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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:

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Definition of torque:

(1)

Newton’s Second law for rotation:

(2)

Combine (1) and (2):

(3)

Clearly

(4)

I am assuming that

(5)

Now given that:

(6)

And, for a rotating disk:

(7)

And, force due to gravity is given by:

(8)

And, the magnitudes of

(9)

If we plug (6), (7), (8), and (9) into (5), cancel out the

(10)

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)

Since we start at rest,

(12)

∆

(13)

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The law of conservation of energy, where the subscript

All the

Which can be rewritten as:

Noting that

Let

Rearranging and solving for

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**points in the +_{1}*z*direction and**r**×_{2}**F**points in the –_{2}*z*direction, so we can say that:(4)

*r*_{1}*F*_{1}sin*ϕ*_{1}–*r*_{2}*F*_{2}sin*ϕ*_{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**pointing to the left and_{1}**F**pointing down, and_{1}**r**pointing to the right and_{2}**F**pointing down. Anyways, if I can say that_{2}*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}systemSince 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#### Attachments

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