Solution

Question 1.1. To find the moment of inertia \(C\) of body 1 about its axis \(Gz\), we consider the cylinder of mass \(m_2\), radius \(r_2\) and axis \(Gz\) and the cylinder of mass \(m_1\), radius \(r_1\) and axis \(Gz\): we have \[ \frac{1}{2} m_2 r_2^2 = \frac{1}{2} m_1 r_1^2 + C \] The masses of these 3 bodies are given by \[ m_2 = \rho (\pi r_2^2 h), \qquad m_1 = \rho (\pi r_2^2 h), \qquad m= m_2 -m_1 \] We can then express \(m_1\) and \(m_2\) in terms of mass \(m\): \[ m_1 = m \frac{r_1^2}{r_2^2- r_1^2}, \qquad m_2 = m \frac{r_2^2}{r_2^2- r_1^2} \] Then moment of inertia \(C\) is found to be \[ C= \frac{1}{2} (m_2 r_2^2 - m_1 r_1^2)= \frac{1}{2} m (\frac{r_2^4}{r_2^2- r_1^2} - \frac{r_1^4}{r_2^2- r_1^2} ) \] or \[ \boxed{ C= \frac{1}{2} m (r_2^2 + r_1^2) } \]

Figure 1

Question 1.2. To find the kinematic screw of body 1 resolved about mass center \(G\), first we find the velocity of \(G\): with \(\boldsymbol{r}_{OG} = -(r_1-r) \boldsymbol{\hat{e}}_r\), we find \[ \boldsymbol{v}_G = -(r_1-r) \dot{\theta}\boldsymbol{\hat{e}}_\theta \] This gives the expression \[ \{ {\cal V}_{1/0} \} = \begin{Bmatrix} \omega_{1/0} \boldsymbol{\hat{z}}_0 \\ -(r_1-r) \dot{\theta}\boldsymbol{\hat{e}}_\theta \end{Bmatrix}_G \] where \(\omega_{1/0}\) is found by imposing the no-slip condition at \(I\): \[ \boldsymbol{v}_{I\in 1/0} = -(r_1-r) \dot{\theta}\boldsymbol{\hat{e}}_\theta+ \omega_{1/0} \boldsymbol{\hat{z}}_0\times r_1 \boldsymbol{\hat{e}}_r = \boldsymbol{0} \] This gives \[ \boxed{ \omega_{1/0} = \frac{r_1-r}{r_1} \dot{\theta} } \]

The body has 1 degree of freedom (angle \(\theta\)).

Question 1.3. We apply the FTD \(\{{\cal D}_{1/0}\}= \{{\cal A}_{ext\to 1}\}\) to find \[ \begin{Bmatrix} -m(r_1-r) (\ddot{\theta}\boldsymbol{\hat{e}}_\theta-\dot{\theta}^2 \boldsymbol{\hat{e}}_r) \\ C \dot{\omega}_{1/0} \boldsymbol{\hat{z}}_0 \end{Bmatrix}_G = \begin{Bmatrix} -mg \boldsymbol{\hat{x}}_0 \\ \boldsymbol{0} \end{Bmatrix}_G + \begin{Bmatrix} N \boldsymbol{\hat{e}}_r + F\boldsymbol{\hat{e}}_\theta \\ \boldsymbol{0} \end{Bmatrix}_I \] where the contact force is decomposed in terms of a normal component \(N\) and a tangential component \(F\) (friction). This gives 3 equations: \[\begin{align*} m(r_1-r) \dot{\theta}^2 = N - mg \cos\theta& \qquad (1) \\ - m(r_1-r) \ddot{\theta}= F + mg \sin\theta& \qquad (2) \\ C \frac{r_1-r}{r_1} \ddot{\theta}= r_1 F & \qquad (3) \end{align*}\] If we combine (2) and (3) we find the equation of motion \[ \boxed{ \frac{r_1-r}{2r_1} (3r_1^2 + r_2^2) \ddot{\theta}= -g r_1 \sin\theta \qquad\qquad (4) } \] (after using the expression of \(C\) found in q. 1.1). Equation (1) gives the normal reaction \(N\) as a function of \(\theta\). After pre-multiplying (4) by \(\dot{\theta}\), we find that (4) in the form \[ \frac{d}{dt}(\frac{\dot{\theta}^2}{2}) + Cst \frac{d}{dt}(\cos\theta)= 0 \] gives a a first-integral of motion \[ \boxed{ \frac{r_1-r}{4r_1} (3r_1^2 + r_2^2) \dot{\theta}^2 = g r_1 \cos\theta+ C_0 \qquad\qquad (5) } \] where the constant \(C_0\) is found from the initial conditions. The body remains in contact with its support as long as \(N >0\) during the motion. With \[ N= m(r_1-r) \dot{\theta}^2 + mg \cos\theta \] and upon using (5), we find that \(N>0\) is equivalent to \(\cos\theta> A\), where \(A\) is a constant (which depends on the geometric parameters and constant \(C_0\)). If \(|A|< 1\), then the body will lose contact at some angle \(\theta_1 = \cos^{-1} A\). Whether this takes place or not depends on the initial conditions (through constant \(C_0\)).

Question 1.4. If the angle \(\theta\) remains small, then (4) can be written in the form \[ \frac{r_1-r}{2r_1} (3r_1^2 + r_2^2) \ddot{\theta}+ g r_1 \theta=0 \] This equation shows that body 1 exhibits small amplitude oscillations about the position \(\theta=0\), of period given by \[ \boxed{ T = 2\pi \sqrt{\frac{3k (r_1-r)}{2g}} } \] with \(k= 1+ r_2^2/3r_1^2\).

Solution

Denote by \(\Sigma\) the system comprised of the massless frame and the two cylinders 1 and 2 and introduce angle \(\phi\) to define the orientation of the frame at an arbitrary time. The corresponding unit vectors are \((\boldsymbol{\hat{e}}_r,\boldsymbol{\hat{e}}_\phi)\). See Figure 2.

At any given time, the angular momentum of system \(\Sigma\) about axis \(\Delta = (O,\boldsymbol{\hat{e}}_z)\) is conserved since \[ \boldsymbol{M}_{O, \bar{\Sigma}\to\Sigma} \cdot \boldsymbol{\hat{e}}_z = \underbrace{\boldsymbol{M}_{O, \bar{\Sigma}\to\Sigma}^c \cdot \boldsymbol{\hat{e}}_z}_{\text{frictionless pivots}} + \underbrace{\boldsymbol{M}_{O, \bar{\Sigma}\to\Sigma}^g \cdot \boldsymbol{\hat{e}}_z}_{\text{gravity}} = 0 \] With \(\boldsymbol{D}_{O, \Sigma}\cdot \boldsymbol{\hat{e}}_z = {d\over dt} \boldsymbol{H}_{O,\Sigma} \cdot \boldsymbol{\hat{e}}_z\), this implies \[ \boxed{ \boldsymbol{H}_{O,1} \cdot \boldsymbol{\hat{e}}_z +\boldsymbol{H}_{O,2} \cdot \boldsymbol{\hat{e}}_z = \text{constant} } \] where the constant is defined by the initial conditions: prior to blocking the rotation of the cylinders, both cylinders rotate at the angular velocity \(\omega_0\boldsymbol{\hat{e}}_z\). Once the rotation of cylinder 2 is blocked, its angular momentum about \(\Delta\) is altered. To maintain conservation of \(\boldsymbol{H}_{O,\Sigma} \cdot \boldsymbol{\hat{e}}_z\) of the system, the frame must necessarily rotate.

We can find the angular momentum of the system at any given time, denoting the angular velocities of the cylinders and the frame as \(\Omega_1 \boldsymbol{\hat{e}}_z\), \(\Omega_2 \boldsymbol{\hat{e}}_z\) and \(\dot{\phi}\boldsymbol{\hat{e}}_z\), respectively: \[ \text{Body 1: } \boldsymbol{H}_{O,1} \cdot \boldsymbol{\hat{e}}_z = (\boldsymbol{H}_{G_1,1} + \boldsymbol{r}_{OG_1} \times m \boldsymbol{v}_{G_1} ) \cdot \boldsymbol{\hat{e}}_z = I \Omega_1 + (-l \boldsymbol{\hat{e}}_r) \times (-ml \dot{\phi}\boldsymbol{\hat{e}}_\phi) \cdot \boldsymbol{\hat{e}}_z = I \Omega_1 + ml^2 \dot{\phi} \] \[ \text{Body 2: } \boldsymbol{H}_{O,2} \cdot \boldsymbol{\hat{e}}_z = (\boldsymbol{H}_{G_2,2} + \boldsymbol{r}_{OG_2} \times m \boldsymbol{v}_{G_2} ) \cdot \boldsymbol{\hat{e}}_z = I \Omega_2 + (l \boldsymbol{\hat{e}}_r) \times (ml \dot{\phi}\boldsymbol{\hat{e}}_\phi) \cdot \boldsymbol{\hat{e}}_z = I \Omega_2 + ml^2 \dot{\phi} \] Conservation of angular momentum of the system about \(\Delta\) can be stated as \[ \boxed{ I (\Omega_1 + \Omega_2) + 2 ml^2 \dot{\phi}= 2 I\omega_0 } \] since at \(t=0\), \(\Omega_1=\Omega_2 = \omega_0\) and \(\dot{\phi}=0\).

Also, in the absence of friction on the axle of the cylinders, the angular momentum of each cylinder is conserved, as long as their rotation is free: this leads to \(\Omega_i = \text{constant}\).

Figure 2

Question 2.1. After cylinder 2 is immobilized, the frame and cylinder 2 rotate at angular velocity \(\boldsymbol{\omega}_1= \dot{\phi}_1 \boldsymbol{\hat{e}}_z\) (since the cylinder is now rigidly attached to the frame), while cylinder 1 continues to rotate at angular velocity \(\Omega_1 \boldsymbol{\hat{e}}_z= \omega_0\boldsymbol{\hat{e}}_z\) (the angular momentum of cylinder 1 is still conserved). With \(\Omega_2= \dot{\phi}_1\) and \(\Omega_1= \omega_0\), conservation of angular momentum of the system about axis \(\Delta\) now gives \[ I (\omega_0 + \dot{\phi}_1) + 2 ml^2 \dot{\phi}_1 = 2I \omega_0 \] leading to \[ \boxed{ \dot{\phi}_1 = \frac{I}{ 2ml^2 +I } \omega_0 } \] This gives the angular velocity \(\boldsymbol{\omega}_1 =\dot{\phi}_1\boldsymbol{\hat{e}}_z\) of the frame.

Question 2.2. Now both cylinders are immobilized relative to the frame. The angular velocities of the cylinders are both equal to that of the frame: \(\Omega_1 =\Omega_2 = \dot{\phi}_2\). Now conservation of angular momentum of the system about \(\Delta\) gives \[ (2I+ 2ml^2) \dot{\phi}_2 = 2I \omega_0 \] leading to \[ \boxed{ \dot{\phi}_2 = \frac{I}{I+m l^2} \omega_0 } \] This gives the final angular velocity \(\boldsymbol{\omega}_2 =\dot{\phi}_2 \boldsymbol{\hat{e}}_z\) of the frame.

If the both cylinders initially rotate in the counterclockwise direction (\(\omega_0 >0\)), then the frame rotates in the counterclockwise direction in both situations, since we found \(\dot{\phi}_2 > \dot{\phi}_2 >0\).

© 2026  R. Valéry Roy.