Ask questions and get helpful answers.

A 0.230-kg wooden rod is 1.55 m long and pivots at one end. It is held horizontally and then released.

1) What is the angular acceleration of the rod after it is released? three sf with units
2) What is the linear acceleration of a spot on the rod that is 1.02 m from the axis of rotation?
3) At what location along the rod should a die be placed so that the die just begins to separate from the rod as it falls?

  1. 👍
  2. 👎
  3. 👁
  4. ℹ️
  5. 🚩
1 answer
  1. T = moment at pivot = m g (1.55/2)

    alpha = T/moment of inertia

    moment of inertia about end = (1/3)mL^2
    = (1/3)(m)(1.55^2)
    so

    alpha = g(1.55/2) /[(1/3)(1.55^2)]
    that is part 1

    2) 1.02 alpha

    3| g = alpha * x

    1. 👍
    2. 👎
    3. ℹ️
    4. 🚩

Answer this Question

Related Questions

Still need help?

You can ask a new question or browse existing questions.