Recall your experimental setup from Lab 05A: a constant force was applied to a disc by attaching a mass to a light string wrapped around a m

Question

Recall your experimental setup from Lab 05A: a constant force was applied to a disc by attaching a mass to a light string wrapped around a mass-less pulley and hanging the mass over the edge of the apparatus. In the lab, you used energy conservation arguments to derive an expression for the angular velocity of the disc after the mass had fallen a distance x . Your goal now is to use kinematics and dynamics to confirm your expression. Use the following symbols throughout this question: m is the mass of the hanging mass, M is the mass of the disc, r is the radius of the pulley, R is the radius of the disc, x is the distance the mass has fallen, and g is the acceleration due to gravity. What is the linear acceleration of the mass after it has fallen a distance x

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Adela 1 year 2021-08-21T23:54:31+00:00 1 Answers 3 views 0

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    2021-08-21T23:55:42+00:00

    Answer:

      w = [tex]\sqrt{\frac{2gy}{r^2 + \frac{1}{2} R^2 } }[/tex]

    Explanation:

    For this exercise let’s start by applying Newton’s second law to the mass with the string

                    W – T = m a

    In this case, as the system is going down, we will assume the vertical directional down as positive.

                    T = W – m a

    Now we apply Newton’s second law for rotational motion to the pulley of radius r. We will assume the positive counterclockwise rotations

                    ∑ τ = I α

                    T r = I α

    the moment of inertia of the disk is

                   I = ½ M R²

    angular and linear acceleration are related

                   a = α r

    we substitute

                   T r = (½ m R²) (a / r)

                   T = ½ m ([tex]\frac{R}{r}[/tex] )² a

    we write our two equations

                   T = W – m a

                   T = ½ m ([tex]\frac{R}{r}[/tex] )² a

    we solve the system of equations

                  W – m a = ½ m (\frac{R}{r} )² a

                  m g = m a [ 1 + ½ (\frac{R}{r} )² ]

                 a = [tex]\frac{g}{ 1 + \frac{1}{2} (\frac{R}{r})^2 }[/tex]

    this acceleration is constant throughout the trajectory, so with the angular and lineal kinematics relations

                 w² = w₀² + 2 α θ

                 v² = v₀² + 2 a y

    as the system is released its initial angular velocity is zero

                  w² = 0 + 2 α θ

                  v² = 0 + 2 a y

    we look for the angular acceleration

                  a =α r

                  α = a / r

                  α = [tex]\frac{g}{r (1 + \frac{1}{2} (\frac{R}{r})^2 }[/tex]

    we look for the angle, remember that they must be measured in radians

                 θ = s / r

    in this case we approximate the arc to the distance

                s = y

                θ = y / r

    we substitute

                w = [tex]\sqrt{2 \frac{g}{ r( \frac{1}{2} (\frac{R}{r})^2 } \frac{y}{r} }[/tex]

                w = [tex]\sqrt{\frac{2gy}{r^2 + \frac{1}{2} R^2 } }[/tex]

        for the simple case where r = R

                w = [tex]\sqrt{ \frac{2gy}{ \frac{3}{2} R^2 } }[/tex]

                w = [tex]\sqrt{ \frac{4}{3} \frac{gy}{R^2} }[/tex]

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