A block of mass m rests on a frictionless, horizontal surface. A cord attached to the block passes over a pulley whose radius is R, to a han

Question

A block of mass m rests on a frictionless, horizontal surface. A cord attached to the block passes over a pulley whose radius is R, to a hanging bucket with mass m . The system is released from rest, and the block is observed to move a distance d over a time interval of T. What is the moment of inertia of the pulley about its axis? Make sure to give your answer in terms of the given quantities.

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Khánh Gia 2 months 2021-07-15T09:02:50+00:00 1 Answers 1 views 0

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    2021-07-15T09:03:59+00:00

    Answer:

    Explanation:

    Let the tension in the cord be T₁ and T₂  .

    for motion of block placed on horizontal table

    T₁ = m a  , a is acceleration of the whole system .

    for motion of hanging bucket of mass m

    mg – T₂ = ma

    adding the two equation

    mg + T₁- T₂ = 2ma

    for rotational motion of the pulley

    torque = moment of inertia x angular acceleration

    (T₂ – T₁) R = I x α , I is moment of inertia of pulley , α is angular acceleration .

    (mg – 2ma ) R = I x α

    (mg – 2ma ) R = I x a / R

    (mg – 2ma ) R² = I x a

    mgR² =  2ma R² + I x a

    a = mgR² / (2m R² + I )

    Since body moves by distance d in time T

    d = 1/2 a T²

    a = 2d / T²

    mgR² / (2m R² + I ) = 2d / T²

    mgR²T² = 2d x (2m R² + I )

    mgR²T² –  4dm R² =  2dI

    m R² ( gT² – 4d ) = 2dI

    I =  m R² ( gT² – 4d ) ] / 2d .

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