Pointers (2020)
Pointers (2020)
Question 1: Please ignore the statement “of your choice” in this question. At this point in the course, we know of only one method to solve this question vectorially.
Question 2: you cannot take the time-derivative of the quantities found in question 1 (understand why…) Also you cannot set \(\boldsymbol{a}_{I\in 1} ={\bf 0}\) (you should know the reason why this is so).
Instead, use the fact that the path of center \(C\) is known. You can then identify the unit tangent and normal vectors associated to this path at the time of the configuration. Then, use the NT (intrinsic) expression of the acceleration of center \(C\).
This is an example which shows how the knowledge of chapter 1 (particle kinematics) is relevant to chapter 2 (rigid body kinematics).
Question 3: You should be able to recover the same answers found in question 1.
We see that there exist 2 independent closed loops in this graph: the loops \(0-2-4-0\) and \(1-3-2-4-1\). The following kinematic loops must be imposed:
first for angular velocities \[ \boldsymbol{\omega}_{0/2}+\boldsymbol{\omega}_{2/4}+\boldsymbol{\omega}_{4/0}= \boldsymbol{0} \] \[ \boldsymbol{\omega}_{1/3}+\boldsymbol{\omega}_{3/2}+\boldsymbol{\omega}_{2/4}+ \boldsymbol{\omega}_{4/1}= \boldsymbol{0} \]
then for the velocities of a particular point \(P\) (any point can be chosen) \[ \boldsymbol{v}_{P\in 0/2}+\boldsymbol{v}_{P\in 2/4}+\boldsymbol{v}_{P\in 4/0}= \boldsymbol{0} \] \[ \boldsymbol{v}_{P\in 1/3}+\boldsymbol{v}_{P\in 3/2}+\boldsymbol{v}_{P\in 2/4}+ \boldsymbol{v}_{P\in 4/1}= \boldsymbol{0} \]
Each term in these equations must be found according to the nature of the particular joint at play between any pair of bodies. It is thus key to carefully identify each joint.
For instance, if the joint between bodies \(i\) and \(j\) is a slider along direction \(\boldsymbol{\hat{e}}\), then you must impose \[ \boldsymbol{v}_{P\in i/j} = v_{i/j} \boldsymbol{\hat{e}} \] for any point \(P\), and \(v_{i/j}\) may or may not be known.
If the joint between bodies \(i\) and \(j\) is a (possibly instantaneous) rotation about a particular point \(K\) at angular velocity \(\omega_{i/j}\boldsymbol{\hat{k}}\), then you must impose \[ \boldsymbol{v}_{P\in i/j} = \omega_{i/j} \boldsymbol{\hat{k}}\times \boldsymbol{r}_{KJ} \] for any point \(P\), and \(\omega_{i/j}\) may or may not be known.
6.1 Repeat the steps taken in Lecture 12 when we solved this problem in the absence of friction. Now add a tangential force due to friction between the 2 blocks. Assume relative motion between the block. Then apply Coulomb laws and Newton 2nd law.
6.2 To find the equation which governs angle \(\theta\), apply Newton’s 2nd law for \(B\) along the direction normal to \(AB\). This eliminates the force (tension) exerted on \(B\) by the connecting rod. Since the rod is assumed massless, this force acts along \(AB\).
6.3 Use polar coordinates. There is no force in the radial direction. This leads to a simple linear ODE in the radial coordinate. You can easily integrate it.
6.4 You can answer q.a from q.b after you find the accelerations. The motions are rectilinear. Simply apply Newton 2nd law for each particle. Include friction (there is slip) and the tension of the rope.
6.5 If the angle \(\phi\) reaches an equilibrium value, then the path is circular. You could choose NT axes to solve the problem.
6.6 Use polar coordinates \((r=R, \theta)\) for the contact point of the string with the disk. Then find the position of \(P\) as a function of \(\theta\) and \(l= l_0 -R\theta\) (the length of the unwound section of the string). Take \(d/dt\) twice to find \(\boldsymbol{a}_P\) and apply Newton 2nd law. You should then get an ODE governing angle \(\theta\). The other equation gives the tension in the string.
7.1 Consult Lecture 7 for the graphical setup and parametrization. You would have to rederive the kinematics since this solution was done at time \(t=0^+\). Start with \(\boldsymbol{r}_{OG}= L \boldsymbol{\hat{u}}+ l\boldsymbol{\hat{u}}_1\) and differentiate twice. Then find \(\boldsymbol{a}_G \cdot \boldsymbol{\hat{u}}\) and \(\boldsymbol{a}_G \cdot \boldsymbol{\hat{v}}\). This is useful because you want to apply \(m \boldsymbol{a}_G = {\bf T} + m {\bf g}\) and project on \(\boldsymbol{\hat{u}}\) and \(\boldsymbol{\hat{v}}\) (this decouples tension \(T\) from one of the ODE). You can then easily modify equations [1-3].
7.2 Choose Cartesian coordinate axes \(Oxyz\) with \(Oz\) normal to the plane directed upward, \(Ox\) along the steepest incline directed downward. Apply the angular momentum form of Newton 2nd law about \(O\): so find angular momentum \({\bf H}_O\) and moment \({\bf M}_O\) of the applied forces. This eliminates the tension. You will obtain an ODE of the type \(\ddot{\theta} + {\omega}^2 \theta=0\) after using the small angle approximation \(\sin\theta \approx \theta\). This linear ODE yields periodic solutions of period \(T= 2\pi/{\omega}\). Use the numerical answer to check your work.
7.3 In both cases you will apply Euler’s 2nd principle about \(O\). One of the method is incorrect. You will again obtain ODEs of the type \(\ddot{\theta} + {\omega}^2 \theta=0\) after using the small angle approximation \(\sin\theta \approx \theta\). In the first case, \({\omega}^2 = 3g/5l\). In the second case, \({\omega}^2 = 2g/3l\).
7.4 This is a 1 DOF motion: use angle \(\theta\). Then find \(\boldsymbol{v}_G\) and \(\boldsymbol{a}_G\) in terms of \(\theta\) and its derivatives. The reaction force at \(O\) has 2 components, horizontal and vertical. You need to find these components as functions of \(\theta\). To do this, you will need both \(\ddot{\theta}\) and \(\dot{\theta}\) versus \(\theta\). So apply the 2 principles to get 3 equations to solve for 3 unknowns: the reaction at \(O\) and angle \(\theta\) (the equation of motion). The ODE is of the type \(\ddot{\theta} + f(\theta) =0\): it can be integrated as \[ \frac{1}{2} \dot{\theta}^2 + \int f(\theta) d\theta= Cst \] Now you have \(\dot{\theta}\) and \(\ddot{\theta}\) vs \(\theta\). The rest is algebraic: you should be able to show that the components of the reaction force can become zero. The horizontal component reaches the zero value at the angle \(\theta_1 = \cos^{-1} (2/3)\). Conclude.
Note that the body is in rotation about \(O\). In that case, it is advisable to apply Euler’s principle about \(O\). This has the advantage of eliminating the moment of the reaction force, and of reducing algebraic manipulations.
In 3.14, assume a translational motion for the crate down the inclined support. Apply the principles, and solve for the acceleration and the location of the point where the contact force is applied along the contact. Impose the sign of the acceleration and the upper bound for the point of application: you end up with two conditions for slip (without tipping) to take place. There is an example in the textbook which follows the same steps.
In 3.15, you want the crate to rotate (tip) about a corner point without slipping. Now assume that the crate is about to tip with angular acceleration \(\alpha_0\) and angular velocity \(\omega_0 =0\). Apply the principles. Then impose \(\alpha_0 >0\) and the condition for no-slip. You will get two conditions to be satisfied by the parameters.
8.1 Proceed as we did in Lecture 16. Make sure the friction force opposes the slip velocity.
8.2 You should consult example 4.11 of the textbook. Before sketching FBDs, you must start the solution with a kinematics setup (as you must alsways do). Identify the kinematic unknowns. The no-slip condition will give a relationship between the velocity of the plank, the velocity of mass center \(G\) and the angular velocity of the cylinder. You will need 7 equations since the problem has 7 unknowns (4 forces+3 kinematic unknowns). Then you must impose the conditions (inequalities) for the motion to take place.
8.3 Find 4 equations for the unknowns \((a_G, \alpha, F_I, N_I)\) (where \(F_I\) and \(N_I\) are the components of the reaction force at \(I\)). The sign of \(a_G\) gives the direction of the motion. Then impose \(N_I >0\) (no loss of contact) and \(|F_I|<\mu N_I\) (no slip).
8.4 This is a multibody problem: enumerate the unknowns. There are 13 of them. To simplify, you could use a symmetry argument: \(N_1=N_2\), \(F_1= F_2\), \(F_A=F_B\), \(\alpha_1= \alpha_2\). Also you necessarily have \(a_A= a_B = R \alpha_1\). This simplifies the solution quite a bit.
9.1 This problem was addressed in recitation 7 (problem 2). Question a was solved by applied conservation of linear momentum. Solve question b by applying the WEP: this will give you the relationship \(\dot{\theta}^2\) vs \(\theta\). Treating the system as a uniform slender rod simply changes the mass distribution: find the position of mass center \(G\) and the mass moment of inertia \(I_{Gzz}\).
9.2 First you would need to find the location of the mass center \(G\) and the mass moment of inertia \(I_{Azz}\) of the (composite) body. Then apply the WEP between \(t=0\) (\(AB\) horizontal) and \(t=t_1\) (\(AB\) vertical). Take into account the no-slip condition. The method does not let us know the minimum value of \(\mu\) required for the motion to take place (if \(\mu\) is too small, the body would slip). In question b, define an angle \(\theta\) for the orientation (relative to the vertical, so you can let \(\theta\) small) of the rod \(AB\): use conservation of energy to find the ODE governing \(\theta\).
9.3 Use the kinematics results found in homework 8. Then apply the WEP for the system (wait for Lecture 21, however the WEP takes the same form as for systems of particles). Does this lead to the same ODE found previously (equation 10)?