A disk-shaped merry go round (I=1/2 MR2 M=500 kg R=2m) is initially at rest. A ring (i=mr2 m=300 kg r=0.5 m) rotating at 5 rad/s is dropped

Question

A disk-shaped merry go round (I=1/2 MR2 M=500 kg R=2m) is initially at rest. A ring (i=mr2 m=300 kg r=0.5 m) rotating at 5 rad/s is dropped onto the non-rotating merry go round. The tricky bit is that it was not dropped onto the center of the merry go round. Rather it is dropped at 1.5 m from the merry go round’s axis of rotation. After the collision the merry go round is now rotating with the ring riding on it off center. What is the angular velocity of the merry go round?

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Farah 1 week 2021-07-22T15:06:15+00:00 1 Answers 2 views 0

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    2021-07-22T15:07:16+00:00

    Answer:

    \omega = 0.214\,\frac{rad}{s}

    Explanation:

    The situation can be modelled by the Principle of Angular Momentum Conservation and the Parallel Axis Theorem:

    (300\,kg)\cdot (0.5\,m)^{2}\cdot (5\,\frac{rad}{s} ) = \left[\frac{1}{2}\cdot (500\,kg)\cdot (2\,m)^{2} +(300\,kg)\cdot (0.5\,m)^{2}+(300\,kg)\cdot (1.5\,m)^{2}\right]\cdot \omegaThe final angular velocity is:

    \omega = 0.214\,\frac{rad}{s}

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