Question

To practice Problem-Solving Strategy 12.1 for rotational dynamics problems. 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.

Answer 1

Answer:

α = 68 rad / s²

Explanation:

For this exercise we will use Newton's second law for rotational movement

     τ = I α

Where τ is the torque, I the moment of inertia and α the angular acceleration

Torque is the vector product of the distance perpendicular to the axis of rotation and the force that is the weight of the pencil (W); the distance (horizontal) is found with trigonometry

    sin 10 = x / (L / 2)

    x = L / 2 sin 10

    τ = W L / 2 sin  10

    τ = m g L / 2 sin 10

The moment of inertia of a pencil can be approximated to a thin rod with an axis of rotation at one end

     I = 1/12 m L²

We substitute in the first equation

    mg L / 2 sin 10 = (1/12 m L²) α

    g / 2 sin 10 = 1/12 L α

    α = 6g / L sin 10

Let's calculate

    α = 6 9.8 / 0.150 sin 10

    α = 68 rad / s²