Ask Question
29 July, 02:49

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?

+1
Answers (1)
  1. 29 July, 03:13
    0
    f = 8 N

    Explanation:

    from the question we are given the following values

    radius of the pulley (r) = 0.05 m

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

    angular acceleration (∝) = 2 rad/sec

    force (f) = ?

    Torque = force x radius = f x r Torque = moment of inertia x angular acceleration = I x ∝ therefore torque = f x r = I x ∝

    we can apply f x r = I x ∝ to get our force

    f x 0.05 = 0.2 x 2

    0.05f = 0.4

    f = 8 N
Know the Answer?
Not Sure About the Answer?
Find an answer to your question 👍 “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 ...” in 📗 Physics if the answers seem to be not correct or there’s no answer. Try a smart search to find answers to similar questions.
Search for Other Answers