Finding angular speed by conservation of angular momentum

[Alexander350]

Homework Statement

Homework Equations

Angular momentum = moment of inertia * angular velocity
Change in angular momentum = impulsive moment
v^2=u^2+2as

The Attempt at a Solution

First I used v^2=u^2+2as to find the velocity of the particle the moment the string goes taut. I got v=\sqrt{8ag}. Then I put m\sqrt{8ag}a=\frac{1}{2}ma^2\dot{θ} because I thought the moment of the impulse by the particle should equal the change in angular momentum of the pulley wheel. However the solution I have has m\sqrt{8ag}a=\frac{1}{2}ma^2\dot{θ}+(ma\dot{θ})a with an extra term. I am not sure why; isn't a\dot{θ} equal to the velocity of a point on the rim of the wheel, and not the particle? After the string goes taunt, why does the particle have angular momentum at all if it is not rotating?

 

[TSny]

Does the particle come to rest with respect to the ground when the string becomes taut?

 

[Alexander350]

Does the particle come to rest with respect to the ground when the string becomes taut?

I don't think so, but I can't really imagine what happens to the particle after the string becomes taut.

 

[TSny]

Assume the string cannot stretch. Just after the string becomes taut (that is, just after the string applies an angular impulse to the pulley), does the pulley have an angular velocity? If so, does the particle also have a velocity (assuming the string does not stretch or become slack)? Can you relate the velocity of the particle to the angular velocity of the pulley?

 

[Alexander350]

Assume the string cannot stretch. Just after the string becomes taut (that is, just after the string applies an angular impulse to the pulley), does the pulley have an angular velocity? If so, does the particle also have a velocity (assuming the string does not stretch or become slack)? Can you relate the velocity of the particle to the angular velocity of the pulley?

Does the string remain taut and vertical and so the particle moves just as if it was on the rim of the wheel? Or does the part of the string attached to the rim of the wheel move without it until it becomes taut again?

 

[TSny]

Does the string remain taut and vertical and so the particle moves just as if it was on the rim of the wheel?

Yes, I think they want you to assume that once the string becomes taut it remains taut. Since the string is also assumed to not stretch, the speed of the particle must equal the speed of a point on the rim of the pulley after the string becomes taut.

Think of an extreme case where the particle is a heavy brick and the pulley is lightweight.

 

[Alexander350]

Yes, I think they want you to assume that once the string becomes taut it remains taut. Since the string is also assumed to not stretch, the speed of the particle must equal the speed of a point on the rim of the pulley after the string becomes taut.

Think of an extreme case where the particle is a heavy brick and the pulley is lightweight.

With (ma\dot{θ})a, why have they written it in that order? What is (ma\dot{θ}) by itself multiplied by a rather than multiplying the moment of inertia, ma^2, by the angular velocity?

 

[TSny]

Either way is fine. You can think of the angular momentum of the particle as $L = I\omega = ma^2 \dot \theta$. Or, you can think of the angular momentum of the particle as $L = mvr\sin \phi$. See http://hyperphysics.phy-astr.gsu.edu/hbase/amom.html

Here, $r$ is the distance from the origin (center of pulley) to the particle. $\phi$ is the angle between the direction $r$ and the direction of $v$. Can you see that $r\sin \phi = a$? So, $mvr \sin \phi = (mv)a$. If you now express $v$ in terms of $\dot \theta$ (of the pulley), you can see how they got their expression $(ma \dot \theta)a$.

EDIT: I would say that the second way is preferable. In the first method it might appear odd to assign an angular velocity $\omega = \dot \theta$ to the particle which is traveling in a straight line.