Point P is located on the outer surface of the top, a distance h = 29.0 mm h=29.0 mm above the ground. The angle that the outer surface of t

Question

Point P is located on the outer surface of the top, a distance h = 29.0 mm h=29.0 mm above the ground. The angle that the outer surface of the top makes with the rotation axis of the top is θ = 19.0 ∘ . θ=19.0∘. If the final tangential speed v t vt of point P is 1.45 m/s, 1.45 m/s, what is the top’s moment of inertia I ?

in progress 0
Vân Khánh 7 mins 2021-07-22T20:27:14+00:00 1 Answers 0 views 0

Answers ( )

    0
    2021-07-22T20:28:57+00:00

    Answer:

    Explanation:

    Given that,

    Height h = 29mm = 0.029m

    Makes an angle of θ = 19.0°

    Then, the radius is

    h = rCosθ

    r = h/Cosθ

    r = 0.029/Cos19

    r = 0.031m

    Linear speed v = 1.45m/s

    The relationship between linear speed and angular speed is

    v = ωr

    ω = v/r

    ω = 1.45/0.031

    ω = 47.28 rad/sec

    Angular momentum is given as

    L = Iω

    Therefore, I = L / ω

    So, the question is not complete check attachment for the first part of the question, from the first part we will get the angular momentum

    Given that,

    r = 3.46mm = 0.00346m

    Tension in string F = 3.15N

    Time taken t = 0.62s.

    Then, Torque is given as

    T = dL/dt

    Tdt = dL

    Integrate both sides

    T•t + C = L(t)

    At t = 0 L=0

    Therefore C = 0

    T•t = L(t)

    The torque from the two string is given as

    T = 2Fr

    Therefore,

    L(t) = 2•F•r•t

    L(t) = 2 × 3.15 × 0.00346 × 0.62

    L(0.62) = 0.0135 kgm²/s

    Then, back to our question,

    I = L / ω

    I = 0.0135/47.28

    I = 2.86 × 10^-4 kgm²

Leave an answer

Browse

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