Problem 8.1.
We choose a basis \((\boldsymbol{\hat{\imath}},\boldsymbol{\hat{\jmath}},\boldsymbol{\hat{k}})\) as shown in Figure 1. We denote \(\boldsymbol{v}_G = v(t)\boldsymbol{\hat{\imath}}\) and \(\boldsymbol{\omega}= -\omega (t) \boldsymbol{\hat{k}}\) the velocity of mass center \(G\) and angular velocity of the body at time \(t\geq 0\). At time \(t=0^+\), we have \(v(0)=0\) and \(\omega (0)=\omega_0 > 0\). At any time \(t \geq 0\), the slip velocity is given by \[ \boldsymbol{v}_{I \in 1/0} = \boldsymbol{v}_G + \boldsymbol{\omega}\times \boldsymbol{r}_{GI} = (v- r\omega) \boldsymbol{\hat{\imath}} \] At \(t=0^+\), we have \(\boldsymbol{v}_{I\in 1/0} = -r\omega_0 \boldsymbol{\hat{\imath}}\). Hence we expect that, in some initial phase of motion, the slip velocity is directed to the left.
We then sketch a FBD (see Figure 1): to oppose the slip velocity, the friction force must be directed to the right. The role of \({\bf F}\) is to cause the slip of the sphere to eventually vanish.
Now we can apply Euler’s Principles: \(m \dot{v}\boldsymbol{\hat{\imath}}= F\boldsymbol{\hat{\imath}}+ (N-mg)\boldsymbol{\hat{\jmath}}\) and \(\frac{d{\bf H}_G}{dt} = {\bf M}_G\). This gives 3 equations \[ m \dot{v}= F, \qquad 0= N-mg, \qquad - \frac{2}{5}mr^2 \dot{\omega}= rF \qquad[1-3] \] after using \({\bf H}_G = - \frac{2}{5}mr^2 \omega \boldsymbol{\hat{k}}\). Since the body slips, we must also impose \[ |F|= F = \mu N \qquad[4] \] We can see that \(v\) increases and \(\omega\) decreases: it is straightforward to obtain \(v(t)\) and \(\omega (t)\) versus time: \[ v(t) = \mu g t, \qquad r\omega (t) = r \omega_0 - \frac{5}{2}\mu gt \] which then gives \[ \boldsymbol{v}_{I \in 1/0}\cdot \boldsymbol{\hat{\imath}}= v- r\omega = -r\omega_0 + \frac{7}{2}\mu gt \] We see that the body stops slipping at time \(t=t_1\) given by \[ \boxed{ t_1 = \frac{2r\omega_0}{7\mu g} } \qquad [5] \] At time \(t=t_1\), the velocity \(v\boldsymbol{\hat{\imath}}\) of the mass center is given by \[ \boxed{ v= v_1 = r \omega_1 = \frac{2}{7}r\omega_0 } \qquad [6] \] We conclude that for \(t \geq t_1\), the body rolls without slipping to the right (\(v_1 >0\)). During this no-slip phase, equation [1-3] are still valid and equation [4] must be replaced by \[ v = r \omega, \qquad |F| < \mu N \qquad [7-8] \] (assuming \(\mu_s=\mu_k = \mu\)). We can see that the solution gives \(\dot{v}=0\), \(\dot{\omega}= 0\), and \(F=0\): \[ \boxed{ v(t) = v_1, \qquad \omega (t)= \omega_1 \qquad \text{for all } t\geq t_1 } \qquad [9] \] The body rolls forward indefinitely at constant angular velocity.
Conclusion (Quiz):
Answer A is correct: The body necessarily starts to slip relative to the support.
Answer B is incorrect as shown by equation [5]. We expect \(t_1\) to be dependent on \(\mu\). The larger \(\mu\) is, the smaller \(t_1\) is!
Answer C is incorrect as shown by equation [6]: the body rolls forward after \(t=t_1\) and cannot be observed to return to its initial position.
Answer D? we did not apply the 2nd principle by choosing contact point \(I\). This would now take the form: \(d{\bf H}_I /dt + \boldsymbol{v}_I \times m \boldsymbol{v}_G = {\bf M}_I\). We see that \({\bf M}_I = {\bf 0}\). The extra term \(\boldsymbol{v}_I \times m \boldsymbol{v}_G\) would make us think that answer D is incorrect. On second thought, we can easily see that \(\boldsymbol{v}_I = d\boldsymbol{r}_{OI}/dt = d\boldsymbol{r}_{OG}/dt = \boldsymbol{v}_G\). We can then conclude that \({\bf H}_I\) is conserved! (This is not expected!). Answer D is correct.
Answer E is correct as shown by equation [6] and [9].
Problem 8.2.
Step 1: We choose a fixed basis as shown in Figure 2 and define coordinates \(x_1\) and \(x_G\) to define the position of the plank 1 and mass center \(G\) of cylinder 2. We then express the no-slip condition of the cylinder relative to the plank: \[ \boldsymbol{v}_{I\in 1/0}=\boldsymbol{v}_{I\in 2/0} \Longrightarrow \dot{x}_1 = \dot{x}_G + R \omega \qquad{[1]} \] after expressing the angular velocity \(\boldsymbol{\omega}_{2/0}=\omega\boldsymbol{\hat{k}}\). Equation [1] is differentiable.
Step 2: After sketching a FBD, we apply Euler’s 1st principle to system \(\{1, 2\}\) to obtain two equations: \[ M \ddot{x}_1 + m \ddot{x}_G = F-F_1, \qquad N_1 = (m+M)g \qquad{[2-3]} \] where the contact force exerted by the support on the plank is denoted \({\bf R}_{0\to 1}= F_1 \boldsymbol{\hat{\imath}}+ N_1 \boldsymbol{\hat{\jmath}}\) with \(F_1 >0\) so as to oppose the slip of the plank. Also note that \((M+m)\boldsymbol{a}_{G_{syst}}= M\boldsymbol{a}_{plank}+m \boldsymbol{a}_G\) and that the internal forces vanish.
Step 3: After sketching a FBD, we apply Euler’s 1st and 2nd principles to body \(2\) to obtain 3 equations \[ F_2 = m \ddot{x}_G , \qquad N_2 =mg, \qquad \frac{1}{2}mR^2 \alpha = RF_2 \qquad{[4-6]} \] where the contact force exerted by the plank on the cylinder is denoted \({\bf R}_{1\to 2}= F_2 \boldsymbol{\hat{\imath}}+ N_2 \boldsymbol{\hat{\jmath}}\) acting through contact point \(I\). Also in [6], the angular acceleration of the cylinder is denoted \(\boldsymbol{\alpha}=\alpha\boldsymbol{\hat{k}}\) and we use the moment of inertia \(I_{Gzz}= \frac{1}{2}mR^2\).
We have obtained 6 equations for 7 unknowns \((\ddot{x}_1,\ddot{x}_G, \alpha, N_1, N_2, F_1, F_2)\). We find an additional equation by including the equation \[ F_1 = \mu N_1 = \mu (m+M)g \qquad [7] \] since body 1 slips relative to its support.
We can now solve [1-7] for accelerations \(\ddot{x}_1\), \(\ddot{x}_G\) and \(\alpha\) \[ \boxed{ \ddot{x}_1 = \frac{3}{2} R\alpha, \qquad \ddot{x}_G = \frac{1}{2} R\alpha, \qquad R \alpha= 2 \frac{F-\mu(M+m)g}{m+3M} } \] We need to impose \(\ddot{x}_1 >0\) for the plank to slide: this gives the condition \(F > \mu (M+m) g\). We also need to impose \(|F_2 | < \mu N_2\) (the condition necessary for cylinder to roll without slipping): with \(N_2 = mg\) and \(F_2 = \frac{1}{2}m R\alpha\) we find the condition \[ \frac{F-\mu(M+m)g}{m+3M} < \mu g \Longrightarrow F < 2\mu g (2M+m) \] In conclusion the motion will take place if \[ \boxed{\mu (M+m)g < F <2\mu (2M+m)g} \]
Problem 8.3.
a. With the notations of Figure 3, we define the kinematics of the spool as follows \[ \boldsymbol{v}_G =v_G \boldsymbol{\hat{\imath}}, \quad \boldsymbol{\omega}= \omega\boldsymbol{\hat{k}}, \quad \boldsymbol{a}_G = a_G \boldsymbol{\hat{\imath}}, \quad \boldsymbol{\alpha}= \alpha\boldsymbol{\hat{k}} \] No slip at contact point \(I\) imposes \[ \boldsymbol{v}_G = \omega\boldsymbol{\hat{k}}\times \boldsymbol{r}_{IG}= -R \omega\boldsymbol{\hat{\imath}}\Longrightarrow \boxed{a_G = -R \alpha} \qquad{[1]} \]
We sketch a FBD (see Figure 3): the spool is subject to (i) gravity, (ii) the applied force \({\bf F}\), and (iii) the reaction force at \(I\), denoted \({\bf R}= N_I \boldsymbol{\hat{\imath}}+ F_I \boldsymbol{\hat{\jmath}}\). There is no need to figure out the actual sense of the friction force since no-slip has been assumed.
Next we apply the Euler’s first and second principles to obtain 3 equations \[ m a_G = F_I + F \cos\beta, \qquad 0= N_I -mg + F \sin\beta, \qquad mk^2 \alpha= rF + R F_I \qquad{[2-4]} \] We have obtained 4 equations to solve for the 4 unknowns \((\alpha, a_G, F_I, N_I)\): \[ \boxed{ \alpha= \frac{r -R\cos\beta}{k^2 + R^2} \frac{F}{m}, \quad a_G = \frac{R(r -R\cos\beta)}{k^2 + R^2} \frac{F}{m}, \quad N_I= mg -F \sin\beta, \quad F_I = - \frac{k^2 \cos\beta +rR}{k^2 + R^2} F } \qquad{[5-8]} \] Since the spool is initially at rest, the sign of \(a_G\) gives the direction of motion: we see that the spool moves to the right if the condition \(\boxed{\cos\beta > r/R}\) is satisfied.
b. The spool will stay in contact with the support if \(N_I > 0\), that is, \(F\) satisfies the condition \[ \boxed{F < \frac{mg}{\sin\beta}} \] The condition of no slip is valid as long as \(|F_I|< \mu N_I\) (\(\mu\) is the coefficient of static friction). This gives the condition \[ \frac{k^2 \cos\beta +rR}{k^2 + R^2} F < \mu (mg -F \sin\beta) \] or \[ \boxed{ F < \frac{mg}{\sin\beta + \frac{\kappa}{\mu}} } \qquad \text{with} \quad \kappa =\frac{k^2 \cos\beta +rR}{k^2 + R^2}. \]
Problem 8.4.
We choose a basis \((\boldsymbol{\hat{\imath}},\boldsymbol{\hat{\jmath}},\boldsymbol{\hat{k}})\). We denote the disks as bodies 1 and 2, and the rod as body 3. See Figure 4.
a. kinematics. We write \(\boldsymbol{v}_A = v_A \boldsymbol{\hat{\imath}}\), \(\boldsymbol{\omega}_1 = -\omega_1 \boldsymbol{\hat{k}}\). The no-slip condition of disk 1 gives \(v_A = R \omega_1\). Since \(\boldsymbol{r}_{AB} = 2l \boldsymbol{\hat{\imath}}\), we have \(\boldsymbol{v}_B = v_A \boldsymbol{\hat{\imath}}\), and therefore \(\boldsymbol{\omega}_2 = -\omega_1 \boldsymbol{\hat{k}}\). Furthermore, time-differentiation gives \[ \boxed{a_A = a_B = R \alpha_1} \qquad{[1]} \] The acceleration of the mass center \(G\) of the rod is found by differentiating \(x_G = x_A + l+ r\sin\theta\) and \(y_G = R-r\cos\theta\) \[ \boxed{\ddot{x}_G = R\alpha_1 + r \ddot{\theta}\cos\theta-r\dot{\theta}^2 \sin\theta, \qquad \ddot{y}_G = r \ddot{\theta}\sin\theta+r\dot{\theta}^2 \cos\theta} \qquad{[2-3]} \] The rod is in translational motion.
b. kinetics. We sketch FBD for each body. See Figure 4. By symmetry, we may assume that forces \(F_A =F_B= F_{AB}\), \(N_1 = N_2= N\), and \(F_1 = F_2= F\). We also denote \(\alpha_1 = \alpha_2 = \alpha\), and \(a_A= a_B= a\).
We apply Euler’s principles to disk 1 (or disk 2): \[ \boxed{ M R \alpha= F +F_{AB} \sin\theta,\quad N - F_{AB} \cos\theta= Mg, \quad -\frac{1}{2} MR^2 \alpha= R F} \qquad{[4-6]} \] Note that force \(F_{AB}\) is directed along \(AC\) due to the fact that the connecting rod \(AC\) is massless.
Finally we apply Euler’s principles to rod 3 (which is in translational motion): \[ \boxed{ m\ddot{x}_G = - 2F_{AB} \sin\theta, \quad m \ddot{y}_G = -mg +2F_{AB} \cos\theta, \quad 0= -F_{AB} l \cos\theta+F_{AB} l \cos\theta } \qquad{[7-9]} \] The fact that equation [9] is automatically satisfied confirms the assumption that \(F_A = F_B\).
We have found 8 equations for 8 unknowns \((a, \alpha, \ddot{x}_G, \ddot{y}_G, \ddot{\theta}, F, N, F_{AB})\). To solve, we first use [6] in [4] to find \[ \frac{3}{2} MR \alpha= F_{AB} \sin\theta \] Then substitution of \(F_{AB}\sin\theta\) in equation [7] gives \[ (m+3M)R\alpha= -m (r \ddot{\theta}\cos\theta-r\dot{\theta}^2 \sin\theta) \] upon using [2] for the expression of \(\ddot{x}_G\). This expression of \(R\alpha\) can be substituted in [8] to obtain an ODE governing angle \(\theta\): \[ \boxed{ \left( 1- \frac{m}{m+3M}\cos^2\theta\right) \ddot{\theta} + \frac{m}{m+3M}\dot{\theta}^2 \cos\theta\sin\theta = -\frac{g}{r} \sin\theta } \qquad{[10]} \] Equation [10] leads to a 1st integral (conservation of energy?) \[ \left( 1- \frac{m}{m+3M}\cos^2\theta\right) \frac{\dot{\theta}^2}{2} -\frac{g}{r} \cos\theta =Cst \]
d. The small amplitude oscillations are obtained by linearizing [10]: \[ \frac{3M}{m+3M} \ddot{\theta} + \frac{g}{r}\theta=0 \] This a standard pendulum equation \(\ddot{\theta}+\Omega^2 \theta=0\) with corresponding period \[ \boxed{ T = 2\pi \sqrt{\frac{r}{g}\frac{3M}{m+3M}} } \] When \(M\gg m\), we obtain \(T= 2\pi \sqrt{\frac{r}{g}}\): this is the expected result, since the disks are now stationary and the rod oscillates about fixed points \(A\) and \(B\).