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25 January, 01:48

Two boxes hang from a solid disk pulley that is free to rotate as the blocks rise/fall.

The left box has a mass m1 = 4.3 kg and the right box has a mass m2 = 2.1 kg. The pulley has mass m3 = 2.5 kg and radius R = 0.15 m. 1)

1) What is the linear acceleration of the left box? (up is the positive direction) m/s2

2) What is the angular acceleration of the pulley? (let counter-clockwise be positive) rad/s2

3) What is the tension in the string between the left mass and the pulley? N

4) What is the tension in the string between the right mass and the pulley? N

5) The boxes accelerate for a time t = 1.51 s. What distance does each box move in the time 1.51 s? m

6) What is the magnitude of the velocity of the boxes after the time 1.51 s? m/s

7) What is the magnitude of the final angular speed of the pulley? rad/s

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Answers (1)
  1. 25 January, 05:23
    0
    1) a = 3.07 m/s²

    2) α = 2.05 rad/s²

    3) T1 = 28.98 N

    4) T2 = 27.05 N

    5) x = 3.5 meter

    6) vf = 4.64 m/s

    7) ωf = 30.93 rad/s

    Explanation:

    Step 1: Data given

    Mass of box 1 = 4.3 kg

    Mass of box 2 = 2.1 kg

    Mass of pulley = 2.5 kg

    Radius = 0.15m

    1) What is the linear acceleration of the left box?

    A force F at the pulley rim gives an accelerating torque (T = F * r) and an angular acceleration dω/dt (rad/s²)

    T = F * r = I*dω/dt

    Fp = (I/r) * dω/dt

    ⇒ with I = ½mr²

    ⇒ with dω/dt = a/r

    Fp = (½mr²/r) * a/r

    Fn = ½ma = (4.2-1.9) * 9.81 = 22.56 N

    Fn - Fp = (m1 + m2) a

    Fn - ½ma = (m1+m2) a

    Fn = a (m1 + m2 + ½m)

    a = Fn / (m1 + m2 + ½m)

    a = 22.56 N / (4.2 + 1.9 + ½*2.5)

    a = 3.07 m/s²

    2) What is the angular acceleration of the pulley?

    angular acceleration α = a/r

    angular acceleration α = 3.07 / 0.15 = 2.05 rad/s²

    3) What is the tension in the string between the left mass and the pulley?

    T1 = m1 * (g-a)

    T1 = 4.3 * (9.81-3.07)

    T1 = 28.98 N

    4) What is the tension in the string between the right mass and the pulley?

    T2 = m2 (g+a)

    T2 = 2.1 * (9.81 + 3.07)

    T2 = 27.05 N

    5) The boxes accelerate for a time t = 1.51 s. What distance does each box move in the time 1.51 s?

    Initial speed v0 = 0 m/s

    Distance x = v0*t + ½at²

    x = 0*1.51 + ½ * 3.07*1.51²

    x = 3.5 meter

    6) What is the magnitude of the velocity of the boxes after the time 1.51 s?

    vf = the final speed after a movement of 3.5 meter

    vf²=v0²+2ax

    vf² = 2*a*x

    vf² = 2*3.07*3.5

    vf² = 21.49

    vf = √ (21.49) = 4.64 m/s

    7) What is the magnitude of the final angular speed of the pulley? rad/s

    final angular speed ωf = vf / r

    ωf = 4.64 m/s / 0.15m

    ωf = 30.93 rad/s
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