The equation of motion(EOM) of rotation of couple moment of magnet bar which rotation axis is fixed at middle and placed in uniform magnetic fields

Question

The magnet bar which rotation axis itself is fixed at the middle of the bar and the magnet bar is placed in the uniform magnetic fields.

$$ I \left[ \text{kg} \cdot \text{m}^{2} \right] :=\text{motion of inertia} $$

$$ N_{\theta_{} }= - \frac{ \partial U }{ \partial \theta_{} } = -MH \sin\left(\theta_{} \right) \left[ \text{N} \cdot m \right] :=\text{couple moment which acts against the magnet bar} 　 $$

The equation of motion of rotation is given as below.

$$ I \frac{ d ^{2} \theta_{} }{ d t ^{2} } = -MH \sin\left(\theta_{} \right) $$

What actually $~ \frac{ d ^{2} \theta_{} }{ d t ^{2} } \left[ \frac{ 1 }{ \text{s}^{2} } \right] ~$ means?

$$ \left[ \text{kg}\cdot \text{} \text{m}^{2} \right] \cdot \left[? \right]= \left[ \text{N} \cdot m \right]~~ \leftarrow~~ \text{I wanted to find out the unit of } ~\frac{ d ^{2} \theta_{} }{ d t ^{2} } $$

$$ \left[ \text{kg}\cdot \text{} \text{m}^{2} \right] \cdot \left[? \right]= \left[\left( \text{kg} \cdot \frac{ \text{m} }{ \text{s} ^{2} } \right) \cdot m \right] $$

$$ = \left[ \frac{ \text{kg} \cdot \text{m}^{2} }{ \text{s}^{2} } \right] $$

$$ \therefore ~~ \left[ ? \right] =\left[ \frac{ 1 }{ \text{s}^{2} } \right] $$

Answer 1

1. What actually $~ \frac{ d ^{2} \theta_{} }{ d t ^{2} } \left[ \frac{ 1 }{ \text{s}^{2} } \right] ~$ means?

This is angular acceleration, the change rate of the angular velocity. $$ \alpha = \frac{d \omega}{dt} = \frac{d}{dt} \frac{d\theta}{dt} = \frac{d^2\theta}{dt^2} $$ Since the angle (in radians) is of no unit, therefore the angular acceleration has a unit $\frac{1}{s^2}$.

2. Dimension analysis. Equation of motion: $$ I \frac{ d ^{2} \theta_{} }{ d t ^{2} } = -MH \sin\left(\theta_{} \right) $$ The right-hand-side is torque of unit $N-m = \frac{Kg-m^2}{s^2} $. In the left-hand-side, the inertial moment $I = kg-m^2$ Thus \begin{align} I \alpha &= \tau;\\ \left(kg-m^2\right) \alpha &= \frac{kg-m^2}{s^2};\\ \alpha &= \frac{1}{s^2} \end{align}