# Constraints and constrainted motions

# Constraints

One of the most common type of motion is constrained motion, where an object is restricted to move in a certain way.

Generally, a point constrained to move in a specific direction obeys the following equation:

Constraint equation for a point $P$.

\[ \vec{v}_P \cdot \hat{n}_P = 0 \]

where $\hat{n}_P$ is the normal vector to the motion of point $P$.

The above equation is difficult to apply to a system with a collection of points or particles. It is not the only way to derive constraint equations. For example, rigid body kinematics equations (also known as relative motion equations) are derived using the definition of a rigid body. Namely, the distance and angles between all points remain constant. See rigid body relations. It is often best to look at examples of constrained motion to see how constraint equations are differentiated to obtain velocity and acceleration.

We can visualize the motion of several constrained systems, shown on the figure below.

System: | rod-circular | slider-crank |

rod-circular | slider-crank |
---|---|

We will begin with a simple example of a rod leaning against a wall constrained to maintain contact with the wall and ground.

Example Problem: Sliding rod

A uniform rod leaning against a wall has a fixed length of $\ell = 10 \rm\ m$. The bottom of the rod at point $P$ is $8 \rm\ m$ from the wall. Point $P$ is moving away from the wall at a speed of $6 \rm\ m/s$. The rod maintains contact with the wall and ground at all times. What is $\vec{v}_Q$ of point $Q$, the point at the top of the rod?

To solve this, note the question states that the rod's length is fixed, and is not permitted to lose contact with the wall and ground. Therefore, the constraint equation is given by the Pythagorean theorem: \[\begin{aligned} x_P^2 + y_Q^2 = \ell^2 \end{aligned}\]

This becomes a related rates problem from calculus 1.

Differentiating with respect to time, keeping in mind the chain rule: \[\begin{aligned} 2x_P\dot{x}_P + 2y_Q\dot{y}_Q = 0 \end{aligned}\]

The quantity we're interested in is $\dot{y}_Q$, or $v_Q$, and $y_Q = \sqrt{\ell^2 - x_P^2}$.

Solving yields $\vec{v}_Q = -8\hat{\jmath} \rm\ m/s$.

Another way to solve this is using the rigid body relations, knowing that the distance and angles from the top to the bottom of the rod remains constant, which is a rigid body by definition. Using rigid body relations: \[\begin{aligned} \vec{v}_Q &= \vec{v}_P + \vec\omega \times \vec{r}_{PQ} \\ &= \vec{v}_P + \omega_z \vec{r}_{PQ}^{\perp} \end{aligned}\] where we have two unknowns: $\vec{v}_Q$, and $\vec\omega$, and $\vec{r}_{PQ} = -\ell \cos\theta \hat{\imath} + \ell \sin\theta \hat{\jmath} \rm\ m$, where $\theta = \cos^{-1}\left(\frac{x_P}{\ell}\right)$

Therefore, $\vec{r}_{PQ}^{\perp} = -\ell \sin\theta \hat{\imath} - \ell \cos\theta \hat{\jmath} \rm\ m$

Expanding the vector equation into its two scalar equations yields: \[\begin{aligned} v_{Qx} &= v_{Px} - \omega_z \ell \sin\theta \\ v_{Qy} &= v_{Py} - \omega_z \ell \cos\theta \end{aligned}\]

Since the top of the rod can only move vertically, $v_{Qx} = 0$, and $\omega_z = \frac{v_{Px}}{\ell \sin\theta} \rm\ rad/s$

Since the bottom of the rod can only move horizontally, $v_{Py} = 0$, and $v_{Qy} = -\omega_z \ell \cos\theta \rm\ m/s$

Plugging in $\omega_z$ expression: \[ v_{Qy} = -\frac{v_{Px}}{\ell \sin\theta} \ell \cos\theta = -v_{Px}\cot\theta \rm\ m/s \]

Plugging in the numbers yields the same result of $\vec{v}_{Q} = -8\hat{\jmath} \rm\ m/s$.

Note in the rigid body relations method, we implicitly used the constraint equation for a point. Namely, we know that the top of the rod can only move vertically, so $\vec{v}_Q \cdot \hat{n}_Q = 0$, where $\hat{n}_Q = \hat{\imath}$ and $\vec{v}_P \cdot \hat{n}_P = 0$, where $\hat{n}_P = \hat{\jmath}$.

In general, the steps involved in analyzing systems with constrained motions are as follows:

Solution procedure for constrained motion.

\[\begin{aligned} &\text{1. Write down constraint equation in terms of distances/positions.} \\ &\text{2. Differentiate once or twice with respect to time to obtain $\vec{v}$ and $\vec{a}$ relationships.} \\ &\text{3. Solve for the desired quantity.} \end{aligned}\]

Example Problem: Compound pulley system

Consider the compound pulley system shown, consisting of two massless pulleys $A, B$. Pulley $A$ is free to move vertically, and pulley $B$ is fixed to the top. The two pulleys are connected by an inextensible, massless rope. One block of mass $m_1$ is connected to pulley $A$ by a string, and another block of mass $m_2$ is hanging from the rope connecting the two pulleys. Assume the pulleys' radii are negligbile. If $m_2$ moves downwards at a velocity $v_2 = 6 \rm\ m/s$ and acceleration $a_2 = 2 \rm\ m/s^2$, what are the velocity and acceleration of $m_1$?

We will begin by underlining important information given in the question. Namely:

- The rope is inextensible, meaning the total length of the rope remains constant. This is our constraint.
- The pulleys are massless, meaning we do not need to consider any rotational motion they may exhibit.
- The pulleys radii' are negligbile, meaning we do not need to consider their radii as a part of the total length of the rope.

Using the solution procedure highlighted in #rkc-cp:

**1. Write down constraint equation in terms of distances/positions.**
If the rope's length remains constant, we need to choose a reference line to relate the distances of
the masses to that reference line and set it equal to the rope's length. The following figure shows
the reference line, as well as the masses' distances from that line.

Note that $y_1$ is only extended to pulley $A$. Since the pulleys are massless, they will not rotate, and therefore $m_1$ and pulley $A$ will translate vertically together at the same rate.

Now, we can write the total length of the rope in terms of those two distances. The total length of the rope, $L$ will be: \[ 2y_1 + y_2 = L \]

**2. Differentiate once or twice with respect to time to obtain $\vec{v}$ and $\vec{a}$ relationships.**
Now, we can take two derivatives with respect to time to obtain the relationship between the speeds $v_1, v_2$, and
the accelerations $a_1, a_2$.
\[\begin{aligned}
2v_1 + v_2 &= 0 \\
2a_1 + a_2 &= 0
\end{aligned}\]

Notice that the length of the rope does not matter ultimately. This makes sense, as the length of the rope is constant, and therefore does not change with time.

**3. Solve for the desired quantity.**
We are interested in both the velocity and acceleration of $m_1$.

Solving for $\vec{v}_1 = \frac{-\vec{v}_2}{2} = 3\hat{\jmath}\rm\ m/s$ and $\vec{a}_1 = \frac{-\vec{a}_2}{2} = \hat{\jmath}\rm\ m/s^2$

Reflecting on the result obtained, it makes sense. The distance $m_1$ has to travel up the rope is two times more than the length $m_2$ has to travel. We can also draw a free-body diagram to understand this result further. We will return to this example further down when constraint forces are covered.

# Constraint forces

If an object is constrained to move in a specific direction, by Newton's second law, there must be
forces or moments preventing the movement of the object in that direction. Those are called *constraint forces*.

The approach to solving for constraint forces is the same approach to solve for any forces/moments that are not constraint forces. Namely, drawing a free body diagram, and applying Newton's equations. See the solution procedure with Newton's equations.

Let's return to our example of the rod leaning against a wall.

Example Problem: Constraint forces on a sliding rod.

A uniform rod of mass $m$ and length $\ell$ is leaning against a frictionless wall at an angle $\phi$ as shown. The rod starts sliding along a frictionless ground under the influence of gravity $\vec{g}$. What are the expressions of the constraint forces $\vec{F}_P, \vec{F}_Q$ that ensure the rod maintains contact with the wall and ground at all times?

We will begin by drawing a free body diagram on the rod.

Note that due to the constraint placed on the rod, point $P$ can only slide along the ground and therefore by #rkc-pc, $\vec{F}_P$ acts vertically. Similarly for point $Q$, $\vec{F}_Q$ acts horizontally.

Applying Newton's second law: \[ \sum \vec{F}_{ext} = m\vec{a}_C \]

Breaking it up into the scalar equations. Note we will use $\ddot{x}_C, \ddot{y}_C$ to denote the acceleration of the center of mass $C$ in each direction. \[\begin{aligned} F_Q &= m\ddot{x}_C \\ F_P - mg &= m\ddot{y}_C \end{aligned}\]

Now we can apply the solution procedure detailed in #rkc-cp to find the acceleration of point $C$. Due to our constraint, the following applies at all times: \[ x_C = \frac{\ell}{2}\sin\phi \qquad \qquad y_C = \frac{\ell}{2}\cos\phi \]

Differentiating twice, keeping in mind the chain rule: \[\begin{aligned} \ddot{x}_C &= -\frac{\ell\dot\theta^2}{2}\sin\phi + \frac{\ell\ddot\theta}{2}\cos\phi \\ \ddot{y}_C &= -\frac{\ell\dot\theta^2}{2}\cos\phi - \frac{\ell\ddot\theta}{2}\sin\phi \end{aligned}\]

Plugging into Newton's second law equations and solving for each constraint force yields: \[\begin{aligned} F_Q &= -\frac{m\ell\dot\theta^2}{2}\sin\phi + \frac{m\ell\ddot\theta}{2}\cos\phi \\ F_P &= -\frac{m\ell\dot\theta^2}{2}\cos\phi - \frac{m\ell\ddot\theta}{2}\sin\phi + mg \end{aligned}\]

Example Problem: Instance where rod loses contact with a wall.

A uniform rod of mass $m$ and length $\ell$ is leaning against a frictionless wall at rest at an angle $\phi$ as shown. The rod starts sliding on a frictionless ground under the influence of gravity $\vec{g}$. At what angle $\theta$ does the rod lose contact with the wall?

While there is contact with the wall, our constraint still holds. This means our constraint forces expressions from #rkc-cf still hold. We are only interested in the constraint force from the wall on the rod, $\vec{F}_Q$ at $\theta$: \[ F_Q = -\frac{m\ell\dot\theta^2}{2}\sin\theta + \frac{m\ell\ddot\theta}{2}\cos\theta \]

If the rod loses contact with the wall, then $F_Q = 0$. We have two unknowns: the angular velocity $\dot\theta$ and the angular acceleration $\ddot\theta$. However, we know that $\ddot\theta = \frac{d\dot\theta}{dt}$, so if we obtain an expression for $\dot\theta$, then finding $\ddot\theta$ is possible. We can use conservation of energy to obtain an expression for $\dot\theta$.

Using the work-energy theorem: \[ W = \Delta T + \Delta V \]

The rod is sliding along frictionless surfaces, so there are no non-conservative forces present, and constraint forces do no work (See work done by a constraint force)

Therefore: \[\begin{aligned} T_1 + V_1 &= T_2 + V_2 \\ 0 + mgh_{C, 1} &= \frac{1}{2}mv_C^2 + \frac{1}{2}I_{C, \hat{k}}\omega^2 + mgh_{C, 2} \end{aligned}\]

From before: \[\begin{aligned} h_{C, 1} = y_{C, 1} = \ell\cos\phi \qquad h_{C, 2} = y_{C, 2} = \ell\cos\theta \end{aligned}\]

And the velocity of the center of mass at $\theta$: \[\begin{aligned} \vec{v}_C &= \dot{x}_{C,2}\hat{\imath} + \dot{y}_{C, 2}\hat{\jmath} \\ &= \ell\dot\theta\cos\theta \hat{\imath} - \ell\dot\theta\sin\theta \hat{\jmath} \end{aligned}\]

So $v_C$ is simply $l\dot\theta$.

Plugging into our energy conservation, where $I_{C, \hat{k}} = \frac{1}{12}m\ell^2$: \[\begin{aligned} mg\ell\cos\phi &= \frac{1}{2}m(\ell\dot\theta)^2 + \frac{1}{2}\left(\frac{1}{12}m\ell^2 \right) \dot\theta^2 + mg\ell\cos\theta \\ mg\ell\cos\phi &= \frac{1}{2}m\ell^2 \dot\theta^2 + \frac{1}{24}m\ell^2 \dot\theta^2 + mg\ell\cos\theta \\ mg\ell\cos\phi &= \frac{13}{24}m\ell^2 \dot\theta^2 + mg\ell\cos\theta \end{aligned}\]

Solving for $\dot\theta^2$: \[ \dot\theta^2 = \frac{24g}{13\ell}\left(\cos\phi - \cos\theta\right) \]

Differentiating to obtain $\ddot\theta$: \[\begin{aligned} 2\dot\theta\ddot\theta &= \frac{24g}{13\ell}\sin\theta\dot\theta \\ \ddot\theta & = \frac{12g}{13\ell}\sin\theta \end{aligned}\]

Plugging our results into $F_Q$: \[\begin{aligned} F_Q &= -\frac{m\ell\frac{24g}{13\ell}\left(\cos\phi - \cos\theta\right)}{2}\sin\theta + \frac{m\ell\frac{12g}{13\ell}\sin\theta}{2}\cos\theta \\ &= -\frac{12}{13}mg\sin\theta \left(\cos\phi - \cos\theta\right) + \frac{6}{13}mg\cos\theta\sin\theta \\ \end{aligned}\]

Setting $F_Q = 0$: \[\begin{aligned} \frac{12}{13}mg\sin\theta \left(\cos\phi - \cos\theta\right) &= \frac{6}{13}mg\cos\theta\sin\theta \\ 2\left(\cos\phi - \cos\theta\right) &= \cos\theta \\ 2\cos\phi - 2\cos\theta - \cos\theta &= 0 \\ 2\cos\phi - 3\cos\theta &= 0 \\ \end{aligned}\]

Solving for $\theta$ yields $\theta = \arccos\left(\frac{2}{3}\cos\phi\right)$, or when the rod has fallen 2/3 of the original height. Note if the rod was unconstrained, the constraint we used still holds until the rod has fallen 2/3 of the original height. We can use rigid body relations to find the velocity of point $Q$ once the rod loses contact with the wall, and it will have both vertical and horizontal components of velocity, as expected.

Example Problem: Tension in a compound pulley system

Consider the compound pulley system shown, consisting of two massless pulleys $A, B$. Pulley $A$ is free to move vertically, and pulley $B$ is fixed to the top. The two pulleys are connected by an inextensible, massless rope. One block of mass $m_1$ is connected to pulley $A$ by a string, and another block of mass $m_2$ is hanging from the rope connecting the two pulleys. Assume the pulleys' radii are negligbile, and the rope can only slip. What is the tension $T$ in the rope?

We added one more assumption to the ones made in #rkc-xcp. Namely, the rope can only slip. This simply means that the tension is the same at any point in the rope.

We will begin by drawing a free body diagram and applying Newton's second law on three bodies: $m_1$, $m_2$ and pulley $A$. We will neglect pulley $B$ as it is fixed to the wall, and the results obtained from it are trivial.

The free body diagram on pulley $A$ is as follows:

Applying Newton's second law: \[ 2T - T_2 = m_A a_A \]

Since the pulley is massless, that means $T_2 = 2T$.

The free body diagram on $m_1$ is as follows:

Applying Newton's second law: \[ 2T - m_1 g = m_1 a_1 \]

Finally, the free body diagram on $m_2$ is as follows:

Applying Newton's second law: \[ T - m_2 g = m_2 a_2 \]

We have 3 unknowns, 2 equations. We apply our constraint equation from #rkc-xcp to relate the accelerations of $m_1$ and $m_2$: \[ 2a_1 + a_2 = 0 \]

Solving for $T$ from Newton's second law on $m_2$: \[ T = m_2 a_2 + m_2 g \]

Plugging it into Newton's second law on $m_1$: \[ 2m_2 a_2 + 2m_2 g - m_1 g = m_1 a_1 \]

Since $a_2 = -2a_1$: \[\begin{aligned} -4m_2 a_1 + 2m_2 g - m_1g &= m_1 a_1 \\ g\left(2m_2 - m_1 \right) &= a_1\left(m_1 + 4m_2\right) \end{aligned}\]

Solving for $a_1$: \[ a_1 = \frac{g\left(2m_2 - m_1 \right)}{m_1 + 4m_2} \]

Plugging that result into Newton's second law on $m_1$ and solving for $T$: \[ T = \frac{m_1\frac{g\left(2m_2 - m_1 \right)}{m_1 + 4m_2} + m_1g}{2} \]

Simplifying a little by multiplying the top and bottom by $m_1 + 4m_2$: \[ T = \frac{m_1g(2m_2 - m_1) + m_1g(m_1 + 4m_2)}{2(m_1 + 4m_2)} \]

We obtain our final result for the tension in the rope $T$: \[ T = \frac{3m_1m_2g}{m_1 + 4m_2} \]

If the rod were to slide along the ground at constant velocity, the constraint forces from the ground and wall tend to infinity, and it yields large speeds. This motion then becomes unrealistic, as the related rates problem from calculus 1 places no regard on the constraint forces that cause this motion.