A bucket of water of mass 20.0 kg is suspended by a rope wrapped around a windlass in the form of a solid cylinder 0.20 m in diameter, also

Question

A bucket of water of mass 20.0 kg is suspended by a rope wrapped around a windlass in the form of a solid cylinder 0.20 m in diameter, also of mass 20.0 kg. The cylinder is pivoted on a frictionless axle through its center. The bucket is released from rest at the top of a well and falls 20.0 m to the water. Neglect the weight of the rope.

What is the tension in the rope while the bucket is falling?

in progress 0
Nguyệt Ánh 3 months 2021-07-28T14:42:18+00:00 1 Answers 2 views 0

Answers ( )

    0
    2021-07-28T14:43:20+00:00

    Answer:T=65.33 N

    Explanation:

    Given

    mass of bucket with water is m=20\ kg

    Diameter of cylinder d=0.2\ m

    mass of cylinder M=20\ kg

    bucket has fall a distance of h=20\ m

    Net force on bucket

    \sum F_{net}=mg-T=ma\quad \ldots (i)

    Consider downward direction to be positive

    Tension(T) will provide torque to the cylinder

    T\times r=I\times \alpha

    where \alpha =\text{angular acceleration}

    T\times \frac{d}{2}=Mr^2\times \alpha

    T=\frac{Mr\alpha }{2}

    Substitute the value of T in (i)

    mg-\frac{Mr\alpha }{2}=ma\quad \text{[Pure rolling}\ a=\alpha r]

    mg-\frac{Ma}{2}=ma

    mg=(\frac{M}{2}+m)a

    a=\frac{20}{10+20}\times g

    a=\frac{2}{3}\times g

    Substitute the value of a in Tension equation

    T=\frac{Ma}{2}

    T=10\times \frac{2}{3}\times g

    T=65.33\ N

Leave an answer

Browse

Giải phương trình 1 ẩn: x + 2 - 2(x + 1) = -x . Hỏi x = ? ( )