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The moment of inertia is the rotational equivalent of mass in translational motion. In the case of a rod rotating around an axis perpendicular to its own axis, the simplest case occurs when the rotation takes place around the center of mass.

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Iframe

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Description

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When the axis of rotation does not pass through the center of mass (CM), the moment of inertia $I$ can be calculated using the Steiner's Theorem. To do this, we start with the moment of inertia with respect to the center of mass, for example:

• For a bar with a perpendicular axis, it is calculated

• For a cylinder with a perpendicular axis, it is calculated

• For a cylinder with a parallel axis, it is calculated

• For a parallelepiped, it is calculated

• For a cube, it is calculated

• For a sphere, it is calculated

Then, the product of mass and the square of the distance between the axis of rotation and the center of mass is added using

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Image

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A bar with mass $m$ and length $l$ rotating around its center, which coincides with the center of mass:

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Parameters

$m$

m

Body mass

kg

$\alpha_0$

alpha_0

Constant Angular Acceleration

rad/s^2

$d$

d

Distance Center of Mass and Axis

m

$\theta_0$

theta_0

Initial Angle

rad

$\omega_0$

omega_0

Initial Angular Speed

rad/s

$l$

l

Length of the Bar

m

$I_{CM}$

I_CM

Moment of Inertia at the CM of a thin Bar, perpendicular Axis

kg m^2

$I$

I

Moment of inertia for axis that does not pass through the CM

kg m^2

$r$

r

Radio

m

$t_0$

t_0

Start Time

s

$T$

T

Torque

N m

Variables

$\theta$

theta

Angle

rad

$\omega$

omega

Angular Speed

rad/s

$F$

F

Force

N

$t$

t

Time

s

Calculations

• Alternative method
• Traditional method
• Strategy
• 2D
• 3D

Equation

Number

First, select the equation: to , then, select the variable: to

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Calculations

Symbol

Equation

Solved

Translated

Calculations

Symbol

Equation

Solved

Translated

Variable Given Calculate Target : Equation To be used

Equations

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Equation

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The moment of inertia of a rod rotating around a perpendicular ($\perp$) axis passing through the center is obtained by dividing the body into small volumes and summing them:

resulting in

$I_{CM} =\displaystyle\frac{1}{12} m l ^2$

Conditions

Variables

$m$

Body mass

$kg$

6150

$l$

Length of the Bar

$m$

6151

$I_{CM}$

Moment of Inertia at the CM of a thin Bar, perpendicular Axis

$kg m^2$

5323

Relation

.

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Equation

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The moment of inertia for axis that does not pass through the CM ($I$) can be calculated using the moment of Inertia Mass Center ($I_{CM}$) and adding the moment of inertia of the body mass ($m$) as if it were a point mass at the distance Center of Mass and Axis ($d$):

$I = I_{CM} + m d ^2$

Conditions

Variables

$m$

Body mass

$kg$

6150

$d$

Distance Center of Mass and Axis

$m$

5285

$I$

Moment of inertia for axis that does not pass through the CM

$kg m^2$

5315

$I_{CM}$

$I_{CM}$

Moment of Inertia at the CM of a thin Bar, perpendicular Axis

$kg m^2$

5323

Relation

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Equation

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In the case where the moment of inertia is constant, the derivative of angular momentum is equal to

which implies that the torque is equal to

$T = I \alpha_0$

$T = I \alpha$

Conditions

Variables

$\alpha$

$\alpha_0$

Constant Angular Acceleration

$rad/s^2$

5298

$I$

$I$

Moment of inertia for axis that does not pass through the CM

$kg m^2$

5315

$T$

Torque

$N m$

4988

Relation

Development

Since the moment is equal to

$L = I \omega$

it follows that in the case where the moment of inertia doesn't change with time,

$T=\displaystyle\frac{dL}{dt}=\displaystyle\frac{d}{dt}(I\omega) = I\displaystyle\frac{d\omega}{dt} = I\alpha$

which implies that

$T = I \alpha$

.

This relationship is the equivalent of Newton's second law for rotation instead of translation.

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Equation

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Since the relationship between angular momentum and torque is

its temporal derivative leads us to the torque relationship

$T = r F$

Conditions

Variables

$F$

Force

$N$

4975

$r$

Radio

$m$

9884

$T$

Torque

$N m$

4988

Relation

Development

Si se deriva en el tiempo la relación para el momento angular

$L = r p$

para el caso de que el radio sea constante

$T=\displaystyle\frac{dL}{dt}=r\displaystyle\frac{dp}{dt}=rF$

por lo que

$T = r F$

The body's rotation occurs around an axis in the direction of the torque, which passes through the center of mass.

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Equation

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With the constant Angular Acceleration ($\alpha_0$), the angular Speed ($\omega$) forms a linear relationship with the time ($t$), incorporating the variables the initial Angular Speed ($\omega_0$) and the start Time ($t_0$) as follows:

$\omega = \omega_0 + \alpha_0 ( t - t_0 )$

Conditions

Variables

$\omega$

Angular Speed

$rad/s$

6068

$\alpha_0$

Constant Angular Acceleration

$rad/s^2$

5298

$\omega_0$

Initial Angular Speed

$rad/s$

5295

$t_0$

Start Time

$s$

5265

$t$

Time

$s$

5264

Relation

Development

If we assume that the mean Angular Acceleration ($\bar{\alpha}$) is constant, equivalent to the constant Angular Acceleration ($\alpha_0$), then the following equation applies:

$\bar{\alpha} = \alpha_0$

Therefore, considering the difference in Angular Speeds ($\Delta\omega$) along with the angular Speed ($\omega$) and the initial Angular Speed ($\omega_0$):

$\Delta\omega = \omega - \omega_0$

and the time elapsed ($\Delta t$) in relation to the time ($t$) and the start Time ($t_0$):

$\Delta t \equiv t - t_0$

the equation for the mean Angular Acceleration ($\bar{\alpha}$):

$\bar{\alpha} \equiv \displaystyle\frac{ \Delta\omega }{ \Delta t }$

can be expressed as:

$\alpha_0 = \alpha = \displaystyle\frac{\Delta \omega}{\Delta t} = \displaystyle\frac{\omega - \omega_0}{t - t_0}$

Solving this, we obtain:

$\omega = \omega_0 + \alpha_0 ( t - t_0 )$

This equation represents a straight line in the angular velocity versus time plane.

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Equation

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Given that the total displacement corresponds to the area under the angular velocity versus time curve, in the case of a constant Angular Acceleration ($\alpha_0$), it is determined that the displacement the angle ($\theta$) with the variables the initial Angle ($\theta_0$), the time ($t$), the start Time ($t_0$), and the initial Angular Speed ($\omega_0$) is as follows:

$\theta = \theta_0 + \omega_0 ( t - t_0 )+\displaystyle\frac{1}{2} \alpha_0 ( t - t_0 )^2$

Conditions

Variables

$\theta$

Angle

$rad$

6065

$\alpha_0$

Constant Angular Acceleration

$rad/s^2$

5298

$\theta_0$

Initial Angle

$rad$

5296

$\omega_0$

Initial Angular Speed

$rad/s$

5295

$t_0$

Start Time

$s$

5265

$t$

Time

$s$

5264

Relation

Development

In the case of the constant Angular Acceleration ($\alpha_0$), the angular Speed ($\omega$) as a function of the time ($t$) follows a linear relationship with the start Time ($t_0$) and the initial Angular Speed ($\omega_0$) in the form of:

$\omega = \omega_0 + \alpha_0 ( t - t_0 )$

Given that the angular displacement is equal to the area under the angular velocity-time curve, in this case, one can add the contributions of the rectangle:

$\omega_0(t-t_0)$

and the triangle:

$\displaystyle\frac{1}{2}\alpha_0(t-t_0)^2$

This leads us to the expression for the angle ($\theta$) and the initial Angle ($\theta_0$):

$\theta = \theta_0 + \omega_0 ( t - t_0 )+\displaystyle\frac{1}{2} \alpha_0 ( t - t_0 )^2$

This expression corresponds to the general form of a parabola.

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Equation

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In the case of the constant Angular Acceleration ($\alpha_0$), the function of the angular Speed ($\omega$) with respect to the time ($t$), along with additional variables the initial Angular Speed ($\omega_0$) and the start Time ($t_0$), is expressed by the equation:

From this equation, it is possible to calculate the relationship between the angle ($\theta$) and the initial Angle ($\theta_0$), as well as the change in angular velocity:

$\theta = \theta_0 +\displaystyle\frac{ \omega ^2- \omega_0 ^2}{2 \alpha_0 }$

Conditions

Variables

$\theta$

Angle

$rad$

6065

$\omega$

Angular Speed

$rad/s$

6068

$\alpha_0$

Constant Angular Acceleration

$rad/s^2$

5298

$\theta_0$

Initial Angle

$rad$

5296

$\omega_0$

Initial Angular Speed

$rad/s$

5295

Relation

Development

If we solve for time in the equation of the angular Speed ($\omega$) that includes the variables the initial Angular Speed ($\omega_0$), the time ($t$), the start Time ($t_0$), and the constant Angular Acceleration ($\alpha_0$):

$\omega = \omega_0 + \alpha_0 ( t - t_0 )$

we obtain the following expression for time:

$t - t_0 = \displaystyle\frac{\omega - \omega_0}{\alpha_0}$

This solution can be substituted into the equation to calculate the angle ($\theta$) using the initial Angle ($\theta_0$) as follows:

$\theta = \theta_0 + \omega_0 ( t - t_0 )+\displaystyle\frac{1}{2} \alpha_0 ( t - t_0 )^2$

which results in the following equation:

$\theta = \theta_0 +\displaystyle\frac{ \omega ^2- \omega_0 ^2}{2 \alpha_0 }$

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