A thin, horizontal rod with length l and mass M pivots about a vertical axis at one end. A force with constant magnitude F is applied to the

Question

A thin, horizontal rod with length l and mass M pivots about a vertical axis at one end. A force with constant magnitude F is applied to the other end, causing the rod to rotate in a horizontal plane. The force is maintained perpendicular to the rod and to the axis of rotation.

Calculate the magnitude of the angular acceleration of the rod.

in progress 0
Thiên Di 4 years 2021-08-16T13:23:04+00:00 1 Answers 24 views 0

Answers ( )

  1. Ladonna
    0
    2021-08-16T13:24:08+00:00
    Reply

    Answer:

    The magnitude of the angular acceleration is α = (3 * F)/(M * L)

    Explanation:

    using the equation of torque to the bar on the pivot, we have to:

    τ = I * α, where

    I = moment of inertia

    α = angular acceleration

    τ = torque

    The moment of inertia is equal to:

    I = (M * L^2)/3

    Also torque is equal to:

    τ = F * L

    Replacing:

    I * α = F * L

    α = (F * L)/I = (F * L)/((M * L^2)/3) = (3 * F)/(M * L)

Leave an answer

Browse

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