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

Question

quangkhai · Answer

The angular acceleration of the pencil is 17 rad/sec^2

Explanation:

When the object is balanced at its center of mass then it is said to be motionless because the net force acts through the center of mass. If we shift the location of the net force then which results in the rotation of the object about the center of mass. The angular acceleration possessed by the object is proportional to the torque generated by the net force about the center of mass.

τ $= I \alpha$

$I$ is the moment of inertia of a body

α is the angular acceleration

Length of the pencil L = 15 c
m = 0.15 m

Mass of the pencil = 10 g = 10 $\times 10^{-3}$ kg

Let α be the angular acceleration of pencil

θ be the inclination pencil from vertical

θ = 10
°

Assume the pencil as thin rod and moment of inertia of the thin rod with the axis of rotation at one end is

$I = \frac{mL^{2} }{3}$

When the pencil is balanced then the net torque acting on the pencil is zero and when we release the pencil and begin to fall then net torque acts on the pencil due to the weight of the pencil.

τ $= F \times d$

τ $= mgsin\theta \times \frac{L}{2}$

τ $= 10 \times 10^{-3} \times 9.81 \times sin10 \times \frac{0.15}{2}$

τ $= 1.277 \times 10^{-3} Nm$

We know that relation between torque and angular acceleration

τ $= I\alpha$

$\alpha = \frac{\tau}{I}$

$\alpha = \frac{\tau}{\frac{mL^{2} }{3} }$

$\alpha = \frac{1.277 \times 10^{-3} \times 3 }{10 \times 10^{-3} \times 0.15^{2} }$

$\alpha = 17.02$

$\alpha \approx 17 rad/sec^{2}$

The angular acceleration of the pencil is 17 rad/sec^2

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