A Cylindrical Hoop

1. Jul 2, 2004

e(ho0n3

I have a cylindrical hoop of mass M and radius R with string (presumably thin and massless) wrapped around it. I take the end of the string from the hoop and let the hoop fall. (a) What is the torque about the CM of the hoop as a function of time? (b) What is the tension of the string as a function of time?

For (a), I decided to use angular momentum and then using the relation $\tau = dL/dt$ to find the torque. I get L = RMv(t) where v(t) = gt and so dL/dt = RMg. I know something is wrong here since the torque is NOT dependent on time. For (b)...well, I can't do (b) since I need to solve (a) first.

What am I doing wrong?

2. Jul 2, 2004

AKG

I'm not sure, but I'd use conservation of energy. The rotational energy and translational energy should have some constant relationship, and the sum of those energies should give the decrease in gravitational energy. I would imagine with this you could express everything in terms of x(t), x'(t), or x''(t). With x''(t) -- the acceleration -- you can find angular acceleration and thus torque.

3. Jul 2, 2004

turin

Torque is only being applied by the string. Find out how much less is the acceleration than g, and from that, you can find the tension in the string. This tension times the radius is the torque.

4. Jul 2, 2004

e(ho0n3

Good idea. I just get frustrated whenever I employ a method that 'should' work and doesn't.

So by conservation of energy

\begin{align*} 0 &= \frac{1}{2}I\omega^2 + \frac{1}{2}Mv^2 + Mgy \\ &= \frac{Iv^2}{2R^2} + \frac{1}{2}Mv^2 + Mgy \\ &= \frac{v^2}{2}(I/R^2 + M) + Mgy \end{align}

Solving for v

$$v= \sqrt{\frac{-2Mgy}{I/R^2 + M}}$$

This seems awfully wierd and complex (and I'm too lazy to solve such an ugly diff. eq.). There must be a simpler way in my opinion.

5. Jul 2, 2004

AKG

$$k = \sqrt{\frac{2M}{I/R^2 + M}}$$

$$v = k\sqrt{-y}$$

$$y' = k\sqrt{-y}$$

$$\frac{dy}{dt} = k\sqrt{-y}$$

$$(-y)^{-1/2}dy = kdt$$

$$-2\sqrt{-y} = kt$$

$$y = -\left (\frac{kt}{2} \right )^2$$

Find y'', and you're done. Oh, of course there should be a constant in the equation for y (i.e a "+ C" or "+ $y_0$) but you can arbitrarily set that to zero, which I suggest you do.

EDIT: used proper sign convention; was too lazy/sick/tired to do it yesterday.

Last edited: Jul 3, 2004
6. Jul 3, 2004

e(ho0n3

This is all fine and dandy, but what I want to know is why my first attempt didn't work. Argh...

7. Jul 3, 2004

Staff: Mentor

What makes you think the torque is dependent on time?
And what makes you think that the acceleration = g! :yuck:

Just apply Newton's 2nd law (or conservation of energy). There are two forces acting on the hoop: tension and weight.

Also: I hope you realize that the expressions that you and AKG found are not "awfully wierd and complex" and that the resulting differential equation is trivial. (It's just uniformly accelerated motion! )

8. Jul 3, 2004

e(ho0n3

Ah! There is my error. You'll have to forgive my as I am a borderline retard.

Anything with a square root looks ugly to me for some reason, hence my statement. I have to stop judging things by the way they look.

Thank you.