A string passing over a pulley has a 3.85-kg mass hanging from one end and a 2.60-kg mass hanging from the other end. The pulley is a unifor

Question

A string passing over a pulley has a 3.85-kg mass hanging from one end and a 2.60-kg mass hanging from the other end. The pulley is a uniform solid cylinder of radius 4.5 cm and mass 0.79 kg .
A. If the bearings of the pulley were frictionless, what would be the acceleration of the two masses?
B. In fact, it is found that if the heavier mass is given a downward speed of 0.20 m/s , it comes to rest in 6.4 s . What is the average frictional torque acting on the pulley?

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Nem 5 years 2021-08-13T14:27:24+00:00 1 Answers 20 views 0

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  1. Acacia
    0
    2021-08-13T14:28:32+00:00
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    Answer:

    Explanation:

    Let the acceleration be a of the system

    T₁ and T₂ be the tension in the string attached with 3.85 and 2.6 kg of mass

    for motion of 3.85 kg , applying newton’s law

    3.85g – T₁ = 3.85 a

    for motion of 2.6 kg

    T₂ – 2.6g = 2.6 a

    T₂ – T₁ + 1.25 g = 6.45 a

    T₁ – T₂ = 1.25 g –  6.45 a

    for motion of pulley

    (T₁ – T₂ ) x R = I x α where R is radius of pulley , I is its moment of inertia and α is angular acceleration

    (T₁ – T₂ ) x R = 1 /2  m R² x a / R

    (T₁ – T₂ )  =   m  x a / 2 = .79 x a / 2 = . 395 a

    1.25 g –  6.45 a = .395 a

    1.25 g = 6.845 a

    a = 1.79 m /s²

    B )

    When heavier mass is given speed of .2 m /s , it comes to rest in 6.4 s

    Average deceleration = .2 / 6.4 = .03125 m /s²

    Total deceleration created by frictional torque = 1.79 + .03125

    = 1.82125 m /s²

    If R be the average frictional torque  acting on the pulley

    angular deceleration of pulley = a / R

    = 1.82125 / .045

    = 40.47 rad /s²

    Now  R = I x 40.47 , I is moment of inertia of pulley

    = 1 /2 x .79 x .045² x 40.47

    = .0323 N.m

    Torque created = .0323 Nm

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