1. Use conservation of energy: mgh = 0.5 * I * omega^2
where h = L/2 because it is the change in
location of the center of mass that is important,
I = 1/3 *
m * L^2 [determined" above" on worksheet]
I solved for the angular velocity: omega = SQRT[3g / L] =
5.42 rad/s
2. Again, start with conservation of energy:
(a) Ei = Ef
initial PE of m2 = KE of m1 + KE of m2 + rotational KE of pulley
+ increase in PE of m1
Note: that the two masses do have the same speed, v, and move the same distance,
h, because they are attached to a "string" that does not stretch;
also, the speed of the masses is the tangential speed of the pulley, so the
angular speed of the pulley, omega = v / R
m2 * gh = 1/2 * m1 * v^2 + 1/2 * m2 * v^2
+ 1/2 * I * omega^2 + m1 * gh
v^2 * 0.5 * [ m1 + m2 + (I / R^2) ] = (m2 - m1) * gh
Note: I / R^2 = [(.5 M * R^2) / R^2 = 0.5 M
When I substituted numbers:
v^2 * 0.5 * [ .2kg + .3kg + (.5*.4 kg) ] = (.3 - .2)
* g * 0.25 m,
my final answer was v = 0.837 m/s
(b) v^2 = v0^2 + 2*a*h ... a = 1.4 m/s^2 DOWN
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(c) Your force diagrams should look like the one linked to the right. T1 - m1*g = m1*a and T2 - m2*g = -m2*a T1 = 2.16 N T2 = 2.65 N |
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This page maintained by Anne G. Young. Last modified 11-Mar-2005 .