What Is the Tension in the String as the Hoop Descends?

[jjlandis]

Homework Statement

A string is wrapped several times around the rim of a small hoop with radius 0.0800 meters and mass 0.344 kg. If the free end of the string is held in place and the hoop is released from rest calculate the tension in the string while the hoop descends as the string unwinds. Give your answer in Newtons to the third decimal place.

Homework Equations

\Sigma Torque = (Tension)*(Radius) = (Impulse)*(angular acceleration) = (1/2)*(Mass)*(Radius²)*(angular acceleration)

The Attempt at a Solution

I tried to derive the solution down to, Tension = (1/3)*(mass)*(g) but I am not returned with the correct answer. I calculated 1.12 N and the correct answer states 1.686 N

 

[LowlyPion]

First of all I is not impulse, it's the moment of Inertia. And the moment of inertia for a hoop would be m*R² not 1/2*m*R² because the mass is at the rim, not evenly distributed across a solid disk.

Secondly you have the right idea.

T*R = I*α

But T here will be m*(g - a) where a is the translational acceleration (vertical).
And α is the angular acceleration that can be expressed as a/R.

Rewriting then and substituting

m*(g - a)*R = m*R²*a/R

m*g*R - a*m*R = a*m*R

So ...

a = g/2

And so ...

T = m*(g - g/2) = m*g/2