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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