When I first started learning mechanics, I began with Fundamentals of Physics by Halliday, Resnick, and Walker. It was one of the canonical texts for introductory university physics, and it was pitched at students who had just learned calculus.
Eventually I sought out a more advanced text, and landed upon Introduction to Mechanics by Kleppner and Kolenkow. This text was the one used in MIT’s Advanced Intro to Mechanics course. It was written in the 1970s and featured problems much more difficult than those in standard physics texts.
The final book I used in my study of introductory mechanics was Morin’s Introduction to Classical Mechanics. Like Kleppner and Kolenkow’s book it was presented as an Advanced Introductory Mechanics text, and featured many difficult problems. One of the most difficult chapters was called “Angular Momentum, Part II” and was concerned with understanding the rotational motion of extended physical bodies. What makes this topic so difficult is that it involves modeling systems within rotating coordinate systems for which every-day intuition is typically not reliable.
Take for example the following problem
How should you approach this problem? Modeling the angular momentum is simpler if you start from the coordinate system that is rotating with the coin (i.e., the \(\textbf{x}_1, \textbf{x}_2, \textbf{x}_3\) system), but if you the gravitational force which sustains the rotational motion can has a simpler expression in the inertial coordinate system of the surrounding lab. These tradeoffs (and the work needed to go from one coordinate system to another) are some of the things that make such rigid body problems difficult.
But working through the problems was a great exercise in understanding these systems and was generally very rewarding.
In this post, we discuss a problem that extends Morin’s problems by combining two ideas from classical mechanics: pendulum motion and rigid body motion.
Table of Contents
- Introduction
- Rolling without Sliding
- Sliding without Rolling
- Pendulum on an Incline
- Stranger than Friction
Introduction
Say that we have a lollipop whose end is connected to the plane of a ramp with incline angle \(\alpha\). The lollipop is idealized as a solid sphere of mass \(m\) and radius \(r\) that is attached to a massless rod of length \(\ell\) which runs to the sphere’s center. The lollipop is rolled away from its stable equilibrium and then released. The lollipop (and adjoining rod) rolls across the incline plane due to the frictional force at the point of contact between the lollipop and the plane. We call this “rolling without sliding.”
We want to to describe the motion and the forces that are exerted on the lollipop. Describing the motion will amount to deriving the differential equation for \(\phi\) the angle that defines the lollipop’s rotation about an axis perpendicular to the plane. From this differential equation we could also infer a small-angle period associated with the motion.
Describing the forces will amount to computing the normal and frictional forces exerted on the lollipop. After computing these results for this case of a lollipop that is “rolling without sliding,” we will repeat the problem for the case where there is no frictional force and the lollipop just slides along the plane (i.e., “sliding without rolling”). Finally, we will consider the simpler case of a pendulum on an incline and discuss why the limiting cases of the previous problem setups are different from this final result.
In an outline the questions we aim to answer are
- Derive a differential equation for the angle \(\phi\) which the lollipop rotates about an axis perpendicular and make a small-angle approximation to find the period at which the lollipop rolls back and forth. Solve for the normal and frictional forces on the lollipop. Assume that the lollipop rolls without sliding.
- Repeat 1. but analyze the case of frictionless sliding without rolling. (Note: No frictional force needs to be calculated in this case)
- Consider the analogous problem for a massive point particle attached to a rod. Explain the discrepancy between this analogous problem and the corresponding limiting case of the main problem.
Rolling without Sliding
We want to find the differential equation and the period of a lollipop oscillating on an inclined plane. Along the way we want to compute the normal force exerted on the lollipop by the plane and the frictional force that keeps the lollipop rolling without sliding. All of these tasks require us to determine the equations of motion of the lollipop. We would typically have at least two approaches available to us for this determination–Newtonian and Lagrangian–but the frictional forces which induce rolling in the lollipop make the Lagrangian method inapplicable1. For the rolling without sliding case, we proceed with a Newtonian approach.
(Figure 2: Side view of the inclined plane. The \(\mathbf{\hat{x}}\) and \(\mathbf{\hat{x}}_1\) vectors coincide and point out of the plane of the figure.)
We will obtain the equations of motion for the system first determining the angular momentum of the lollipop, differentiating this quantity in time, computing the net torque applied to the lollipop, and, finally, equating the second result to the third.
Before we begin, we state some useful relations between the parameters defining the system. From the given information, we can create Figure. 2. From, the figure we can define new variables \(\theta\) and \(d\) that are related to \(\ell\) and \(r\) through the system of equations
\begin{equation}
r = \ell \sin \theta, \qquad d = \ell \cos \theta . \qquad (1)
\label{eq:lengths}
\end{equation}
The parameter \(\theta\) is the angle the rod makes with the plane, and \(d\) is the distance from the pivot point of the rod to the point of contact with the head of the lollipop. From the figure, we can also write an equation representing the no slip condition for the lollipop. We have chosen to define positive \(\dot{\phi}\) (the angular velocity about the \(\mathbf{\hat{x}}\) axis) in the “left-handed orientation” (which is opposite its typical direction) in order to avoid negative signs in the no-sliding constraint. The quantity \(\dot{\psi}\) (the angular velocity about the spin axis of the lollipop) has the “right-handed orientation” and is defined in the typical positive direction. The relationship between the two quantities is
\begin{equation}
\dot{\psi} r = \dot{\phi} d \quad \longrightarrow \quad \dot{\psi} = \dot{\phi}\cot\theta \qquad \text{[Rolling without sliding condition]} \qquad (2)
\label{eq:no_slip}
\end{equation}
where we used Eq.(1) in the final arrow.
Angular Momentum and its Time Derivative
Before we begin, we describe the two coordinate systems relevant to the problem. The “coordinate axes” are defined by the standard \(\mathbf{\hat{z}}\), \(\mathbf{\hat{y}}\), \(\mathbf{\hat{x}}\) directions. From Figure 2, we see that we define these axes relative to the plane of the incline rather than the plane of the figure. The “principal axes” are formally defined by the eigenvectors of the moment of inertia tensor for the rigid body. In the case of the rolling lollipop, the principal axes vectors comprise three vectors: one vector that goes through the symmetry axis of the rod and the lollipop ( \(\mathbf{\hat{x}}_3\)); a vector that is perpendicular to the first and which is parallel to the plane of the incline ( \(\mathbf{\hat{x}}_1\)); a final vector that is perpendicular to the first two ( \(\mathbf{\hat{x}}_2\)). In the subsequent discussion we label the principal axes frame as \(S\) and the coordinate axes frame as \(R\).
Following the discussion in Chapter 9 of Morin’s Introduction to Classical Mechanics, we will work in the “principal axes” frame of the lollipop (rather than the coordinate axes) because the moment of inertia tensor is diagonal in this frame. Now, to compute the angular momentum, we must first compute the angular velocity of the lollipop. This angular velocity exists in two parts: There is the spin angular velocity of the lollipop, \(\dot{\psi}\, \mathbf{\hat{x}}_3\), and there is the angular velocity of the frame principal axes frame \(S\) relative to the coordinate axes frame \(R\), \(\boldsymbol{\omega}_S\)2. Thus, the total angular velocity is
\begin{equation}
\boldsymbol{\omega} = \dot{\psi}\,\mathbf{\hat{x}}_3 + \boldsymbol{\omega}_S. \qquad (3)
\end{equation}
From Figure 2, we can write the angular velocity \(\boldsymbol{\omega}_S\) as
\begin{equation}
\boldsymbol{\omega}_S =-\dot{\phi} \, \mathbf{\hat{z}} + \dot{\theta}\, \mathbf{\hat{x}}_1, \qquad (4)
\label{eq:omegaS}
\end{equation}
where, for generality, we assumed that \(\theta\) is not constant. We want to express \(\boldsymbol{\omega}\) in the principal axes basis, and so we need to find \(\mathbf{\hat{z}}\) in terms of these axes. From Figure 2, we can infer that
\begin{equation}
\mathbf{\hat{z}} = -\cos\theta\, \mathbf{\hat{x}}_2 + \sin\theta\, \mathbf{\hat{x}}_3. \qquad (5)
\label{eq:zdef}
\end{equation}
Thus, we have for the final expression of the angular velocity \(\boldsymbol{\omega}\)
\begin{equation}
\boldsymbol{\omega} = (\dot{\psi} – \dot{\phi} \sin\theta)\, \mathbf{\hat{x}}_3 + \dot{\phi} \cos\theta \,\mathbf{\hat{x}}_2 + \dot{\theta}\, \mathbf{\hat{x}}_1. \qquad (6)
\end{equation}
We now have the angular velocity, and to compute the angular momentum the last things we need are the principal axes. For the lollipop (consisting of a rod of length \(\ell\) attached to a solid sphere of radius \(r\)) these moments are
\begin{equation}
I_1 = I_2 \equiv I = m \ell ^2 + \frac{2}{5} mr^2 , \, \qquad I_3 = \frac{2}{5} m r^2, \qquad (7)
\label{eq:inertias}
\end{equation}
where \(I_1\) and \(I_2\) are computed through the principal axis theorem. Thus, with Eq.(7), we find that the \textit{general} angular momentum of our system is
\begin{align}
\boldsymbol{L} & = I_3 (\dot{\psi} – \dot{\phi} \sin\theta)\, \mathbf{\hat{x}}_3 + I \dot{\phi} \mathbf{\hat{x}}_1, \qquad \text{[General result]}. \qquad (8)
\end{align}
For our particular case, we have \(\theta\) as a constant, and “rolling without sliding” which requires \(\dot{\psi}\) and \(\dot{\phi}\) to be related via Eq.(2). Thus the angular momentum for the rolling without sliding case is
\begin{equation}
\boldsymbol{L} = I_3 \dot{\phi} (\cot \theta – \sin\theta)\,\mathbf{\hat{x}}_3 + I \dot{\phi} \cos\theta \,\mathbf{\hat{x}}_2, \qquad \text{[Rolling without sliding]}, \qquad (9)
\label{eq:Ldef}
\end{equation}
Next, we need to compute the time derivative of Eq.(9) in preparation for equating the result to the net torque. The time dependence in Eq.(9) comes from two sources: the change in the magnitude of each axis direction, and the change in the direction of the axes themselves. Since the axes are unit vectors, they do not change in magnitude and they simply rotate in time. Their rate and direction of rotation are set by Eq.(4). Thus, with Eq.(4) and Eq.(5) (and with \(\dot{\theta} =0\)) we find
\begin{equation}
\frac{d}{dt} \mathbf{\hat{x}}_3 = \boldsymbol{\omega}{S} \times \mathbf{\hat{x}}_3 = \dot{\phi} \cos \theta\, \mathbf{\hat{x}}_1, \qquad \frac{d}{dt} \mathbf{\hat{x}}_2 = \boldsymbol{\omega}{S} \times \mathbf{\hat{x}}_2 = \dot{\phi} \sin \theta\, \mathbf{\hat{x}}_1. \qquad (10)
\end{equation}
Therefore, the time derivative of Eq.(9) is
\begin{align}
\frac{d}{dt} \textbf{L} & = I_3 \ddot{\phi} (\cot \theta – \sin\theta)\, \mathbf{\hat{x}}_3 + I \ddot{\phi} \cos\theta \,\mathbf{\hat{x}}_2 +\dot{\phi}^2 \Big[I_3 \cot\theta \cos\theta + (I-I_3) \cos\theta \sin\theta \Big] \mathbf{\hat{x}}_1. \qquad (11)
\label{eq:dLdt}
\end{align}
Next, we need to compute the torque on the lollipop, and then, finally, equate the result to Eq.(11) to find the equations of motion.
Computing Torque
(Figure 3: Top view of the inclined plane. We depict (for pictorial clarity) the geometrically impossible case of \(\theta = 0\) for which \(\mathbf{\hat{x}}_3\) exists within the plane of \(\mathbf{\hat{x}}\) and \(\mathbf{\hat{y}}\). In this figure \(\mathbf{\hat{z}}\) is pointing outside the plane of the figure, and \(\mathbf{\hat{x}}_2\) is pointing into the plane of the figure.)
To compute the torque, we need to determine the forces acting on the lollipop, their points of contact, and the distances from the points of contact to a chosen origin. For the lollipop that is rolling without sliding, there are three forces: a gravitational force, a normal force, and a frictional force. The gravitational force always points in the same direction relative to the coordinates axes (defined by \(\mathbf{\hat{z}}\), \(\mathbf{\hat{y}}\), and \(\mathbf{\hat{x}}\)) as does the normal force. However, the frictional force points in a direction opposite the motion of the lollipop. From Figure 2 and Figure 3, we see that the vector decompositions of these forces are
\begin{equation}
\textbf{F}_{g} = mg \left(\sin \alpha\, \mathbf{\hat{y}} – \cos \alpha\, \mathbf{\hat{z}}\right), \qquad \textbf{F}_{N} = F_N \, \mathbf{\hat{z}} , \qquad \textbf{F}_{f} = F{f} \,\mathbf{\hat{x}}_1. \qquad (12)
\label{eq:forces}
\end{equation}
The gravitational force acts from the center of mass of the lollipop, but the normal and frictional forces are contact forces acting from the point of contact between the head of the lollipop and the plane. Taking the pivot point to be the origin, the position of the center of mass is \(\ell \mathbf{\hat{x}}_3\), while the point of contact is \(d(\cos\theta \,\mathbf{\hat{x}}_3 + \sin\theta\, \mathbf{\hat{x}}_2)\)3. The net torque acting on the lollipop is therefore,
\begin{equation}
\boldsymbol{\tau}_{\text{net}} = \ell \mathbf{\hat{x}}_3 \times \textbf{F}{g} + d(\cos\theta \, \mathbf{\hat{x}}_3 +\sin\theta\, \mathbf{\hat{x}}_2) \times \textbf{F}{N} + d(\cos\theta \, \mathbf{\hat{x}}_3 +\sin\theta\, \mathbf{\hat{x}}_2) \times \textbf{F}{f}. \qquad (13)
\label{eq:nettau}
\end{equation}
Eq.(11), the time derivative of the angular momentum, is expressed in the principal axes basis, so we would like to do the same for Eq.(13). We can already express the third term in Eq.(13) in the appropriate basis, and with Eq.(5) we can do the same for the second term, but we need additional transformations to express the first term in the appropriate basis. Namely, we need an expression for \(\mathbf{\hat{y}}\) in terms of \(\mathbf{\hat{x}}_3\), \(\mathbf{\hat{x}}_2\), and \(\mathbf{\hat{x}}_1\).
Finding the necessary transformations amounts to a careful application of rotation matrices. First, we note that we can obtain the principal axes vectors by first rotating the coordinate axes vectors by \(\gamma = -(\pi/2-\theta)\) about the \(\mathbf{\hat{x}}\) axis (i.e., a negative rotation by the complementary angle to $\theta$) and then rotating the resulting vector by \(\phi\) about the \(\mathbf{\hat{z}}\) axis (with our chosen positive direction opposite the typical “right-hand rule” direction) . For example, the vector \(\mathbf{\hat{x}}_3\) can be obtained as
\begin{align}
\mathbf{\hat{x}}_3 = \left(\begin{array}{c c c} \cos \phi& \sin\phi & 0\\ -\sin\phi & \cos\phi &0 \\ 0 &0 & 1 \end{array} \right)\left(\begin{array}{c c c} 1& 0 & 0\\ 0 & \cos \gamma & -\sin \gamma \\0 &\sin \gamma & \cos\gamma \end{array} \right) \left(\begin{array}{c} 0 \\ 0 \\ 1 \end{array}\right)_R = \left(\begin{array}{c} -\sin\phi \sin\gamma \\ -\cos\phi \sin\gamma \\ \cos\gamma \end{array}\right)_R \qquad (14)
\end{align}
where \(R\) denotes a vector written in the coordinate axes basis. We can write this result explicitly in terms of the coordinate axes basis (with \(\gamma = -(\pi/2 – \theta)\)) as
\begin{equation}
\mathbf{\hat{x}}_3 = \sin \theta \, \mathbf{\hat{z}} + \cos \theta \left( \cos\phi\, \mathbf{\hat{y}}+ \sin\phi\, \mathbf{\hat{x}}\right). \qquad (15)
\end{equation}
Performing analogous calculations for \(\mathbf{\hat{x}}_2\) and \(\mathbf{\hat{x}}_1\), we can find that the change of bases between the two coordinate system is
\begin{equation}
\left(\begin{array}{c} \mathbf{\hat{x}}_1 \\ \mathbf{\hat{x}}_2 \\ \mathbf{\hat{x}}_3 \end{array} \right) = \left(\begin{array}{c c c} \cos \phi & – \sin \phi & 0 \\ \sin \theta \sin\phi & \sin\theta \cos\phi &- \cos\theta \\ \cos\theta \sin \phi &\cos\theta\cos\phi &\sin\theta \end{array} \right) \left(\begin{array}{c} \mathbf{\hat{x}} \\ \mathbf{\hat{y}} \\ \mathbf{\hat{z}} \end{array}\right) \qquad (16)
\label{eq:firstcob}
\end{equation}
Eq.(16) gives us the principal axes basis in terms of the coordinate axes basis, but in order to compute the torque in terms of the principal axes basis we need the reverse transformation. The matrix in Eq.(16) is a rotation matrix which has the defining property of being orthonormal. This means that the inverse of the matrix is its transpose, and thus to find the reverse transformation we need only transpose the existing one. Therefore, the coordinate axes basis expressed in terms of the principal axes basis is
\begin{equation}
\left(\begin{array}{c} \mathbf{\hat{x}} \\ \mathbf{\hat{y}} \\ \mathbf{\hat{z}} \end{array}\right) = \left(\begin{array}{c c c} \cos \phi &\sin\theta \sin\phi &\cos\theta \sin \phi \\ – \sin \phi& \sin\theta \cos\phi&\cos\theta\cos\phi \\ 0 &- \cos\theta & \sin\theta\end{array} \right) \left(\begin{array}{c} \mathbf{\hat{x}}_1 \\ \mathbf{\hat{x}}_2 \\ \mathbf{\hat{x}}_3 \end{array} \right). \qquad (17)
\label{eq:seccob}
\end{equation}
We note that the bottom row of this matrix reproduces the Eq.(5) result we obtained from inspection of Figure 2. With the system of equations obtained from this matrix, we find that our force vectors Eq.(12) can be expressed as
\begin{align}
\textbf{F}_{g}& = mg \Big[\sin \alpha\, \left(- \sin \phi\, \mathbf{\hat{x}}_1 + \cos\phi(\sin\theta \, \mathbf{\hat{x}}_2 + \cos\theta\, \mathbf{\hat{x}}_3)\right) \\ \qquad (18)
& \qquad \quad – \cos \alpha\, (-\cos\theta\, \mathbf{\hat{x}}_2 + \sin\theta \, \mathbf{\hat{x}}_3)\Big] \\ \qquad (19)
\textbf{F}_{N} & = F_N \, (- \cos \theta \, \mathbf{\hat{x}}_2 + \sin\theta \, \mathbf{\hat{x}}_3) \\
\textbf{F}_{f} & = F_f \mathbf{\hat{x}}_1, \qquad (20)
\end{align}
and the net torque is
\begin{align}
\boldsymbol{\tau}_{\text{net}} & = \ell \mathbf{\hat{x}}_3 \times \textbf{F}_{g} + d(\cos\theta \, \mathbf{\hat{x}}_3 +\sin\theta\, \mathbf{\hat{x}}_2) \times \textbf{F}_{N} + d(\cos\theta \, \mathbf{\hat{x}}_3 +\sin\theta\,\mathbf{\hat{x}}_2) \times \textbf{F}_{f} \\[.5em]
& = – F_f d \sin \theta\, \mathbf{\hat{x}}_3 + \Big\{ F_f d\cos \theta – m g \ell \sin \alpha \sin \phi \Big\}\, \mathbf{\hat{x}}_2 \\[.5em]
& \quad + \Big\{ F_N d – m g \ell \big[\sin\alpha \cos\phi \sin \theta + \cos\alpha \cos\theta \big]\Big\}\, \mathbf{\hat{x}}_1 \qquad (21)
\end{align}
Equations of Motion
We have now computed \(d\textbf{L}/dt\) and \(\boldsymbol{\tau}_{\text{net}}\) in terms of the principal axes basis. Equating the two results, yields the following system of equations:
\begin{align}
I_3 \ddot{\phi} (\cot \theta – \sin\theta) & = – F_f d \sin \theta \qquad (22) \\
I \ddot{\phi} \cos\theta & = F_f d\cos \theta – m g \ell \sin \alpha \sin \phi \qquad (23)\\
I_3 \dot{\phi}^2 \cot\theta + (I-I_3) \dot{\phi}^2\cos\theta \sin\theta & = F_N d – m g \ell \big[\sin\alpha \cos\phi \sin \theta + \cos\alpha \cos\theta \big] \qquad (24)
\end{align}
With Eq.(7) and Eq.(1), this system can be rewritten as
\begin{align}
\frac{2}{5}m \ell^2\ddot{\phi} (\cos\theta- \sin^2\theta) & = – F_f d \qquad (25) \\
m \ell^2\ddot{\phi}\left(1 + \frac{2}{5} \sin^2\theta\right) \cos\theta & = F_f d\cos \theta – m g \ell \sin \alpha \sin \phi \qquad (26)\\
\frac{2}{5}m \ell^2 \dot{\phi}^2 \cos\theta \sin\theta+ m\ell^2 \dot{\phi}^2\cos\theta \sin\theta & = F_N d – m g \ell \big[\sin\alpha \cos\phi \sin \theta + \cos\alpha \cos\theta \big] \qquad (27)
\end{align}
Solving these equations for \(\ddot{\phi}\), \(F_N\), and \(F_f\) for completeness, we find
\begin{equation}
\ddot{\phi} = – \frac{g}{\ell} \dfrac{\sin\alpha \sin \phi}{\cos\theta \left(1 + \dfrac{2}{5} \cos\theta\right)}\qquad \text{[EOM; Rolling without sliding]} \qquad (28)
\label{eq:eom_rolling}
\end{equation}
\begin{equation}
F_N = mg \cos \alpha \big( 1 + \cos\phi \tan \alpha\tan\theta\big)+ m\ell \sin \theta \dot{\phi}^2\left( 1+ \frac{2}{5} \cos \theta\right) \quad \text{[\(F_N\); Rolling without sliding]} \qquad (29)
\label{eq:FNroll}
\end{equation}
\begin{equation}
F_f = – \frac{2}{5}mg \sin\alpha \sin \phi \dfrac{ 1 – \sin\theta \tan\theta}{\cos\theta \left(1 + \dfrac{2}{5} \cos\theta\right)} \qquad (30)
\label{eq:fric_soln}
\end{equation}
From Eq.(28), we can make a small angle approximation and find that the period of the resulting simple harmonic motion is
\begin{equation}
T = 2\pi \sqrt{\frac{\ell \cos\theta}{g \sin\alpha } \left( 1+ \frac{2}{5} \cos\theta\right)} \qquad \text{[Period; Rolling without sliding]} \label{eq:period_roll} \qquad (31)
\end{equation}
With our collected results we can now begin considering various limiting cases.
For \(\alpha = 0\), Eq.(29) becomes
\begin{equation}
F_N = mg + mr\dot{\phi}^2 \left(1 + \frac{2}{5} \cos\theta\right) \qquad (32)
\end{equation}
which is a solution to the “Rolling Lollipop” problem in Chapter 9 of Morin’s Introduction to Classical Mechanics. Moreover \(\alpha = 0\) yield \(\ddot{\phi}=0\) indicating that when the lollipop is not on an incline, it rotates uniformly about the \(\mathbf{\hat{z}}\) axis.
The \(\theta =0\) limiting case (corresponding to a point particle a distance \(\ell\) from the pivot) is stranger. It yields \(F_N = mg\cos\alpha\) as one might expect, but applying the limit to Eq.(31) yields
\begin{equation}
T = 2\pi \sqrt{\frac{7 \ell}{5 g \sin\alpha }}.\qquad (33)
\end{equation}
As we will later show, this is not the period that one would obtain from a direct analysis of a slanted pendulum problem does not result in the same equation that one would obtain from a direct analysis of this case. The reason comes down to the incorporation of “frictional force” as a means of generating rotation. We will see that our stated “frictional force” does not really behave like friction at all, a fact that we could infer from the angular dependence of Eq.(30): Intuition leads us to expect this frictional force to be a function of \(\dot{\phi}\), but it is instead a function of \(\phi\).
Sliding without rolling
Now, we want to repeat this problem but under frictionless conditions. The main difference between this case and the former one is that for this case \(\textbf{F}_f =0\), and \(\dot{\psi}\) and \(\dot{\phi}\) are independent variables (i.e., Eq.(2) no longer applies). Going through the previous analysis with these changes yields the equations of motion
\begin{align}
\frac{d}{dt}I_3(\dot{\psi}- \dot{\phi}\sin\theta) & = 0 \quad \longrightarrow \quad L_3 \equiv I_3(\dot{\psi}- \dot{\phi}\sin\theta) \label{eq:eom21} \qquad (34) \\[.5em]
I \ddot{\phi} \cos\theta & = – m g \ell \sin \alpha \sin \phi \label{eq:eom22} \qquad (35) \\[.5em]
L_3 \dot{\phi}\cos\theta+ I \dot{\phi}^2\cos\theta \sin\theta & = F_N d – m g \ell \big[\sin\alpha \cos\phi \sin \theta + \cos\alpha \cos\theta \big] \qquad (36)
\label{eq:eom23}
\end{align}
which when solved for \(\ddot{\phi}\) gives us
\begin{equation}
\ddot{\phi} = – \frac{g}{\ell} \dfrac{\sin\alpha \sin \phi}{\cos\theta \left(1 + \dfrac{2}{5} \sin^2\theta\right)} \qquad \text{[EOM; Sliding without rolling]} \label{eq:eom_sliding} \qquad (37)
\end{equation}
and solved for \(F_N\), give us
\begin{equation}
F_N = mg \cos \alpha \big( 1 + \cos\phi \tan \alpha\tan\theta\big)+ m\ell \sin \theta \dot{\phi}^2\left( 1+ \frac{2}{5} \sin^2\theta\right) + \frac{L_3 \dot{\phi}}{\ell}\quad \text{[Sliding without rolling]}.
\label{eq:FNslide}\qquad (38)
\end{equation}
Taking the small angle approximation of Eq.(37), we find that the period of the oscillatory motion is
\begin{equation}
T = 2\pi \sqrt{\frac{\ell \cos\theta}{g \sin\alpha } \left( 1+ \frac{2}{5} \sin^2\theta\right)} \qquad \text{[Sliding without rolling]}.\qquad (39)
\label{eq:Tslide}
\end{equation}
Computing the \(\alpha =0\) limiting case of \rfw{FNslide}, we find
\begin{equation}
F_N = mg + mr\dot{\phi}\left(\dot{\phi} + \frac{2}{5}\dot{\psi}\sin\theta\right)\qquad (40)
\end{equation}
which (for \(\dot{\psi}=0\)) reproduces the solution to the “Sliding lollipop” exercise in Morin Chapter 9.
The \(\theta =0\) limit (unlike that for the rolling lollipop) produces the expected dynamical results: Applying the limit to Eq.(39) yields
\begin{equation}
{T = 2\pi \sqrt{\frac{\ell}{g \sin\alpha } }}, \qquad(41)
\end{equation}
which is what we will expect for a point-particle pendulum on an incline.
Lagrangian Approach
Before moving on to analyze the limiting cases suggested in the prompt, we obtain the sliding without rolling equations of motion through the Lagrangian approach. First, we note that the kinetic energy of the system is
\begin{equation}
T = \frac{1}{2} I_3 (\dot{\psi}- \dot{\phi} \sin\theta)^2 + \frac{1}{2} I \left(\dot{\phi}^2 \cos^2 \theta+ \dot{\theta}^2\right). \qquad(42)
\end{equation}
The potential energy can be found by integrating the gravitational force in Eq.(12):
\begin{align}
V & = – \int d\boldsymbol{\ell} \cdot \textbf{F}_g = – mg (\sin \alpha \, y – \cos\alpha \,z). \qquad(43)
\end{align}
With \(y = \ell \mathbf{\hat{x}}_3 \cdot \mathbf{\hat{y}} \), \(z = \ell \mathbf{\hat{x}}_3 \cdot \mathbf{\hat{y}} \), and the system of equations defined by Eq.(17) we find
\begin{equation}
V = -mg\ell \sin\alpha \cos\phi \cos\theta + mg \ell \cos\alpha \sin \theta. \qquad(44)
\end{equation}
Finally, we can incorporate the geometric constraint on \(\theta\) by including a Lagrange multiplier \(\lambda\) in the Lagrangian. The resulting Lagrangian is then
\begin{align}
L_{\text{lagrange}} & = T-V – \lambda(\ell \sin \theta – r) \\[.5em]
& = \frac{1}{2} I_3 (\dot{\psi}- \dot{\phi} \sin\theta)^2 + \frac{1}{2} I \left(\dot{\phi}^2 \cos^2 \theta+ \dot{\theta}^2\right) + mg\ell \sin\alpha \cos\phi \cos\theta \\[.5em]
& \qquad- mg \ell \cos\alpha \sin \theta – \lambda(\ell \sin \theta – r) \qquad(45) \label{eq:lagrange_sliding}
\end{align}
Applying the Euler Lagrange equations to Eq.(45) yields the equations
\begin{align}
\frac{d}{dt}I_3(\dot{\psi}- \dot{\phi}\sin\theta) & = 0 \quad \longrightarrow \quad L_3 \equiv I_3(\dot{\psi}- \dot{\phi}\sin\theta) \qquad(46)\\[.5em]
\frac{d}{dt}\left[I \dot{\phi} \cos^2\theta – L_3 \sin\theta \right] & = – m g \ell \sin \alpha \sin \phi\cos\theta \qquad(47)\\[.5em]
I \ddot{\theta} = -L_3 \dot{\phi}\cos\theta- I \dot{\phi}^2\cos\theta \sin\theta – \lambda \ell \cos\theta & – m g \ell \big[\sin\alpha \cos\phi \sin \theta + \cos\alpha \cos\theta \big] \qquad(48)\\[.5em]
\ell \sin\theta = r \qquad(49)\label{eq:theta_const}
\end{align}
Imposing the constraint \(\ell \sin\theta = r\) on the rest of the equations and identifying the Lagrange multiplier \(\lambda\) with \(F_N\), we see that these equations are identical to Eq.(34), Eq.(35), and Eq.(36), and thus they produce the equation of motion Eq.(37) and the normal force Eq.(40).
Pendulum on an Incline
In this final section, we consider the analogous problem for a point particle of mass \(m\) attached to a rod of length \(\ell\). In this case, \(r=0\) and \(\theta =0\) since the rod lies flat on the plane of the incline, and the point particle (we will assume) has no spin, (i.e., \(\dot{\psi} =0\)). (You can imagine Figure 2 with \(\theta =0\) and no spin).
The expressions for the coordinate axes in terms of the principal axes are
\begin{align}
\mathbf{\hat{z}} &= -\mathbf{\hat{x}}_2 \qquad(50)\\
\mathbf{\hat{y}} &= \cos\phi\,\mathbf{\hat{x}}_3 – \sin \phi\,\mathbf{\hat{x}}_1 \qquad(51)\\
\mathbf{\hat{x}} &= \cos\phi\, \mathbf{\hat{x}}_1 + \sin\phi\, \mathbf{\hat{x}}_3 \qquad(52)
\end{align}
and thus the force due to gravity is
\begin{equation}
\textbf{F}_g = mg\left(\sin \alpha \, \mathbf{\hat{y}}-\cos\alpha\, \mathbf{\hat{z}} \right) = mg \left[ \cos\alpha \, \mathbf{\hat{x}}_2+ \sin\alpha \cos\phi\, \mathbf{\hat{x}}_3 – \sin\alpha \sin\phi\, \mathbf{\hat{x}}_1\right]. \qquad(53)
\end{equation}
The time derivative of the angular momentum is
\begin{equation}
\frac{d}{dt}\textbf{L} = \frac{d}{dt} \Big[I \dot{\phi} \,\mathbf{\hat{x}}_2\Big] = I \ddot{\phi} \, \mathbf{\hat{x}}_2. \qquad(54)
\end{equation}
The net torque on the point particle is
\begin{align}
\boldsymbol{\tau}_{\text{net}} &= \ell \mathbf{\hat{x}}_3 \times \textbf{F}_g + \ell \mathbf{\hat{x}}_3 \times \textbf{F}_N + \ell \mathbf{\hat{x}}_3 \times \textbf{F}_f \\
& = – mg\ell \cos\alpha\, \mathbf{\hat{x}}_1 – mg\ell \sin\alpha\sin\phi\, \mathbf{\hat{x}}_2 + \ell F_N \, \mathbf{\hat{x}}_1 + \ell F_f \mathbf{\hat{x}}_2, \qquad(55)
\end{align}
where we used \(\textbf{F}N =- F_N \, \mathbf{\hat{x}}_2\) and \(\textbf{F}_f = F_f \, \mathbf{\hat{x}}_1\) (friction is assumed for generality). Setting \(\boldsymbol{\tau}{\text{net}}\) equal to \(d\textbf{L}/dt\), and noting that \(I = m\ell^2\) we find
\begin{equation}
F_N = mg \cos \alpha, \qquad \ddot{\phi} = -\frac{g}{\ell} \sin \alpha \sin\phi + \frac{\ell F_f}{I}.
\label{eq:pendulum_eom} \qquad(56)
\end{equation}
Comparing the equation of motion result to Eq.(28) and Eq.(37), we see that Eq.(56) is more similar to Eq.(37) than to Eq.(28) even though it is the latter that incorporates friction. In particular, the \(\theta =0\) limiting case of Eq.(28) does not at all produce the right coefficient to be consistent with Eq.(56), where as Eq.(37) does. Why is this?
A hint comes from the comparison of two different problems. Say we have a massive point particle sliding down an incline which makes an angle of \(\alpha\) with the horizontal. Assuming there is a retarding frictional force \(F_f\), the acceleration of the point particle is \(\ddot{x} = g \sin\alpha – F_f\). If instead of point particle, we had a massive sphere rolling (without slipping) down the incline, we would find an acceleration of \(\ddot{c} = \frac{5}{7} g \sin \alpha\) a result which is not only independent of the size of the sphere, is also independent of the frictional force generating the rolling. Thus it would not be possible to take a limiting case from the latter rolling sphere to the former sliding point particle. The two situations make incommensurately different assumptions about the constraints of their respective motion.
In our calculation of the lollipop rolling without slipping we have a similar incommensurability with the pendulum on an incline limiting case. Thus taking the \(r\to 0\) limiting case of the rolling lollipop does not produce a result we can compare to anything involving the motion of a point particle.
Stranger than Friction
The comparison between the rolling sphere and the sliding point particle reveals something special about the “friction” responsible for generating rolling. It is a friction which is entirely different from informal notions of friction. For one, this friction does not result in a loss of energy in the system and is not proportional to the normal force where the proportionality constant is defined by a velocity dependent constant. Thus the friction responsible for locking \(\dot{\psi}\) and \(\dot{\phi}\) together in the rolling lollipop problem is different from the kinetic friction often considered in mechanics problems.
Footnotes
- Though we could use a Lagrange multiplier approach to impose the desired constraints on the system ↩︎
- Though we are writing our final results in terms of the principal axes, we still need to incorporate the non-inertial movement of the \(S\) axis relative to \(R\) in a calculation of the angular velocity. ↩︎
- We could also write this point of contact as \(d(\sin\phi \, \mathbf{\hat{x}} + \cos\phi \, \mathbf{\hat{y}})\) but it is more useful to express this vector in terms of the principal axes basis. ↩︎