A uniform-density 8 kg disk of radius 0.25 m is mounted on a nearly frictionless axle. Initially it is not spinning. A string is wrapped tig

Question

A uniform-density 8 kg disk of radius 0.25 m is mounted on a nearly frictionless axle. Initially it is not spinning. A string is wrapped tightly around the disk, and you pull on the string with a constant force of 41 N through a distance of 0.9 m. Now what is the angular speed

in progress 0
Doris 3 years 2021-07-19T16:37:40+00:00 1 Answers 27 views 0

Answers ( )

  1. Adela
    0
    2021-07-19T16:38:59+00:00
    Reply

    Answer:

    Explanation:

    moment of inertia of the disc I = 1/2 m r²

    = .5 x 8 x .25²

    = .25 kg m²

    The wok done by force will be converted into rotational kinetic energy

    F x d = 1/2 I ω²

    F is force applied , d is displacement , I is moment of inertia of disc and ω

    is angular velocity of disc

    41 x .9 = 1/2 x .25 ω²

    ω² = .25

    ω = 17.18  rad / s

Leave an answer

Browse

Giải phương trình 1 ẩn: x + 2 - 2(x + 1) = -x . Hỏi x = ? ( )