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    http://filestore.aqa.org.uk/sample-p...W-QP-JUN11.PDF

    Could someone please explain the last part of the last question please. I'm kind of lost and don't understand why 3l is in the momentum for the clay before colliding.
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    (Original post by AspiringUnderdog)
    http://filestore.aqa.org.uk/sample-p...W-QP-JUN11.PDF

    Could someone please explain the last part of the last question please. I'm kind of lost and don't understand why 3l is in the momentum for the clay before colliding.
    At the instant right before collision, the clay is at a perpendicular distance of 3l from P, which is the point about which the pendulum will turn in a circular path, and since angular momentum about the centre of a circle is given by (mass) x (radius distance) x (perp. velocity to the radius), we get the that the angular momentum is 3mvl right before collision.
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    (Original post by RDKGames)
    At the instant right before collision, the clay is at a perpendicular distance of 3l from P, which is the point about which the pendulum will turn in a circular path, and since angular momentum about the centre of a circle is given by (mass) x (radius distance) x (perp. velocity to the radius), we get the that the angular momentum is 3mvl right before collision.
    Okay so the perpendicular distance makes sense to me. I didn't know about that equation though. I thought angular momentum is equal to Inertia x angular speed so that would be mass x radius^2 x velocity x radius again.
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    (Original post by AspiringUnderdog)
    Okay so the perpendicular distance makes sense to me. I didn't know about that equation though. I thought angular momentum is equal to Inertia x angular speed so that would be mass x radius^2 x velocity x radius again.
    Yes it is L = I \omega but note that \omega = \frac{v}{r} on circular motion and I= mr^2 as well. Together we get L=mrv as mentioned.
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    (Original post by RDKGames)
    Yes it is L = I \omega but note that \omega = \frac{v}{r} on circular motion and I= mr^2 as well. Together we get L=mrv as mentioned.
    oh wow I got that confused velocity x radius isn't anything I should have done velocity over radius thanks. I'll try to do the question again and ask if I need more help.
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    (Original post by RDKGames)
    Yes it is L = I \omega but note that \omega = \frac{v}{r} on circular motion and I= mr^2 as well. Together we get L=mrv as mentioned.
    Is the angular momentum after collision for the clay coming from the pendulum moving with the clay stuck on it?
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    (Original post by AspiringUnderdog)
    Is the angular momentum after collision for the clay coming from the pendulum moving with the clay stuck on it?
    It is its own thing but it is moving with the pendulum after so we use I = mr^2 still and \omega \neq \frac{v}{r} anymore as the speed is changed due to encountering the pendulum.
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    (Original post by RDKGames)
    It is its own thing but it is moving with the pendulum after so we use I = mr^2 still and \omega \neq \frac{v}{r} anymore as the speed is changed due to encountering the pendulum.
    Ah okay that makes sense thanks!
 
 
 
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