b. A string is wrapped around a pulley of radius 0.05 m and moment of inertia 0.2 kg  m2. If the string is pulled with a force F, the resul

Question

b. A string is wrapped around a pulley of radius 0.05 m and moment of inertia 0.2 kg  m2. If the string is pulled with a force F, the resulting angular acceleration of the pulley is 2 rad/s2. Determine the magnitude of the force F.

in progress 0
Nho 5 years 2021-08-22T03:03:45+00:00 1 Answers 176 views 0

Answers ( )

  1. giabao
    0
    2021-08-22T03:04:48+00:00
    Reply

    Answer:

    f = 8 N

    Explanation:

    Data provided in the question

    Radius of the pulley  = r = 0.05 m

    Moment of inertia = (I) = 0.2 kg.m^{2}

    Angular acceleration = ∝ = 2 rad/sec

    Based on the above information

    As we know that

    Torque is

    = force \times radius

    = f \times r

    And,

    Torque is also

    = moment\ of\ inertia \times angular\ acceleration

    = I \times \alpha

    So,

    We can say that

    f \times r = I \times \alpha

    f \times 0.05 = 0.2 \times 2

    0.05f = 0.4

    f = 8 N

    We simply applied the above formulas

Leave an answer

Browse

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