A uniform rod rotates in a horizontal plane about a vertical axis through one end. The rod is 3.46 m long, weighs 12.8 N, and rotates at 226

Question

A uniform rod rotates in a horizontal plane about a vertical axis through one end. The rod is 3.46 m long, weighs 12.8 N, and rotates at 226 rev/min. Calculate (a) its rotational inertia about the axis of rotation and (b) the magnitude of its angular momentum about that axis.

in progress 0
Verity 4 years 2021-08-21T14:37:18+00:00 1 Answers 49 views 0

Answers ( )

  1. Delwyn
    0
    2021-08-21T14:38:36+00:00
    Reply

    Answer:

    a. Rotational inertia: 5.21kgm²

    b. Magnitude of it’s angular momentum: 123.32kgm²/s

    Explanation:

    Length of the rod = 3.46m

    Weight of the rod = 12.8 N

    Angular velocity of the rod= 226 rev/min

    a. Rotational Inertia (I) about its axis

    The formula for rotational inertia =

    I = (1/12×m×L²) + m × ( L ÷ 2)²

    Where L = length of the rod

    m = mass of the rod

    Mass of the rod is calculated by dividing the weight of the rod with the acceleration due to gravity.

    Acceleration due to gravity = 9.81m/s²

    Mass of the rod = 12.8N/ 9.81m/s²

    Mass of the rod = 1.305kg

    Rotational Inertia =

    (1/12× 1.305 × 3.46²)+ 1.305 ( 3.46÷2)²

    Rotational Inertia =  1.3019115 + 3.9057345

    Rotational Inertia = 5.207646kgm²

    Approximately = 5.21kgm²

    b. The magnitude of the rod’s angular momentum about the rotational axis is calculated as

    Rotational Inertia about its axis × angular speed of the rod.

    Angular speed of the rod is calculated as= (Angular velocity of the rod × 2π)/60

    = (226×2π) /60

    = 23.67 rad/s

    Rotational Inertia = 5.21kgm²

    The magnitude of the rod’s angular momentum about the rotational axis

    = 5.21kgm²× 23.67 rad/s

    = 123.3207kgm²/s

    Approximately = 123.32kgm²/s

Leave an answer

Browse

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