A disk-shaped merry-go-round of radius 2.83 m and mass 185 kg rotates freely with an angular speed of 0.701 rev/s . A 63.4 kg person running

Question

A disk-shaped merry-go-round of radius 2.83 m and mass 185 kg rotates freely with an angular speed of 0.701 rev/s . A 63.4 kg person running tangential to the rim of the merry-go-round at 3.51 m/s jumps onto its rim and holds on. Before jumping on the merry-go-round, the person was moving in the same direction as the merry-go-round’s rim. Part A What is the final angular speed of the merry-go-round

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Thành Đạt 5 months 2021-08-23T20:28:39+00:00 1 Answers 10 views 0

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    2021-08-23T20:30:26+00:00

    Answer:

    The final angular speed of the merry-go-round is 3.118\,\frac{rad}{s} \left(0.496\,\frac{rev}{s} \right).

    Explanation:

    Given the absence of external forces, the final angular speed of the merry-go-round can be determined with the resource of the Principle of Angular Momentum Conservation, which is described in this case as:

    I_{g, m} \cdot \omega_{o,m} + I_{g, p}\cdot \omega_{o,p} = (I_{g,m} + I_{g, p})\cdot \omega_{f}

    Where:

    I_{g,m} – Moment of inertia of the merry-go-round with respect to its axis of rotation, measured in kg\cdot m^{2}.

    I_{g,p} – Moment of inertia of the person with respect to the axis of rotation of the merry-go-round, measured in kg\cdot m^{2}.

    \omega_{o, m} – Initial angular speed of the merry-go-round with respect to its axis of rotation, measured in radians per second.

    \omega_{o,p} – Initial angular speed of the merry-go-round with respect to the axis of rotation of the merry-go-round, measured in radians per second.

    \omega_{f} – Final angular speed of the merry-go-round-person system, measured in radians per second.

    The final angular speed is cleared:

    \omega_{f} = \frac{I_{g,m}\cdot \omega_{o,m}+I_{g,p}\cdot \omega_{o,p}}{I_{g,m}+I_{g,p}}

    Merry-go-round is modelled as uniform disk-like rigid body, whereas the person can be modelled as a particle. The expressions for their moments of inertia are, respectively:

    Merry-go-round

    I_{g,m} = \frac{1}{2}\cdot M \cdot R^{2}

    Where:

    M – The mass of the merry-go-round, measured in kilograms.

    R – Radius of the merry-go-round, measured in meters.

    Person

    I_{g,p} = m\cdot r^{2}

    Where:

    m – The mass of the person, measured in kilograms.

    r – Distance of the person with respect to the axis of rotation of the merry-go-round, measured in meters.

    If M = 185\,kg, m = 63.4\,kg, R = r = 2.83\,m, the moments of inertia are, respectively:

    I_{g,m} = \frac{1}{2}\cdot (185\,kg)\cdot (2.83\,m)^{2}

    I_{g,m} = 740.823\,kg\cdot m^{2}

    I_{g,p} = (63.4\,kg)\cdot (2.83\,m)^{2}

    I_{g,p} = 507.764\,kg\cdot m^{2}

    The angular speed experimented by the person with respect to the axis of rotation of the merry-go-round is:

    \omega_{o,p} = \frac{v_{p}}{r}

    \omega_{o,p} = \frac{3.51\,\frac{m}{s} }{2.83\,m}

    \omega_{o,p} = 1.240\,\frac{rad}{s}

    Given that I_{g,m} = 740.823\,kg\cdot m^{2}, I_{g,p} = 507.764\,kg\cdot m^{2}, \omega_{o,m} = 4.405\,\frac{rad}{s} and \omega_{o,p} = 1.240\,\frac{rad}{s}, the final angular speed of the merry-go-round is:

    \omega_{f} = \frac{(740.823\,kg\cdot m^{2})\cdot \left(4.405\,\frac{rad}{s} \right)+(507.764\,kg\cdot m^{2})\cdot \left(1.240\,\frac{rad}{s} \right)}{740.823\,kg\cdot m^{2}+507.764\,kg\cdot m^{2}}

    \omega_{f} = 3.118\,\frac{rad}{s}

    \omega_{f} = 0.496\,\frac{rad}{s}

    The final angular speed of the merry-go-round is 3.118\,\frac{rad}{s} \left(0.496\,\frac{rev}{s} \right).

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