At least, they're difficult for me. I find them very hard to set up, in particular the interaction of the rotation and translation is hard for me to grasp. I've worked on both of these very hard, with only some headway, so be liberal in your help.(adsbygoogle = window.adsbygoogle || []).push({});

(For some reason, all my greek letters are superscripted - ignore that, they should not be)

The first

1. The problem statement, all variables and given/known data

Hanging from one side of an ideal pulley is a mass m. On the other side, there is a cylinder with the pulley-rope wrapped around it. The cylinder has mass M and radius R, and we can assume it to be uniform (though it shouldn't matter to the problem - just say it has moment of inertia I if you like). Both objects are acted on by the force of gravity, (gravitational acceleration = g), as well as the tension of the pulley rope. Mass m accelerates downwards at rate a, while cylinder M accelerates at rate A, with the rope around it unwinding at angular acceleration a [tex]\alpha[/tex].

From this, determine the values of a, A and [tex]\alpha[/tex] in terms of m, M, R and g.

2. Relevant equations

General equations, the letters don't stand for the quantities of the problems.

a = g -T/m

Ek = 1/2mv^2

Er = 1/2I[tex]\omega[/tex]^2

Eg = mgh

[tex]\alpha[/tex] = [tex]\omega[/tex]t

a = [tex]\alpha[/tex]R

I = 1/2mR^2

[tex]\tau[/tex] = FXR = [tex]\alpha[/tex]I

Some important constraints (the letters represent the quantities of the problem)

a + A = [tex]\alpha[/tex]*R, due the constraint of the rope length

-g <= a, A <= g

0 <= [tex]\alpha[/tex]*R <= 2g

3. The attempt at a solution

I tried two different ways, both of which ended in dead ends. The first was to set up the following (though they may be incorrect):

Note: T = force of tension in the rope;

1) a + A = [tex]\alpha[/tex]*R

2) a = g - T/m

3) A = g - T/M + [tex]\alpha[/tex]*R

4) [tex]\alpha[/tex]R = [tex]\tau[/tex]R/I = T*R^2/I = T/M*(M*R^2/I)

From this, I get:

2g - T(m+M)/(mM) + [tex]\alpha[/tex]R = [tex]\alpha[/tex]R

Or

2g - T(m+M)/(mM) = 0

Giving T = 2g(mM)/(m+M)

a = g - T/m = g - 2g(M)/(m+M) = g(m+M)/(m+M) - 2gM/(m+M) = g(m-M)/(m+M)

[tex]\alpha[/tex]R = T/M*(M*R^2/I) = 2g(m)/(m+M)*(M*R^2/I)

divide by R to get [tex]\alpha[/tex]

A = [tex]\alpha[/tex]R - a = 2g(m)/(m+M)*(M*R^2/I) - g(m-M)/(m+M)

This looks fine and dandy at first, but for one problem:

I = 1/2(M*R^2)

(M*R^2)/I = 2

Therefore, if m>>M,

[tex]\alpha[/tex]R ~= 2g*1/1*2 = 4g

THIS DEFIES OUR EARLIER CONSTRAINT! It's impossible, in other words, which is a good indication that something is wrong. I suspect I made a false assumption in one of the equations, probably #3. With a good formula for A, or for T, I'm sure I could solve this problem easily. As it is, I'm having a tough time.

I also tried another way. Using work-energy, I derived a quadratic equation for a in terms of A (which, strangely enough, was identical to A in terms of a). Said equation is so complicated however, that I'm practically certain I either messed it up (and I'm pretty sure that's not so, since I did it twice), or energy simply isn't a fruitful path to this solution.

The Second problem:

1. The problem statement, all variables and given/known data

This one is a good deal more straightforward.

A uniform disk of mass m and radius R lies on a surface of frictionless ice in an x-y plane. A boy of mass m (same m) is skating towards it at Vi, directly parallel to the x axis. The boy jumps and lands on the disk outer edge of the disk (at coordinate (0,-R) in other words), sticking to it completely, at t = 0. Give equations for the boy's subsequent path the x coordinate and the y coordinate. Also, give the amount of energy dissipated in the boy's jump.

2. Relevant equations

L = I[tex]\omega[/tex] + mv X r(vector, not radius)

Ek = 1/2mv^2

Er = 1/2I[tex]\omega[/tex]^2

I = 1/2mR^2

3. The attempt at a solution

I figured that angular momentum was conserved. After the boy's jump, the disk is going to have a translational velocity Vf and an angular velocity [tex]\omega[/tex].

1) Li = mViR

2) Lf = 2mVfR + (mR^2 + 1/2mR^2)[tex]\omega[/tex] = 2mVfR + 3/2mR^2[tex]\omega[/tex]

I'm not completely sure about #2

3) Li = Lf; simplifies to

4) Vi = 2Vf + 3/2R[tex]\omega[/tex]

The problem is, I don't know what to do from there - I have no clue how the momentum distributes between linear and rotational motion. I feel like it's 2/3rds rotational, 1/3rd linear, due to I relative to mR^2, but I don't know why or how to show it if so.

The lack of constraints or relations leaves me floundering - I'm also clueless on the second part, about energy dissipation; I'd assume 0, and that energy would be a constraint that would help solve the motion part of the problem, but clearly this isn't so.

To whomever can help me, THANK YOU.

1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

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# Homework Help: Two difficult translational and rotational motion problems

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