Posted by Una on Sunday, December 18, 2011 at 1:46pm.
Since the cylinder's center of mass is not moving up or down, T = m g
to maintain translational equilibrium
The applied torque is T*R = m*g*R.
That equals the rate of change of angular momentum,
d/dt(I*w) = (mR^2/2)dw/dt = m*g*R
m cancels out and
dw/dt = g/(2R) rad/s^2 = alpha
The linear acceleration of the person's hand is alpha*R = g/2
Cute problem!
Well, the posted solution is incorrect.
Te => tension, To => torque, al => alpha
Te = mg
To = I * al = Te * R
0.5mR^2 * al = mgR
al * R = 2g = a
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