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bcca
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I think I'm not understanding something about this problem. Can someone help me out?
A 2.0 kg disk traveling a 3.0 m/s strikes a 1.0 kg stick of length 4.0 m that is lying flat on nearly frictionless ice. The disk strikes the endpoint of the stick, at a distance r = 2.0 m from the stick's center. Suppose the collision is perfectly inelastic so that the disk adheres to the stick at the endpoint at which it strikes. The moment of inertia of the stick about its center of mass is 1.33 kg*m2. Find the angular speed of the system after the collision.
(I think) Parallel axis theorem: I= Icm + mr2
L= r cross p
L= I*omega
m= mass disk
L = L'
r cross p = I(system)*omega
mvr = (I(stick) + I(disk) + mvr2)*omega
mvr = (I(stick) + (1/2)mr2 + mr2)*omega
omega = mvr/(I(stick) + 2/3(mr2))
omega = 2.0kg(3.0m/s)(2.0m)/(1.33kg*m2+1.5(2.0kg)(2.0m)2)
omega= .90 rad/s
The answer is 1.0 rad/s.
Homework Statement
A 2.0 kg disk traveling a 3.0 m/s strikes a 1.0 kg stick of length 4.0 m that is lying flat on nearly frictionless ice. The disk strikes the endpoint of the stick, at a distance r = 2.0 m from the stick's center. Suppose the collision is perfectly inelastic so that the disk adheres to the stick at the endpoint at which it strikes. The moment of inertia of the stick about its center of mass is 1.33 kg*m2. Find the angular speed of the system after the collision.
Homework Equations
(I think) Parallel axis theorem: I= Icm + mr2
L= r cross p
L= I*omega
m= mass disk
The Attempt at a Solution
L = L'
r cross p = I(system)*omega
mvr = (I(stick) + I(disk) + mvr2)*omega
mvr = (I(stick) + (1/2)mr2 + mr2)*omega
omega = mvr/(I(stick) + 2/3(mr2))
omega = 2.0kg(3.0m/s)(2.0m)/(1.33kg*m2+1.5(2.0kg)(2.0m)2)
omega= .90 rad/s
The answer is 1.0 rad/s.