show answer No Attempt 50% Part (b) Now consider a disk of mass 8 kg with radius 0.55 m. Disk rotates around itself once every 0.35 sec. Wha

Question

show answer No Attempt 50% Part (b) Now consider a disk of mass 8 kg with radius 0.55 m. Disk rotates around itself once every 0.35 sec. What is the rotational energy of this disk?

in progress 0
Cherry 4 weeks 2021-08-19T13:28:22+00:00 2 Answers 0 views 0

Answers ( )

    0
    2021-08-19T13:29:27+00:00

    Answer:

    Explanation:

    mass of disc, m = 8 kg

    radius of disc, r = 0.55 m

    Time period, t = 0.35 s

    Angular velocity, ω = 2π/T

    ω = 2 x 3.14 / 0.35 = 17.94 rad/s

    moment of inertia of the disc, I = 0.5 mr²

    I = 0.5 x 8 x 0.55 x 0.55 = 1.21 kgm²

    The rotational kinetic energy is given by

    K = 0.5 x I x ω²

    K = 0.5 x 1.21 x 17.94 x 17.94 = 194.72 J

    0
    2021-08-19T13:29:50+00:00

    Answer:

    KE=194.9750\ J

    Explanation:

    Given:

    mass of disk, m=8\ kg

    radius of the disk, r=0.55\ m

    time taken by the disk to complete on rotation, T=0.35\ s

    We know that rotational kinetic energy is given as:

    KE=\frac{1}{2}\times I.\omega^2 ……………………….(1)

    Now the angular speed of the disk:

    \omega=\frac{2\pi}{T}

    \omega=\frac{2\pi}{0.35}

    \omega=17.952\ rad.s^{-1}

    The moment of inertia of a disk is given as:

    I=\frac{1}{2} m.r^2

    I=0.5\times 8\times 0.55^2

    I=1.21\ kg.m^2

    Now using eq. (1):

    KE=\frac{1}{2} \times 1.21\times 17.952^2

    KE=194.9750\ J

Leave an answer

Browse

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