Expressions have been found for the vertical acceleration of the cylinder ay and the angular acceleration α of the cylinder in the k^ direct

Question

Expressions have been found for the vertical acceleration of the cylinder ay and the angular acceleration α of the cylinder in the k^ direction; both expressions include an unknown variable, namely, the tension T in the vertical section of string. The string constrains the rotational and vertical motions, providing a third equation relating ay and α. Solve these three equations to find the vertical acceleration, ay, of the center of mass of the cylinder. Express ay in terms of g, m, r, and I; a positive answer indicates upward acceleration.

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Mộc Miên 4 years 2021-08-14T11:37:31+00:00 2 Answers 445 views 0

Answers ( )

  1. thachthao
    0
    2021-08-14T11:38:32+00:00
    Reply

    The first part of this question is missing and it says;

    A cylinder with a moment of inertia I (about its axis of symmetry), mass m, and radius r has a massless string wrapped around it which is tied to the ceiling(Figure 1) .

    At time t=0 the cylinder is released from rest at height h above the ground. Use g for the magnitude of the acceleration of gravity. Assume that the string does not slip on the cylinder. Let v⃗ represent the instantaneous velocity of the center of mass of the cylinder, and let ω⃗ represent the instantaneous angular velocity of the cylinder about its center of mass. Note that there are no horizontal forces present, so for this problem v⃗ =−vj^ and ω⃗ =−ωk^.

    Answer:

    ay = -mg / (m + I/r²)

    Explanation:

    Therefore we need to find an expression for T in terms of g, m, r, and I.

    Equation of motion says;

    may = T – mg

    The relationship between tension and moment of inertia is

    Iα = -Tr

    Tr = -Iα

    T = -Iα/r

    Since

    Angular acceleration is given as;

    α = ay/r

    Thus;

    Since, T = T = -Iα/r

    We now have,

    T = -I(ay/r)/r = -(Iay)/r²

    Now, since may = T – mg

    We now have;

    may = -(Iay)/r² – mg

    may + (Iay)/r² = -mg

    ay(m + I/r²) = -mg

    ay = -mg / (m + I/r²)

  2. Helga
    0
    2021-08-14T11:38:54+00:00
    Reply

    Complete Question

    A cylinder with a moment of inertia I (about its axis of symmetry), mass m, and radius r has a mass-less string wrapped around it which is tied to the ceiling .

    At time t=0 the cylinder is released from rest at height h above the ground. Use g for the magnitude of the acceleration of gravity. Assume that the string does not slip on the cylinder. Let v  represent the instantaneous velocity of the center of mass of the cylinder, and let \omega  represent the instantaneous angular velocity of the cylinder about its center of mass. Note that there are no horizontal forces present, so for this problem \= v =- v\r j \ and \ \omega =-\omega \r k.

    Expressions have been found for the vertical acceleration of the cylinder a_y and the angular acceleration α of the cylinder in the k^ direction; both expressions include an unknown variable, namely, the tension T in the vertical section of string. The string constrains the rotational and vertical motions, providing a third equation relating a_y and α. Solve these three equations to find the vertical acceleration, a_y, of the center of mass of the cylinder. Express a_y in terms of g, m, r, and I; a positive answer indicates upward acceleration.

    Answer:

    The vertical acceleration is  a_y = \frac{mg}{[m+ \frac{I}{r^2} ]}

    Explanation:

    The equation of motion is mathematically represented as

            ma_y = T -mg ---(1)

    The relation between the tension and the moment of inertia is

            I \alpha = -Tr

             T = \frac{-I\alpha }{r} ---(2)

    Now angular acceleration can be mathematically represented as

                     \alpha = \frac{a_y}{r}

    Now substituting this into equation 2

                  T = \frac{-Ia_y}{r^2} ---(3)

    Now substituting these into equation 1

              ma_y = \frac{-Ia_y}{r^2} - mg

                a_y [m + \frac{I}{r^2} ] = -mg

    Hence the vertical acceleration is evaluates as

                a_y = \frac{mg}{[m+ \frac{I}{r^2} ]}

     

       

               

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