Question

Suppose that you are holding a pencil balanced on its point. If you release the pencil and it begins to fall, what will be the angular acceleration when it has an angle of 10.0 degrees from the vertical? A typical pencil has an average length of 15.0 cm and an average mass of 10.0 g . Assume the tip of the pencil does not slip as it falls.What is the angular acceleration of the pencil when it makes an angle of 10.0 degrees with the vertical?

Answer 1

Answer:

The angular acceleration of the pencil α  = 17 rad·s⁻²

Explanation:

Using Newton's second angular law or torque to find angular acceleration, we get the following expressions:

    τ = I α                              (1)

    W r = I α                          (2)

The weight is that the pencil has is,

   sin 10 = r / (L/2)

   r = L/2(sin(10))

 

The shape of the pencil can be approximated to be a cylinder that rotates on one end and therefore its moment of inertia will be:

    I = 1/3 M L²

Thus,

   mg(L / 2)sin(10) = (1/3 m L²)(α) 

   α(f) = 3/2(g) / Lsin(10)

   α  = 3/2(9.8) / 0.150sin(10)

   α  = 17 rad·s⁻²

Therefore, the angular acceleration of the pencil is 17 rad·s⁻²