it did, the masses would be in equilibrium.) You must solve for the tensions in the rope. (That's why you need additional equations besides that for the pulley.)

The weight of the masses acts on the masses, not directly on the pulley.
Ooh okay got it! That makes sense. Is it possible to solve for the tension forces? (If so I will try it)

Is it possible to solve for the tension forces?
Of course.

(If so I will try it)
Good!