# Physics

posted by on .

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 - ,

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 - ,

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 - ,

okovko is right

• Physics - ,

okovko is correct