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3 May, 12:20

A rod of mass M = 154 g and length L = 35 cm can rotate about a hinge at its left end and is initially at rest. A putty ball of mass m = 11 g, moving with speed V = 9 m/s, strikes the rod at angle θ = 29° a distance D = L/3 from the end and sticks to the rod after the collision. Calculate the rotational kinetic energy, in joules, of the system after the collision.

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  1. 3 May, 15:35
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    moment of inertia of the rod = 1/3 mL², m is mass and L is length of rod.

    1/3 x. 154 x. 35²

    =.00629

    moment of inertia of putty about the axis of rotation

    = m d², m is mass of putty and d is distance fro axis

    =.011 x (.35 / 3) ²

    =.00015

    Total moment of inertia I =.00644 kgm²

    angular momentum of putty about the axis of rotation

    = mvRsinθ

    m is mass, v is velocity, R is distance where it strikes the rod and θ is angle with the rod at which putty strikes

    =.011 x 9 x. 35 / 3 x sin 29

    =.0056

    Applying conservation of angular momentum

    angular momentum of putty = angular momentum of system after of collision

    .0056 =.00644 ω where ω is angular velocity of the rod after collision

    ω =.87 rad / s.

    Rotational energy

    = 1/2 I ω²

    I is total moment of inertia

    =.5 x. 00644 x. 87²

    = 2.44 x 10⁻³ J.
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