Question

Solve for the tension T in the vertical section of string.
Express T in terms of the known variables I, m, r, and g.

Answer 1

yes

Answer 2

no

Answer 3

college

Answer 4

ok

Answer 5

yea

Answer 6

so T=-mg?

Answer 7

ok

Answer 8

Alright I may have gotten it this time!!

Start again with both equations. One for translational motion and two for rotational motion.
\[m a = mg - T\]\[N=I \frac{d \omega}{dt}\]\[N = r x T=rTsin(\theta) = rT\]
Equate the rotational equations and use time derivative of v = (omega) * r:\[I \frac{a}{r}=rT\]
Use I = 1/2 m r^2 for a solid disc, then solve for T:\[\frac{1}{2}mr^{2} \frac{a}{r} = r T\]\[T = \frac{1}{2}ma\]
Now plug this T into the translational equation:\[ma = mg - \frac{1}{2} m a\]\[a = g - \frac{1}{2}a\]\[a = \frac{2}{3}g\]This acceleration makes sense. If the disk was just falling, it would be a=g. But the string tension will make this value less than 'g', which is what we found here.

Now we can use this acceleration to find the tension in terms of known variables.\[T = \frac{1}{2}ma = \frac{1}{2}m \frac{2}{3}g=\frac{2}{6}mg\]

Therefore:
\[T = \frac{1}{3} mg\]