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scZoUnD [109]
3 years ago
7

To practice Problem-Solving Strategy 12.1 for rotational dynamics problems. Suppose that you are holding a pencil balanced on it

s 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.
Physics
1 answer:
Cloud [144]3 years ago
4 0

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

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