Problem 6.1.
We use basis \((\boldsymbol{\hat{\imath}}, \boldsymbol{\hat{\jmath}})\). The blocks are in translation. We have \(\boldsymbol{a}_1 = \ddot{x}_1 \boldsymbol{\hat{\imath}}= a_1 \boldsymbol{\hat{\imath}}\) and \(\boldsymbol{a}_2 = \ddot{y}_2 \boldsymbol{\hat{\jmath}}= a_2 \boldsymbol{\hat{\jmath}}\).
The motions are not independent: we have the constraint equation \[ \dot{y}_2 = - \dot{x}_1 \tan\alpha \qquad{(1)} \] obtained by imposing \(\boldsymbol{v}_{P\in 2/1}\cdot \boldsymbol{\hat{n}}= 0\), where \[ \boldsymbol{v}_{P\in 2/1} = \boldsymbol{v}_{P\in 2/0}-\boldsymbol{v}_{P\in 1/0} = \dot{y}_2 \boldsymbol{\hat{\jmath}}-\dot{x}_1 \boldsymbol{\hat{\imath}} \] with \(\boldsymbol{\hat{n}}\) the unit vector normal to the contact line.
We sketch the FBDs for both blocks. On these diagrams we draw friction force \(F_{12}\) applied to block \(1\) opposed to the slip velocity \(\boldsymbol{v}_{P\in 1/2}\) of block \(1\) relative to block \(2\) (we assume that as \(F\) is applied, block \(2\) moves upward).
Newton 2nd law applied to block \(1\) gives \[ m_1 \ddot{x}_1 = -F + N_{12} (\sin\alpha+ \mu \cos\alpha) , \quad 0= N_1-m_1 g +N_{12} (-\cos\alpha+\mu \sin\alpha) \qquad{(2-3)} \] where we have used \(F_{12} = \mu N_{12}\).
Newton 2nd law applied to block \(2\) gives
\[
0=N_2 -N_{12} (\sin\alpha+ \mu \cos\alpha) , \quad m_2 \ddot{y}_2 = -m_2 g - N_{12} (-\cos\alpha+\mu \sin\alpha)
\qquad{(4-5)}
\]
We have obtained 5 equations for 5 unknowns \((\ddot{x}_1, \ddot{y}_2, N_{12}, N_1, N_2)\): we solve for \(\ddot{y}_2\) to find
\[
\boxed{\ddot{y}_2 = \frac{\kappa F- m_2 g}{m_2 + \kappa m_1\cot\alpha}}
\]
We must require \(\ddot{y}_2 >0\): this gives the condition
\(\boxed{\kappa F> m_2 g}\),
where \(\kappa\) is defined as
\[
\boxed{
\kappa= \frac{\cos\alpha-\mu \sin\alpha}{\sin\alpha+ \mu \cos\alpha}
}
\]
Problem 6.2.
When the slider \(A\) is no longer in contact with the backstop, the system has 2 DOF described by \((x,\theta)\). The kinematics of \(A\) and \(B\) is now given by
\[
\boldsymbol{v}_A = \dot{x}\boldsymbol{\hat{\imath}}, \qquad \boldsymbol{v}_B = \dot{x}\boldsymbol{\hat{\imath}}+ l\dot{\theta}\boldsymbol{\hat{v}}
\]
and
\[
\boldsymbol{a}_A = \ddot{x}\boldsymbol{\hat{\imath}}, \qquad \boldsymbol{a}_B = \ddot{x}\boldsymbol{\hat{\imath}}+ l\ddot{\theta}\boldsymbol{\hat{v}}- l\dot{\theta}^2 \boldsymbol{\hat{u}}
\]
Since the system \(\Sigma = \{ A, B, rod\}\) is not subject to external forces in the \(\boldsymbol{\hat{\imath}}\)-direction, we have \({\bf F}_{ext\to \Sigma} \cdot \boldsymbol{\hat{\imath}}= 0 = (m_A \boldsymbol{a}_A + m_B \boldsymbol{a}_B) \cdot \boldsymbol{\hat{\imath}}\) which implies that \((m_A \boldsymbol{v}_A + m_B \boldsymbol{v}_B) \cdot \boldsymbol{\hat{\imath}}= Cst\): \[ (m_A + m_B) \dot{x}+ m_B l\dot{\theta}\cos\theta= Cst \qquad{(1)} \]
To find the other equation, we apply Newton 2nd law for \(B\) along the \(\boldsymbol{\hat{v}}\)-direction: \(m_B \boldsymbol{a}_B \cdot \boldsymbol{\hat{v}}= - m_B g\sin\theta\). This eliminates the tensile force due to the connecting rod which acts along \(\boldsymbol{\hat{u}}\) (this is a consequence of neglecting its mass). This gives the equation \[ \ddot{x}\cos\theta+ l \ddot{\theta}= - g\sin\theta \] We can substitute \(\ddot{x}\) given by \[ (m_A + m_B) \ddot{x}=- m_B l\ddot{\theta}\cos\theta+ m_B l \dot{\theta}^2 \sin\theta \] to obtain the ODE governing angle \(\theta\): \[ \boxed{ \left( 1- \kappa \cos^2 \theta\right) \ddot{\theta}+ \kappa \dot{\theta}^2 \sin\theta\cos\theta= - \frac{g}{l} \sin\theta } \qquad{(2)} \] with the parameter \(\boxed{\kappa= m_B/(m_A+m_B)}\).
Problem 6.3.
The motion relative to disk \({\cal F}\) is rectilinear with relative velocity \(\boldsymbol{v}_{P/{\cal F}} = \dot{r}\boldsymbol{\hat{e}}_r\).
Referential \({\cal F}\) is not Newtonian. We must apply Newton’s 2nd law relative to lab referential
\({\cal E}\) relative to which \({\cal F}\) rotates at angular velocity \(\boldsymbol{\omega}= \omega\boldsymbol{\hat{e}}_z\). The velocity of \(P\)
in \({\cal E}\) is given by
\[
\boldsymbol{v}_{P/{\cal E}} = \dot{r}\boldsymbol{\hat{e}}_r + \omega\boldsymbol{\hat{e}}_z \times r\boldsymbol{\hat{e}}_r =\dot{r}\boldsymbol{\hat{e}}_r + r\omega\boldsymbol{\hat{e}}_\theta\qquad{[1]}
\]
Time-differentiation (relative to \({\cal E}\)) of [1] gives the acceleration of \(P\):
\[
\boldsymbol{a}_{P/{\cal E}} = (\ddot{r}-r\omega^2) \boldsymbol{\hat{e}}_r + 2\dot{r}\omega\boldsymbol{\hat{e}}_\theta\qquad{[2]}
\]
[1-2] can also be obtained in polar coordinates.
A FBD is shown in Figure 3: \(P\) is subject to gravity and to the support’s reaction \(N_\theta\boldsymbol{\hat{e}}_\theta+ N_z \boldsymbol{\hat{e}}_z\) normal to the surfaces of contact. Application of Newton second law gives on \((\boldsymbol{\hat{e}}_r, \boldsymbol{\hat{e}}_\theta,\boldsymbol{\hat{e}}_z)\) \[ m (\ddot{r}-r\omega^2) =0, \qquad 2m \dot{r}\omega= N_\theta, \qquad 0= N_z-mg \qquad{[3-5]} \] Equation [3] gives the equation of motion: it is easily integrated with the initial conditions \(\dot{r}=0\) and \(r=R/2\) at \(t=0\). \[ r = a \cosh(\omega t) + b \sinh (\omega t) \] \(\dot{r}=0\) at \(t=0\) gives \(b=0\). Then \(a= R/2\) is obtained by imposing \(r=R/2\) at \(t=0\). The final solution is given by \[ \boxed{r =\frac{R}{2} \cosh(\omega t)} \] Time \(T\) is found by imposing \(r(T)= R\): \[ \boxed{\omega T = \cosh^{-1}(2)= 1.32} \] The reaction force at time \(t=T\) is found by using [4-5]: \[ \boxed{ {\bf R} = \sqrt{3} mR\omega^2 \boldsymbol{\hat{e}}_\theta+ mg \boldsymbol{\hat{e}}_z } \] where we used \(\dot{r}= \frac{\sqrt{3}}{2} R\omega\).
Problem 6.4.
We choose an axis \(Ox\) directed downward along the inclined plane.
We can find the condition for the two blocks to remain in equilibrium
by finding the acceleration of block \(A\).
If we write \(\boldsymbol{a}_A = a_A \boldsymbol{\hat{e}}_x\), the motion takes place only if \(a_A >0\) is satisfied
(assuming \(m_A > m_B\)). Otherwise, the system remains in equilibrium.
Both blocks are in rectilinear motion. The position of \(A\) and \(B\) are defined by their coordinate along the \(x\)-axis: \[ \boldsymbol{r}_{OA} = x_A \boldsymbol{\hat{e}}_x, \qquad \boldsymbol{r}_{OB} = x_B \boldsymbol{\hat{e}}_x \] with \(x_A + x_B = Constant\) since the cable connecting the blocks must be of constant length. The accelerations are then given by \[ \boldsymbol{a}_A = \ddot{x}_A \boldsymbol{\hat{e}}_x, \qquad \boldsymbol{a}_B = - \ddot{x}_A \boldsymbol{\hat{e}}_x \] For \(m_A > m_B\) we expect \(\ddot{x}_A >0\).
The FBDs are shown in Figure 4. The blocks are subject to gravity, the rope’s tension and the support’s reaction force. We then apply Newton 2nd law and project on unit vectors \((\boldsymbol{\hat{e}}_x,\boldsymbol{\hat{e}}_z)\) to obtain 4 equations \[ m_A \ddot{x}_A = m_A g \sin\theta-T - \mu_A N_A, \qquad N_A -m_A g \cos\theta=0 \qquad{[1-2]} \] \[ -m_B \ddot{x}_A = m_B g \sin\theta-T + \mu_B N_B, \qquad N_B -m_B g \cos\theta=0 \qquad{[3-4]} \] To find \(\ddot{x}_A\) we subtract [3] from [1] to eliminate \(T\): \[ (m_A+m_B)\ddot{x}_A= (m_A-m_B)g \sin\theta- (\mu_A m_A + \mu_B m_B) g \cos\theta \] leading to \[ \boxed{ \frac{\ddot{x}_A}{g} = \frac{m_A-m_B}{m_A+m_B} \sin\theta- \frac{\mu_A m_A + \mu_B m_B}{m_A+m_B} \cos\theta } \] We conclude that the system stays in equilibrium if the condition \[ \boxed{ \tan\theta < \frac{\mu_A m_A + \mu_B m_B}{m_A -m_B} } \] is satisfied. For \(m_A/m_B =1.25\) and \(\mu_A = \mu_B =0.05\) and \(\theta= 30^\circ\) we find \(a_A = 0.012 \, g\).
Problem 6.5.
- a. We use the basis \((\boldsymbol{\hat{e}}_N,\boldsymbol{\hat{e}}_T,\boldsymbol{\hat{e}}_z)\) as shown in the figure: when angle \(\phi\) reaches an equilibrium value, \(A\) has a circular trajectory of radius \(\rho= l\sin\phi\), with \((\boldsymbol{\hat{e}}_T,\boldsymbol{\hat{e}}_N)\) the tangential/normal unit vectors associated with the motion of \(A\).
Particle \(A\) moves at constant speed \(v=\rho \omega\) and has (normal) acceleration \(\boldsymbol{a}= (v^2/\rho) \boldsymbol{\hat{e}}_N\). Its motion is found by applying Newton 2nd law: \[ m \frac{v^2}{l\sin\phi} \boldsymbol{\hat{e}}_N = {\bf T} + mg\boldsymbol{\hat{e}}_z \] which gives two equations on \((\boldsymbol{\hat{e}}_N,\boldsymbol{\hat{e}}_z)\): \[ \frac{v^2}{l\sin\phi}= T \sin\phi, \qquad T\cos\phi =mg \] with \(v = \rho \omega\). This gives \(T\) and angle \(\phi\) \[ \boxed{\cos\phi = \frac{g}{l\omega^2}, \qquad T = ml\omega^2} \]
- b. When \(T=20 mg\) we obtain \(\omega^2 = 20 g/l\).
Problem 6.6.
- a. Step 1: To define the position of \(P\) at any given time, we introduce the two fixed unit vectors \(\boldsymbol{\hat{\imath}}\) and \(\boldsymbol{\hat{\jmath}}\) as shown in Figure 6 and the two moving unit vectors \(\boldsymbol{\hat{e}}_r\) and \(\boldsymbol{\hat{e}}_\theta\) associated with the circular motion of point \(Q\) defined as the point of contact of the string with the cylinder.
We also introduce the parameters \(R\) and \(l_0\). Then we can express the position of \(P\) as \[ \boldsymbol{r}_{OP} = \boldsymbol{r}_{OQ} + \boldsymbol{r}_{QP} = R\boldsymbol{\hat{e}}_r + l \boldsymbol{\hat{e}}_\theta \qquad{(1)} \] where \(l\) is the length of the unwound part of the string. This length can be found as a function of angle \(\theta\) and total length \(l_0 = s + l = R\theta + l\): \[ l = l_0 - R\theta\qquad{(2)} \]
Step 2: To find the expression of velocity \(\boldsymbol{v}_P\) on \((\boldsymbol{\hat{e}}_r , \boldsymbol{\hat{e}}_\theta)\) we differentiate the position vector \(\boldsymbol{r}_{OP}= R\boldsymbol{\hat{e}}_r + l \boldsymbol{\hat{e}}_\theta\) w.r.t.~ time, using $ dr /dt = $ and \(d\boldsymbol{\hat{e}}_\theta/dt = - \dot{\theta}\boldsymbol{\hat{e}}_r\): \[ \boldsymbol{v}_P = R \dot{\theta}\boldsymbol{\hat{e}}_\theta+ \dot{l} \boldsymbol{\hat{e}}_\theta- l \dot{\theta}\boldsymbol{\hat{e}}_r = R \dot{\theta}\boldsymbol{\hat{e}}_\theta-R \dot{\theta}\boldsymbol{\hat{e}}_\theta- (l_0 -R\theta) \dot{\theta}\boldsymbol{\hat{e}}_r \Longrightarrow \boxed{ \boldsymbol{v}_P =- (l_0 -R\theta) \dot{\theta}\boldsymbol{\hat{e}}_r } \qquad{(3)} \] Note that the velocity of \(P\) cannot be found by differentiating vector \(\boldsymbol{r}_{QP}\) since point \(Q\) is in motion. Likewise, to find \(\boldsymbol{a}_P\) we differentiate \(\boldsymbol{v}_P = -l\dot{\theta}\boldsymbol{\hat{e}}_r\): \[ \boldsymbol{a}_P = - \frac{d}{dt}(l\dot{\theta}) \boldsymbol{\hat{e}}_r - l\dot{\theta}^2 \boldsymbol{\hat{e}}_\theta \qquad{(4)} \]
Step 3: We sketch a FBD. \(P\) is subjected to tension \(-T\boldsymbol{\hat{e}}_\theta\) and gravity \(m {\bf g} = m g (-\cos\theta\boldsymbol{\hat{e}}_r +\sin\theta\boldsymbol{\hat{e}}_\theta)\).
Step 4: We apply Newton 2nd law \(m\boldsymbol{a}_P = -T\boldsymbol{\hat{e}}_\theta+ m {\bf g}\): \[\begin{align*} \frac{d}{dt}(l\dot{\theta}) = g \cos\theta& \qquad\qquad (5) \\ -m l \dot{\theta}^2 = -T+ mg \sin\theta& \qquad\qquad (6) \end{align*}\] Equation (5) is the equation of motion: it is an ODE which governs angle \(\theta\).
Step 5: We can derive a first-integral from (5) by pre-multiplying both sides by \(l\dot{\theta}\): then \[ l\dot{\theta}\frac{d}{dt}(l\dot{\theta}) = \frac{d}{dt}( \frac{1}{2} l^2 \dot{\theta}^2 ) \] Now (5) becomes after integration \[ \frac{1}{2} l^2 \dot{\theta}^2 = \int_0^\theta g l \sin\theta d\theta+ Cst \] where we find the constant \(Cst\) to be \(0\) by imposing \(\theta=0\) and \(\dot{\theta}= 0\) at \(t=0\). This leads to (using \(\int \theta\cos\theta d\theta= \cos\theta+\theta\sin\theta\)) \[ \boxed{\frac{1}{2} (l_0-R\theta)^2 \dot{\theta}^2 = g (l_0-R\theta) \sin\theta+ gR (1-\cos\theta) } \qquad{(7)} \] Note: This first-integral expresses the conservation of energy of the system.
- b. The unwound length \(l\) is given by equation (2). The speed of \(P\) is given by \(v_P = l \dot{\theta}\) and equation (7): \[ \boxed{ v_P^2 = \frac{2 g}{ l} \sin\theta+ \frac{2gR}{l^2} (1-\cos\theta) } \] The tension \(T\) is given by equation (6): \[ \boxed{T = m l \dot{\theta}^2 + mg \sin\theta= 3mg \sin\theta+ 2 mg \frac{R}{l} (1-\cos\theta)} \] To find out if the rope becomes slack, we seek roots of the equation \(T=0\): we look for values of angle \(\theta_*\) solution of \[ \boxed{ 3 (\frac{l_0}{R} -\theta_*) \sin\theta_* + 2 (1-\cos\theta_*) = 0 } \] Numerically, we find \(\theta_* = 208^\circ\) for \(l_0/R = 2\pi\). We can show that \(\dot{\theta}_* > 0\) at \(\theta=\theta_*\): \[ l^2 \dot{\theta}^2_* = \frac{2R}{3}(1-\cos\theta_*) > 0 \] For \(\theta> \theta_*\), particle \(P\) behaves like a projectile.