Question

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

Answer 1

Answer:

The initial angular acceleration is $16.8 s^{-2}$

Explanation:

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

$I = \dfrac{1}{12}ML^2+MD^2$

since $L = 1m$ & $D = 0.25m$, we have

$I = \dfrac{1}{12}M(1)^2+M(0.25m)^2$

$I = \dfrac{7}{48} M$

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

$\tau = I\alpha$

since $\tau = Mg D$

$MgD =I \alpha$

$MgD = \dfrac{7}{48} M \alpha$

solving for $\alpha$ we get

$\boxed{\alpha = \dfrac{48}{7} gD}$

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

$\boxed{\alpha = 16.8\: s^{-2}.}$

which is the initial angular acceleration.

Answer 2

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:

$I_{O} = I_{g} + M\cdot r^{2}$ (1)

Where:

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

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

$I_{O} = \frac{1}{12}\cdot M + \frac{1}{16}\cdot M$

$I_{O} = \frac{7}{48}\cdot M$ (2)

We know that initial angular acceleration ($\alpha$), 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:

$\Sigma M = M\cdot g\cdot r = I_{O}\cdot \alpha$ (3)

Where:

• $g$ - Gravitational acceleration, in meters per square second.

• $r$ - Distance between the point of rotation and the center of mass, in meters.

 

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

$M\cdot g \cdot r = \frac{7}{48} \cdot M\cdot \alpha$

$\alpha = \frac{48}{7}\cdot g \cdot r$ (4)

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

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

$\alpha = 16.812\,\frac{rad}{s^{2}}$

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

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