Question
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?
Answers
drwls
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!
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!
okovko
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
Te => tension, To => torque, al => alpha
Te = mg
To = I * al = Te * R
0.5mR^2 * al = mgR
al * R = 2g = a
asdfdd
okovko is right
a.o.k.
okovko is correct
Jacky
But why is I 1/2mr^2?
If we are looking at the perspective of the hand, I should be 3/4 mr^2 right? (Parallel axis theorem
If we are looking at the perspective of the hand, I should be 3/4 mr^2 right? (Parallel axis theorem