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A uniform meter stick (with a length of 1 m) is pivoted to rotate about a horizontal axis through the 25 cm mark on the stick. The stick is released from rest in a horizontal position. The moment of inertia of a uniform rod about an axis perpendicular to the rod and through the center of mass of the rod is given by 1 12 ML2. Determine the magnitude of the initial angular acceleration of the stick

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Poltergeist

Answer:

The initial angular acceleration is [tex]16.8 s^{-2}[/tex]

Explanation:

From the parallel axis theorem the moment of inertia about 25cm mark is

[tex]I = \dfrac{1}{12}ML^2+MD^2[/tex]

since [tex]L = 1m[/tex] & [tex]D = 0.25m[/tex], we have

[tex]I = \dfrac{1}{12}M(1)^2+M(0.25m)^2[/tex]

[tex]I = \dfrac{7}{48} M[/tex]

Now, the gravitational force (equal to the weight of the object) acts as a torque [tex]\tau[/tex] on the center of mass of the rod, which induces angular acceleration [tex]\alpha[/tex]according to

[tex]\tau = I\alpha[/tex]

since [tex]\tau = Mg D[/tex]

[tex]MgD =I \alpha[/tex]

[tex]MgD = \dfrac{7}{48} M \alpha[/tex]

solving for [tex]\alpha[/tex] we get

[tex]\boxed{\alpha = \dfrac{48}{7} gD}[/tex]

putting in [tex]g= 9.8m/s^2[/tex] and [tex]D = 0.25m[/tex] we get:

[tex]\boxed{\alpha = 16.8\: s^{-2}.}[/tex]

which is the initial angular acceleration.

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The initial angular acceleration of the stick is 16.812 radians per square second.

First, we calculate the resulting moment of inertia by Steiner theorem:

[tex]I_{O} = I_{g} + M\cdot r^{2}[/tex] (1)

Where:

  • [tex]I_{g}[/tex] - Moment of inertia with respect to center of mass, in kilogram-square meters.
  • [tex]M[/tex] - Mass of the stick, in meters.
  • [tex]r[/tex] - Distance between the point of rotation and the center of mass, in meters.
  • [tex]I_{O}[/tex] - Moment of inertia with respect to the point of rotation, in meters.

If we know that [tex]I_{g} = \frac{1}{12} \cdot M\cdot L^{2}[/tex], [tex]L = 1\,m[/tex] and [tex]r = 0.25\,m[/tex], then the formula for the moment of inertia is:

[tex]I_{O} = \frac{1}{12}\cdot M + \frac{1}{16}\cdot M[/tex]

[tex]I_{O} = \frac{7}{48}\cdot M[/tex] (2)

We know that initial angular acceleration ([tex]\alpha[/tex]), in radians per square second, is solely due to gravity. By the second Newton's law and the D'Alembert principle, we derive an expression for the initial angular acceleration of the stick:

[tex]\Sigma M = M\cdot g\cdot r = I_{O}\cdot \alpha[/tex] (3)

Where:

  • [tex]g[/tex] - Gravitational acceleration, in meters per square second.
  • [tex]r[/tex] - Distance between the point of rotation and the center of mass, in meters.

 

By (2) and (3) we have the following simplified expression:

[tex]M\cdot g \cdot r = \frac{7}{48} \cdot M\cdot \alpha[/tex]

[tex]\alpha = \frac{48}{7}\cdot g \cdot r[/tex] (4)

If we know that [tex]g = 9.807\,\frac{m}{s^{2}}[/tex] and [tex]r = 0.25\,m[/tex], then the initial angular acceleration of the stick is:

[tex]\alpha = \frac{48}{7}\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (0.25\,m)[/tex]

[tex]\alpha = 16.812\,\frac{rad}{s^{2}}[/tex]

The initial angular acceleration of the stick is 16.812 radians per square second.

To learn more on moment of inertia, we kindly invite to read this verified question: https://brainly.com/question/6953943

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