Consider a pulley of mass mp and radius R that has a moment of inertia 1/2mpR2. The pulley is free to rotate about a frictionless pivot at i

Question

Consider a pulley of mass mp and radius R that has a moment of inertia 1/2mpR2. The pulley is free to rotate about a frictionless pivot at its center. A massless string is wound around the pulley and the other end of the rope is attached to a block of mass m that is initially held at rest on frictionless inclined plane that is inclined at an angle β with respect to the horizontal. The downward acceleration of gravity is g. The block is released from rest .
How long does it take the block to move a distance d down the inclined plane?
Write your answer using some or all of the following: R, m, g, d, mp,

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Ladonna 3 years 2021-08-31T22:55:13+00:00 1 Answers 9 views 0

Answers ( )

  1. Orla
    0
    2021-08-31T22:56:13+00:00
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    Answer:

      a = \frac{m}{m+ \frac{1}{2} m_p} \ g \ sin \beta  ,       t = \sqrt{ \frac{2d}{a} }

    Explanation:

    To solve this exercise we must use Newton’s second law

    For the block

    let’s set a reference system with the x axis parallel to the plane

    X axis

             Wₓ – T = m a

    Y axis  

             N- W_y = 0

             N = W_y

    for pulley

              ∑τ = I α

               T R = (½ m_p R²) α

             

    let’s use trigonometry for the weight components

             sin β = Wₓ / W

             cos β = W_y / W

             Wx = W sin β

    angular and linear variables are related

              a = α R

              α = a / R

    we substitute and group our equations

             W sin β – T = m a

             T R = ½ m_p R² (a / R)

             

             W sin β – T = m a

                            T = ½ m_p a

    we solve the system of equations

             W sin β = (m + ½ m_p) a

              a = \frac{m}{m+ \frac{1}{2} m_p} \ g \ sin \beta

    let’s find the time to travel the distance (d) through the block

              x = v₀ t + ½ a t²

              d = 0 + ½ a t²

              t = \sqrt{ \frac{2d}{a} }

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