10.3: Example - Mass on a String

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Franz S. Hover & Michael S. Triantafyllou
Massachusetts Institute of Technology via MIT OpenCourseWare

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Consider a mass on a string, being swung around around in a circle at speed \(U\), with radius \(r\). The centrifugal force can be computed in at least three different ways. The vector equation at the start is

\[ \vec{F} \, = \, m \left( \dfrac{\partial \vec{v}_o}{\partial t} + \vec{\omega} \times \vec{v}_o + \dfrac{d \vec{\omega}}{dt} \times \vec{r}_G + \vec{\omega} \times (\vec{\omega} \times \vec{r}_G) \right). \]

Moving Frame Affixed to Mass

Affixing a reference frame on the mass, with the local \(x\) oriented forward and \(y\) inward towards the circle center, gives

\begin{align*} \vec{v}_o \,\, &= \,\, \{ U, \, 0, \, 0 \}^T \\[4pt] \vec{\omega} \,\, &= \,\, \{ 0, \, 0, \, U/r \}^T \\[4pt] \vec{r}_G \,\, &= \,\, \{ 0, \, 0, \, 0 \}^T \\[4pt] \dfrac{\partial \vec{v}_o}{\partial t} \,\, &= \,\, \{ 0, \, 0, \, 0 \}^T \\[4pt] \dfrac{\partial \vec{\omega}}{\partial t} \,\, &= \,\, \{ 0, \, 0, \, 0 \}^T , \end{align*}

such that \[ \vec{F} \, = \, m \vec{\omega} \times \vec{r}_o \, = \, m \{ 0, \, U^2 /r, \, 0 \}^T. \]

The force of the string pulls in on the mass to create the circular motion.

Rotating Frame Attached to Pivot Point

Affixing the moving reference frame to the pivot point of the string, with the same orientation as above but allowing it to rotate with the string, we have

\begin{align*} \vec{v}_o \,\, &= \,\, \{ 0, \, 0, \, 0 \}^T \\[4pt] \vec{\omega} \,\, &= \,\, \{0, \, 0, \, U/r \}^T \\[4pt] \vec{r}_G \,\, &= \,\, \{ 0, \, r, \, 0 \}^T \\[4pt] \dfrac{\partial \vec{v}_o}{\partial t} \,\, &= \,\, \{ 0, \, 0, \, 0 \}^T \\[4pt] \dfrac{\partial \vec{\omega}}{\partial t} \,\, &= \,\, \{ 0, \, 0, \, 0 \}^T , \end{align*}

giving the same result: \[ \vec{F} \, = \, m \vec{\omega} \times (\vec{\omega} \times \vec{r}_G) \, = \, m \{ 0, \, U^2/r, \, 0 \}^T. \]

Stationary Frame

A frame fixed in inertial space, and momentarily coincident with the frame on the mass, can also be used for the calculation. In this case, as the string travels through a small arc \(\delta \psi\), vector subtraction gives \[ \delta \vec{v} \, = \, \{ 0, \, U \sin \delta \psi, \, 0 \}^T \, \simeq \, \{ 0, \, U \delta \psi, \, 0 \}^T. \]Since \(\psi = U / r\), it follows easily that in the fixed frame \(d \vec{v} / dt = \{ 0, \, U^2 / r, 0 \}^T \), as before.