Posted by Una on Sunday, December 18, 2011 at 1:46pm.
I need help with this question please. A solid cylinder of mass m and radius R has a string wound around it. A person holding the string pulls it vertically upward such that the cylinder is suspended in midair for a brief time interval (change in)t and its center of mass does not move. The tension in the string is T, and the rotational inertia of the cylinder about its axis is (1/2)mR^2.
What is the linear acceleration of the person's hand during the time interval change in t?

Physics  drwls, Sunday, December 18, 2011 at 3:09pm
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!

Physics  okovko, Monday, December 10, 2012 at 2:46am
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

Physics  asdfdd, Sunday, February 9, 2014 at 2:39am
okovko is right

Physics  a.o.k., Thursday, January 19, 2017 at 11:22pm
okovko is correct
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