A 120-kg hollow spherical ball 1 m in diameter accelerates at a constant rate from rest to 5 rpm in 20 s and then continues to rotate at con

Question

A 120-kg hollow spherical ball 1 m in diameter accelerates at a constant rate from rest to 5 rpm in 20 s and then continues to rotate at constant speed. The ball can be treated as a spherical shell.
a. What is the moment of inertia of the disco ball? (3)
b. What is the angular acceleration of the ball in rad/s?? (5)
c. What is the torque delivered by the motor? (2)
d. Through how many turns does the ball rotate during the 20-s acceleration period? (7)
e. What is the tangential speed at t=20 s of a point on the surface of the ball at a distance 0.5 m (measured perpendicularly) from the axle? (3)

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Đan Thu 5 years 2021-07-26T15:42:09+00:00 1 Answers 15 views 0

Answers ( )

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    2021-07-26T15:43:23+00:00

    Answer:

    Explanation:

    a )

    moment of inertia of hollow ball

    = 2 / 3  mR²  , m is mass and R is radius of the ball

    = 2 / 3 x 120 x .5²

    = 20 kg m²

    b )

    5 rpm =      5 / 60 rps

    n = .0833

    angular velocity ω = 2πn= 2 x 3.14 x .0833=  .523 rad /s

    angular acceleration = increase in angular velocity / time

    = .523 – 0 / 20

    α = .02615 rad /s²

    c )

    Torque = moment of inertia x angular acceleration

    = 20 x .02615

    = .523 Nm

    d )

    θ = 1/2 α t²

    = .5 x  .02615 x 20²

    = 5.23

    2π n = 5.23 where n is required number

    n = .83

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