Explore BrainMass

Explore BrainMass

    dynamics and inertia

    Not what you're looking for? Search our solutions OR ask your own Custom question.

    This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!

    The vehicle s used to transport supplies to and from the bottom of the 25-present grade. Each par of wheels one at A and the other at B has a mass of 140 kg with a radius of gyration of 100 mm. The total mass of the vehicle s 520 g. The vehicle is released from rest with a restraining force T of 500 N n the control cable which passes around the drum and is secured at D. The wheels roll without slipping.
    a) Calculate the inertia of wheel A about its centre in kg.m^2
    b) Calculate the inertia of wheel C about its centre in kg.m^2
    c) Determine the initial acceleration a of the vehicle in m/s^2

    Refer to attachment for diagram

    © BrainMass Inc. brainmass.com December 24, 2021, 11:21 pm ad1c9bdddf
    https://brainmass.com/physics/applied-physics/dynamics-inertia-562122

    Attachments

    SOLUTION This solution is FREE courtesy of BrainMass!

    a)
    mass of A, mA = 140 kg
    Radius of gyration, kA = 150 mm == 0.15 m

    inertia I = m*k^2

    inertia of A,
    IA = mA * kA^2 = 140*0.15^2 = 3.15 kg.m^2

    b)
    mass of C, mC= 40 kg
    radius of gyration of C, kC = 100 mm = 0.1 m

    Hence, inertia of C,
    IC = mC*kC^2 = 40*0.1^2 = 0.4 kg.m^2

    c)
    Vehicle moving downward.
    Let us assume, acceleration of the vehicle = a

    Therefore, cable acceleration = 2*a

    Hence, drum C angular acceleration:
    alphaC = 2*a/rC
    rC (radius of C) = 150 mm = 0.15 m

    Let us assume tension in other side of the cable (towards D) = T2
    Hence
    (T2 - T)*rC = IC*alphaC = IC *2*a/rC
    => T2 = T + 2*IC*a/rC^2

    For wheel A, and B, let us assume friction forces are fA and fB acting upward -- supporting rotation of wheels.
    Angular acceleration of A (alphaA) = angular acceleration of B (alphaB) = a/rA
    rA = rB = 200 mm = 0.2 m

    Hence,
    fA*rA = IA*alphaA
    => fA = IA*a/rA^2

    similarly,
    fB = IB*a/rB^2

    IA = IB = 3.15 kg.m^2

    Hence,
    fA = fB = IA*a/rA^2 = IB*a/rB^2

    Now, for complete system of the vehicle,
    mass of the system, M = 520 kg
    T = 500 N

    M*g*sin(x) - T - T2 - fA - fB = M*a

    where, x: angle of the plane with horizontal.
    tan(x) = 25/100 = 1/4 == sin(x) (For small angle).

    => M*g*sin(x) - T - (T + 2*IC*a/rC^2) - IA*a/rA^2 - IA*a/rA^2 = M*a

    => M*g*sin(x) - 2*T= (M + 2*IC/rC^2 + 2*IA/rA^2)*a

    => a = (M*g*sin(x) - 2*T)/(M + 2*IC/rC^2 + 2*IA/rA^2)

    => a = (520*9.8*1/4 - 2*500)/(520 + 2*0.4/0.15^2 + 2*3.15/0.2^2) = 0.38 m/s^2

    This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!

    © BrainMass Inc. brainmass.com December 24, 2021, 11:21 pm ad1c9bdddf>
    https://brainmass.com/physics/applied-physics/dynamics-inertia-562122

    Attachments

    ADVERTISEMENT