A 432 kg merry-go-round in the shape of a horizontal disk with a radius of 2.3 m is set in motion by wrapping a rope about the rim of the di

Question

A 432 kg merry-go-round in the shape of a horizontal disk with a radius of 2.3 m is set in motion by wrapping a rope about the rim of the disk and pulling on the rope. How large a torque would have to be exerted to bring the merry-go-round from rest to an angular speed of 3.1 rad/s in 2.1 s

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Thu Cúc 3 years 2021-08-22T16:18:23+00:00 1 Answers 9 views 0

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  1. Giakhanh
    0
    2021-08-22T16:19:48+00:00
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    Answer:

    τ = 1679.68Nm

    Explanation:

    In order to calculate the required torque you first take into account the following formula:

    \tau=I\alpha           (1)

    τ: torque

    I: moment of inertia of the merry-go-round

    α: angular acceleration

    Next, you use the following formulas for the calculation of the angular acceleration and the moment of inertia:

    \omega=\omega_o+\alpha t         (2)

    I=\frac{1}{2}MR^2           (3)       (it is considered that the merry-go-round is a disk)

    w: final angular speed = 3.1 rad/s

    wo: initial angular speed = 0 rad/s

    M: mass of the merry-go-round = 432 kg

    R: radius of the merry-go-round = 2.3m

    You solve the equation (2) for α. Furthermore you calculate the moment of inertia:

    \alpha=\frac{\omega}{t}=\frac{3.1rad/s}{2.1s}=1.47\frac{rad}{s^2}\\\\I=\frac{1}{2}(432kg)(2.3)^2=1142.64kg\frac{m}{s}

    Finally, you replace the values of the moment of inertia and angular acceleration in the equation (1):

    \tau=(1142.64kgm/s)(1.47rad/s^2)=1679.68Nm

    The required torque is 1679.68Nm

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