Physics, 04.04.2020 08:19 Jstylez9712

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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Answer from: mercedesamatap21hx0

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.

- Mass of the stick, in meters.

- 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:

- Gravitational acceleration, in meters per square second.

- 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.

To learn more on moment of inertia, we kindly invite to read this verified question: link

A uniform meter stick (with a length of 1 m) is pivoted to rotate about a horizontal axis through th

Answer from: c4thegoat

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 from: Quest

force required to detach it is given by

$f = 4 * 10^{-8} n$

charges are taken to be fundamental charge

$q_1 = q_2 = 1.6 * 10^{-19}c$

now we have

$f = \frac{kq_1q_2}{r^2}$

$4 * 10^{-8} = \frac{9*10^9 * 1.6*10^{-19}* 1.6* 10^{-19}}{r^2}$

$r = 7.6 * 10^{-11} m$

so above is the minimum distance at which it will detach

Answer from: Quest

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