Homework Help: Unwinding Cylinder-Dynamics

1. Dec 3, 2008

soupastupid

Unwinding Cylinder--Dynamics

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

A cylinder with a moment of inertia I (about its axis of symmetry), mass m, and radius r has a massless string wrapped around it which is tied to the ceiling (Intro 1 figure) .

At time t=0 the cylinder is released from rest at height h above the ground. Use g for the magnitude of the acceleration of gravity. Assume that the string does not slip on the cylinder. Let v_vec represent the instantaneous velocity of the center of mass of the cylinder, and let omega_vec represent the instantaneous angular velocity of the cylinder about its center of mass. Note that there are no horizontal forces present, so for this problem

v_vec = -v(y-hat) and omega_vec = - omega(z-hat)

2. Relevant equations

Torque = I(angular_acceleration)
F= m(linear_acceleration)

3. The attempt at a solution

I'm having trouble with part B
T is the tension of the spring
if I goes to zero, the object will be turning faster in the -z direction.

The tension of the spring does less work on its rotation and more work on its linear momentum.

and so

If the T went to 0, then linear a should be g right?
because the tension does no work on either the angular and linear momentum and so the cylinder will just fall right?

the answer is the second choice
right?

if T went to infinity, then there would not be acceleration in the in the -y direction.
so if T went to infinity, a should be zero right?

if T equaled mg then linear acceleration would also be zero because the forces of the weight and T cancel out

right?

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2. Dec 3, 2008

turin

Re: Unwinding Cylinder--Dynamics

Um ... What is the problem asking you to solve? As far as I can see, it is just a bunch of information, but no question.

3. Dec 4, 2008

soupastupid

Re: Unwinding Cylinder--Dynamics

im having trouble with part B

i think when t=0, a = 0 because

if I goes to zero, t no longer does work on rotation but does work on linear acceleration
but if the tension goes to 0 then the object will continue to spin but simply fall in the -y direction due to gravity

right?

4. Dec 4, 2008

Staff: Mentor

Re: Unwinding Cylinder--Dynamics

I don't quite understand your reasoning (and you seem to have contradicted yourself).

Why not apply Newton's law to translation and rotation and come up with some equations. Much easier to understand things when you put Newton to work.

5. Dec 5, 2008

soupastupid

Re: Unwinding Cylinder--Dynamics

ur right

i contradicted myself

anyway

my logic tells me

if t = 0, there is no tension and the object should just fall
and so -y direction its a=g (linear)

but the equations tells me otherwise
that when t = 0

a = 0

(as seen in picture)

that does not make sense to me though

why wouldnt the cylinder fall? gravity is still acting upon it

and since T=0 and I = 0

it cant be spinning right?

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6. Dec 5, 2008

Staff: Mentor

Re: Unwinding Cylinder--Dynamics

OK. (But you have to show that T = 0.)

The equations don't say that. (Please simplify your equation for acceleration. That could be throwing you off.)

7. Dec 5, 2008

soupastupid

Re: Unwinding Cylinder--Dynamics

ok

for vertical a in the -y direction

a_y = ( -gmr^2 ) / (I + mr^2)

i dont see T in the equation

so but it has to affect a_y

but i do see I

and if i make that equal to 0

i get

a_y = -g

WOWOW

thats amazing

thanks!

8. Dec 5, 2008

soupastupid

Re: Unwinding Cylinder--Dynamics

is that right?

now that im looking at the problem more

if T = mg
shouldnt a_y equal to 0?

9. Dec 5, 2008

Staff: Mentor

Re: Unwinding Cylinder--Dynamics

Yep.

Sure, but T ≠ mg.

10. Dec 6, 2008

soupastupid

Re: Unwinding Cylinder--Dynamics

im having trouble with part c

it says its a yoyo

and the string is wrapped around the axis

and the radius casing >>>> radius of axis

but i didnt that would matter because the string is wrapped around the axis

but then the problem says

I >>>> m(r_axis)^2

and then i go ????

this is what i am thinking

using the equations

for then

T = (gIm) / (I + m(raxis)^2)

and so since I = infinity

T = mg

and when i plug that into a_y equation

a_y = -(T(r_axis)^2)/I <--am i using the right r?

and I = infinity

i get

a = 0

is that right??

another question
kinda random

how do you quote previous response?

Attached Files:

• unwindingc.jpg
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11. Dec 6, 2008

Staff: Mentor

Re: Unwinding Cylinder--Dynamics

Good.

Click on the "Quote" button at the bottom right of the post you want to quote. (Just make sure that the part you want to quote is between quote tags. Just try it.)

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