A horizontal 861 N merry-go-round of radius 1.84 m is started from rest by a constant horizontal force of 57.6 N applied tangentially to the

Question

A horizontal 861 N merry-go-round of radius 1.84 m is started from rest by a constant horizontal force of 57.6 N applied tangentially to the merry-go-round. Find the kinetic energy of the merry-goround after 2.54 s. The acceleration of gravity is 9.8 m/s 2 . Assume the merry-go-round is a solid cylinder. Answer in units of J.

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Tài Đức 5 years 2021-07-27T11:06:06+00:00 1 Answers 17 views 0

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  1. Verity
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    2021-07-27T11:07:52+00:00
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    Answer:

    K = 60.861\,J

    Explanation:

    The moment of inertia of the merry-go-round is:

    I = \frac{1}{2}\cdot \left(\frac{861\,N}{9.807\,\frac{m}{s^{2}} } \right)\cdot (1.84\,m)^{2}

    I = 148.618\,kg\cdot m^{2}

    The tangential acceleration experimented by the merry-go-round is:

    a_{t} = \frac{57.6\,N}{\left(\frac{861\,N}{9.807\,\frac{m}{s} }\right) }

    a_{t} = 0.656\,\frac{m}{s^{2}}

    The linear speed after 2.54 seconds is determined hereafter:

    v = 0\,\frac{m}{s} + (0.656\,\frac{m}{s^{2}} )\cdot (2.54\,s)

    v = 1.666\,\frac{m}{s}

    The angular speed of the merry-go-round is:

    \omega = \frac{1.666\,\frac{m}{s} }{1.84\,m}

    \omega = 0.905\,\frac{rad}{s}

    The kinetic energy due to the rotation for the merry-go-round is:

    K =\frac{1}{2}\cdot (148.618\,kg\cdot m^{2})\cdot (0.905\,\frac{rad}{s} )^{2}

    K = 60.861\,J

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