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Posted by on Wednesday, April 9, 2014 at 12:33pm.

A uniform thin rod of length 0.55 m and mass 5.5 kg can rotate in a horizontal plane about a vertical axis through its center. The rod is a rest when a 3.0-g bullet traveling in the horizontal plane of the rod is fired into one end of the rod. As viewed from above, the direction of the bullet velocity makes an angle of 60° with the rod. If the bullet lodges in the rod and the angular velocity of the rod is 11.0 rad/s immediately after the collision, what is the magnitude of the bullet's velocity just before impact?

  • NWU - , Tuesday, June 2, 2015 at 4:45am

    angular momentum:
    l=rmvsinQ

    The rotational inertia on the rod:
    I=ML^2/12

    The parallel-axis theorem:
    I=Irod+mr^2

    Therefore:
    rmvsinQ=(ML^2/12+mr^2)w
    v=(5.5)(0.55)^2+(0.003)(0.275)^2(11.0)/(0.275)(0.003)sin(60.0)
    v=25618.6 m/s

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