General Definitions

This section collects the laws and equations that will be used to build the instance models.

RefnameGD:velocityIX
LabelThe \(x\)-component of velocity of the pendulum
Units\(\frac{\text{m}}{\text{s}}\)
Equation\[{v_{\text{x}}}=ω\,{L_{\text{rod}}}\,\cos\left({θ_{p}}\right)\]
Description
  • \({v_{\text{x}}}\) is the \(x\)-component of velocity (\(\frac{\text{m}}{\text{s}}\))
  • \(ω\) is the angular velocity (\(\frac{\text{rad}}{\text{s}}\))
  • \({L_{\text{rod}}}\) is the length of the rod (\({\text{m}}\))
  • \({θ_{p}}\) is the displacement angle of the pendulum (\({\text{rad}}\))
Source
RefBy

Detailed derivation of the \(x\)-component of velocity:

At a given point in time, velocity may be defined as

\[\boldsymbol{v}\text{(}t\text{)}=\frac{\,d\boldsymbol{p}\text{(}t\text{)}}{\,dt}\]

We also know the horizontal position

\[{p_{\text{x}}}={L_{\text{rod}}}\,\sin\left({θ_{p}}\right)\]

Applying this,

\[{v_{\text{x}}}=\frac{\,d{L_{\text{rod}}}\,\sin\left({θ_{p}}\right)}{\,dt}\]

\({L_{\text{rod}}}\) is constant with respect to time, so

\[{v_{\text{x}}}={L_{\text{rod}}}\,\frac{\,d\sin\left({θ_{p}}\right)}{\,dt}\]

Therefore, using the chain rule,

\[{v_{\text{x}}}=ω\,{L_{\text{rod}}}\,\cos\left({θ_{p}}\right)\]

RefnameGD:velocityIY
LabelThe \(y\)-component of velocity of the pendulum
Units\(\frac{\text{m}}{\text{s}}\)
Equation\[{v_{\text{y}}}=ω\,{L_{\text{rod}}}\,\sin\left({θ_{p}}\right)\]
Description
  • \({v_{\text{y}}}\) is the \(y\)-component of velocity (\(\frac{\text{m}}{\text{s}}\))
  • \(ω\) is the angular velocity (\(\frac{\text{rad}}{\text{s}}\))
  • \({L_{\text{rod}}}\) is the length of the rod (\({\text{m}}\))
  • \({θ_{p}}\) is the displacement angle of the pendulum (\({\text{rad}}\))
Source
RefBy

Detailed derivation of the \(y\)-component of velocity:

At a given point in time, velocity may be defined as

\[\boldsymbol{v}\text{(}t\text{)}=\frac{\,d\boldsymbol{p}\text{(}t\text{)}}{\,dt}\]

We also know the vertical position

\[{p_{\text{y}}}=-{L_{\text{rod}}}\,\cos\left({θ_{p}}\right)\]

Applying this,

\[{v_{\text{y}}}=-\left(\frac{\,d{L_{\text{rod}}}\,\cos\left({θ_{p}}\right)}{\,dt}\right)\]

\({L_{\text{rod}}}\) is constant with respect to time, so

\[{v_{\text{y}}}=-{L_{\text{rod}}}\,\frac{\,d\cos\left({θ_{p}}\right)}{\,dt}\]

Therefore, using the chain rule,

\[{v_{\text{y}}}=ω\,{L_{\text{rod}}}\,\sin\left({θ_{p}}\right)\]

RefnameGD:accelerationIX
LabelThe \(x\)-component of acceleration of the pendulum
Units\(\frac{\text{m}}{\text{s}^{2}}\)
Equation\[{a_{\text{x}}}=-ω^{2}\,{L_{\text{rod}}}\,\sin\left({θ_{p}}\right)+α\,{L_{\text{rod}}}\,\cos\left({θ_{p}}\right)\]
Description
  • \({a_{\text{x}}}\) is the \(x\)-component of acceleration (\(\frac{\text{m}}{\text{s}^{2}}\))
  • \(ω\) is the angular velocity (\(\frac{\text{rad}}{\text{s}}\))
  • \({L_{\text{rod}}}\) is the length of the rod (\({\text{m}}\))
  • \({θ_{p}}\) is the displacement angle of the pendulum (\({\text{rad}}\))
  • \(α\) is the angular acceleration (\(\frac{\text{rad}}{\text{s}^{2}}\))
Source
RefBy

Detailed derivation of the \(x\)-component of acceleration:

Our acceleration is:

\[\boldsymbol{a}\text{(}t\text{)}=\frac{\,d\boldsymbol{v}\text{(}t\text{)}}{\,dt}\]

Earlier, we found the horizontal velocity to be

\[{v_{\text{x}}}=ω\,{L_{\text{rod}}}\,\cos\left({θ_{p}}\right)\]

Applying this to our equation for acceleration

\[{a_{\text{x}}}=\frac{\,dω\,{L_{\text{rod}}}\,\cos\left({θ_{p}}\right)}{\,dt}\]

By the product and chain rules, we find

\[{a_{\text{x}}}=\frac{\,dω}{\,dt}\,{L_{\text{rod}}}\,\cos\left({θ_{p}}\right)-ω\,{L_{\text{rod}}}\,\sin\left({θ_{p}}\right)\,\frac{\,d{θ_{p}}}{\,dt}\]

Simplifying,

\[{a_{\text{x}}}=-ω^{2}\,{L_{\text{rod}}}\,\sin\left({θ_{p}}\right)+α\,{L_{\text{rod}}}\,\cos\left({θ_{p}}\right)\]

RefnameGD:accelerationIY
LabelThe \(y\)-component of acceleration of the pendulum
Units\(\frac{\text{m}}{\text{s}^{2}}\)
Equation\[{a_{\text{y}}}=ω^{2}\,{L_{\text{rod}}}\,\cos\left({θ_{p}}\right)+α\,{L_{\text{rod}}}\,\sin\left({θ_{p}}\right)\]
Description
  • \({a_{\text{y}}}\) is the \(y\)-component of acceleration (\(\frac{\text{m}}{\text{s}^{2}}\))
  • \(ω\) is the angular velocity (\(\frac{\text{rad}}{\text{s}}\))
  • \({L_{\text{rod}}}\) is the length of the rod (\({\text{m}}\))
  • \({θ_{p}}\) is the displacement angle of the pendulum (\({\text{rad}}\))
  • \(α\) is the angular acceleration (\(\frac{\text{rad}}{\text{s}^{2}}\))
Source
RefBy

Detailed derivation of the \(y\)-component of acceleration:

Our acceleration is:

\[\boldsymbol{a}\text{(}t\text{)}=\frac{\,d\boldsymbol{v}\text{(}t\text{)}}{\,dt}\]

Earlier, we found the vertical velocity to be

\[{v_{\text{y}}}=ω\,{L_{\text{rod}}}\,\sin\left({θ_{p}}\right)\]

Applying this to our equation for acceleration

\[{a_{\text{y}}}=\frac{\,dω\,{L_{\text{rod}}}\,\sin\left({θ_{p}}\right)}{\,dt}\]

By the product and chain rules, we find

\[{a_{\text{y}}}=\frac{\,dω}{\,dt}\,{L_{\text{rod}}}\,\sin\left({θ_{p}}\right)+ω\,{L_{\text{rod}}}\,\cos\left({θ_{p}}\right)\,\frac{\,d{θ_{p}}}{\,dt}\]

Simplifying,

\[{a_{\text{y}}}=ω^{2}\,{L_{\text{rod}}}\,\cos\left({θ_{p}}\right)+α\,{L_{\text{rod}}}\,\sin\left({θ_{p}}\right)\]

RefnameGD:hForceOnPendulum
LabelHorizontal force on the pendulum
Units\({\text{N}}\)
Equation\[\boldsymbol{F}=m\,{a_{\text{x}}}=-\boldsymbol{T}\,\sin\left({θ_{p}}\right)\]
Description
  • \(\boldsymbol{F}\) is the force (\({\text{N}}\))
  • \(m\) is the mass (\({\text{kg}}\))
  • \({a_{\text{x}}}\) is the \(x\)-component of acceleration (\(\frac{\text{m}}{\text{s}^{2}}\))
  • \(\boldsymbol{T}\) is the tension (\({\text{N}}\))
  • \({θ_{p}}\) is the displacement angle of the pendulum (\({\text{rad}}\))
Source
RefBy

Detailed derivation of force on the pendulum:

\[\boldsymbol{F}=m\,{a_{\text{x}}}=-\boldsymbol{T}\,\sin\left({θ_{p}}\right)\]

RefnameGD:vForceOnPendulum
LabelVertical force on the pendulum
Units\({\text{N}}\)
Equation\[\boldsymbol{F}=m\,{a_{\text{y}}}=\boldsymbol{T}\,\cos\left({θ_{p}}\right)-m\,\boldsymbol{g}\]
Description
  • \(\boldsymbol{F}\) is the force (\({\text{N}}\))
  • \(m\) is the mass (\({\text{kg}}\))
  • \({a_{\text{y}}}\) is the \(y\)-component of acceleration (\(\frac{\text{m}}{\text{s}^{2}}\))
  • \(\boldsymbol{T}\) is the tension (\({\text{N}}\))
  • \({θ_{p}}\) is the displacement angle of the pendulum (\({\text{rad}}\))
  • \(\boldsymbol{g}\) is the gravitational acceleration (\(\frac{\text{m}}{\text{s}^{2}}\))
Source
RefBy

Detailed derivation of force on the pendulum:

\[\boldsymbol{F}=m\,{a_{\text{y}}}=\boldsymbol{T}\,\cos\left({θ_{p}}\right)-m\,\boldsymbol{g}\]

RefnameGD:angFrequencyGD
LabelThe angular frequency of the pendulum
Units\({\text{s}}\)
Equation\[Ω=\sqrt{\frac{\boldsymbol{g}}{{L_{\text{rod}}}}}\]
Description
  • \(Ω\) is the angular frequency (\({\text{s}}\))
  • \(\boldsymbol{g}\) is the gravitational acceleration (\(\frac{\text{m}}{\text{s}^{2}}\))
  • \({L_{\text{rod}}}\) is the length of the rod (\({\text{m}}\))
Notes
Source
RefByGD:periodPend and IM:calOfAngularDisplacement

Detailed derivation of the angular frequency of the pendulum:

Consider the torque on a pendulum defined in TM:NewtonSecLawRotMot. The force providing the restoring torque is the component of weight of the pendulum bob that acts along the arc length. The torque is the length of the string \({L_{\text{rod}}}\) multiplied by the component of the net force that is perpendicular to the radius of the arc. The minus sign indicates the torque acts in the opposite direction of the angular displacement:

\[\boldsymbol{τ}=-{L_{\text{rod}}}\,m\,\boldsymbol{g}\,\sin\left({θ_{p}}\right)\]

So then

\[\boldsymbol{I}\,α=-{L_{\text{rod}}}\,m\,\boldsymbol{g}\,\sin\left({θ_{p}}\right)\]

Therefore,

\[\boldsymbol{I}\,\frac{\,d\frac{\,d{θ_{p}}}{\,dt}}{\,dt}=-{L_{\text{rod}}}\,m\,\boldsymbol{g}\,\sin\left({θ_{p}}\right)\]

Substituting for \(\boldsymbol{I}\)

\[m\,{L_{\text{rod}}}^{2}\,\frac{\,d\frac{\,d{θ_{p}}}{\,dt}}{\,dt}=-{L_{\text{rod}}}\,m\,\boldsymbol{g}\,\sin\left({θ_{p}}\right)\]

Crossing out \(m\) and \({L_{\text{rod}}}\) we have

\[\frac{\,d\frac{\,d{θ_{p}}}{\,dt}}{\,dt}=-\left(\frac{\boldsymbol{g}}{{L_{\text{rod}}}}\right)\,\sin\left({θ_{p}}\right)\]

For small angles, we approximate sin \({θ_{p}}\) to \({θ_{p}}\)

\[\frac{\,d\frac{\,d{θ_{p}}}{\,dt}}{\,dt}=-\left(\frac{\boldsymbol{g}}{{L_{\text{rod}}}}\right)\,{θ_{p}}\]

Because this equation, has the same form as the equation for simple harmonic motion the solution is easy to find. The angular frequency

\[Ω=\sqrt{\frac{\boldsymbol{g}}{{L_{\text{rod}}}}}\]

RefnameGD:periodPend
LabelThe period of the pendulum
Units\({\text{s}}\)
Equation\[T=2\,π\,\sqrt{\frac{{L_{\text{rod}}}}{\boldsymbol{g}}}\]
Description
  • \(T\) is the period (\({\text{s}}\))
  • \(π\) is the ratio of circumference to diameter for any circle (Unitless)
  • \({L_{\text{rod}}}\) is the length of the rod (\({\text{m}}\))
  • \(\boldsymbol{g}\) is the gravitational acceleration (\(\frac{\text{m}}{\text{s}^{2}}\))
Notes
  • The frequency and period are defined in the data definitions for frequency and period respectively
Source
RefBy

Detailed derivation of the period of the pendulum:

The period of the pendulum can be defined from the general definition for the equation of angular frequency

\[Ω=\sqrt{\frac{\boldsymbol{g}}{{L_{\text{rod}}}}}\]

Therefore from the data definition of the equation for angular frequency, we have

\[T=2\,π\,\sqrt{\frac{{L_{\text{rod}}}}{\boldsymbol{g}}}\]