Solving Torque and Tension for a Cylindrical Hoop with Mass and Radius R

[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?

 

[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.

 

[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.

 

[e(ho0n3]

I'm not sure, but I'd use conservation of energy.

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

So by conservation of energy

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

Solving for v

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

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

 

[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.

 

[e(ho0n3]

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

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. Oh, and I haven't taken sign into account, so it might be $y = - \left (\frac{kt}{2} \right ) ^2$, but I'll leave that to you.

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

 

[Doc Al]

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?

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

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 weird and complex" and that the resulting differential equation is trivial. (It's just uniformly accelerated motion! )

 

[e(ho0n3]

And what makes you think that the acceleration = g!

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

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

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.