uniform disk with mass 40.0 kg and radius 0.200 m is pivoted at its center about a horizontal, frictionless axle that is stationary. The dis

Question

uniform disk with mass 40.0 kg and radius 0.200 m is pivoted at its center about a horizontal, frictionless axle that is stationary. The disk is initially at rest, and then a constant force 퐹=30.0푁is applied tangent to the rim of the disk.a) What is the magnitude 푣of the tangential velocity of a point on the rim of the disk after the disk has turned through 0.200 revolution?b) What is the magnitude 푎of the resultant acceleration of that point in part a?

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Thông Đạt 5 years 2021-07-22T10:42:50+00:00 1 Answers 452 views 0

Answers ( )

  1. binhan
    1
    2021-07-22T10:44:35+00:00
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    Answer:

    The magnitude of the tangential velocity is v= 0.868 m/s

    The magnitude of the resultant acceleration at that point is  a = 4.057 m/s^2

    Explanation:

    From the question we are told that

          The mass of the uniform disk is m_d = 40.0kg

           The radius of the uniform disk is R_d = 0.200m

           The force applied on the disk is F_d = 30.0N

    Generally the angular speed i mathematically represented as

                 w = \sqrt{2 \alpha \theta}

    Where \theta is the angular displacement given from the question as

               \theta = 0.2000 rev = 0.2000 rev * \frac{2 \pi \ rad }{1 rev}

                     =1.257\ rad

       \alpha is the angular acceleration which is mathematically represented as

                        \alpha = \frac{torque }{moment \ of \ inertia} = \frac{F_d * R_d}{I}

        The moment of inertial is mathematically represented as

                         I = \frac{1}{2} m_dR^2_d

    Substituting values

                        I = 0.5 * 40 * 0.200^2

                            = 0.8kg \cdot m^2

    Considering the equation for angular acceleration

                   \alpha = \frac{torque }{moment \ of \ inertia} = \frac{F_d * R_d}{I}

    Substituting values

                   \alph[tex]\alpha = \frac{(30.0)(0.200)}{0.8}[/tex]

                       = 7.5 rad/s^2

    Considering the equation for angular velocity

        w = \sqrt{2 \alpha \theta}

    Substituting values

         w =\sqrt{2 * (7.5) * 1.257}

             = 4.34 \ rad/s

    The tangential velocity of a given point on the rim is mathematically represented as

                     v = R_d w

    Substituting values

                        = (0.200)(4.34)

                         v= 0.868 m/s

    The radial acceleration at hat point  is mathematically represented as

                \alpha_r = \frac{v^2}{R}

                      = \frac{0.868^2}{0.200^2}

                     = 3.7699 \ m/s^2

    The tangential acceleration at that point is mathematically represented as

                   \alpha _t = R \alpha

    Substituting values

               \alpha _t = (0.200) (7.5)

                     = 1.5 m/s^2

    The magnitude of resultant acceleration at that point is

                     a = \sqrt{\alpha_r ^2+ \alpha_t^2 }

    Substituting values

                    a = \sqrt{(3.7699)^2 + (1.5)^2}

                       a = 4.057 m/s^2

             

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