Lesson 7 · 40 min
Equations of Motion
Newton's second law \(\sum\Fvec = m\avec\), written along \(\er\), \(\et\) and \(\ez\), becomes three scalar equations. They link the forces on a particle to the acceleration terms of Lesson 5, so you can find the force a rod, a cord or a slide must supply, or the motion that given forces produce.
Learning objectives
- Write the equations of motion \(\sum F_r = m(\ddot r - r\dot\theta^2)\), \(\sum F_\theta = m(r\ddot\theta + 2\dot r\dot\theta)\) and \(\sum F_z = m\ddot z\).
- Draw a free-body diagram and a kinetic diagram with every force resolved along \(\er\), \(\et\) and \(\ez\).
- Resolve the forces that appear in these problems: weight, smooth rods and slots, springs, cords and friction.
- Solve for unknown forces on particles carried by rotating arms, held by cords or sliding down helical paths.
Newton's second law in cylindrical components
Write the resultant force on the particle and its acceleration in the same unit vectors, \(\sum\Fvec = \sum F_r\,\er + \sum F_\theta\,\et + \sum F_z\,\ez\), and set \(\sum\Fvec = m\avec\) component by component:
Equations of motion in cylindrical coordinates
\[ \begin{aligned} \textstyle\sum F_r &= m a_r = m\left(\ddot r - r\dot\theta^2\right) \\ \textstyle\sum F_\theta &= m a_\theta = m\left(r\ddot\theta + 2\dot r\dot\theta\right) \\ \textstyle\sum F_z &= m a_z = m\ddot z \end{aligned} \]Three equations, so a problem can have at most three unknowns: forces, or acceleration terms such as \(\ddot r\) and \(\ddot\theta\).
The left sides are the components of the forces along the unit vectors at the particle's current position. The right sides come from the kinematics: everything you learned in Lessons 4 to 6. Hibbeler calls this Section 13.6; in Beer & Johnston and in Meriam & Kraige the same equations are called the equations of motion in radial and transverse components.
Free-body and kinetic diagrams
Draw the particle with the unit vectors \(\er\) and \(\et\) at its current position. On the free-body diagram show every force acting on it; on the kinetic diagram show \(m a_r\) along \(\er\) and \(m a_\theta\) along \(\et\) (and \(m a_z\) along \(\ez\)). The two diagrams are equivalent: their components along each unit vector are equal.
| Force | Components | Notes |
|---|---|---|
| Weight, motion in a horizontal plane | \(-mg\,\ez\) | Balanced by a vertical support force when \(z\) is constant. |
| Weight, motion in a vertical plane | \(-mg\sin\theta\,\er - mg\cos\theta\,\et\) | With \(\theta\) measured from the horizontal \(x\)-axis and \(y\) up (Example 3.1). |
| Smooth rod or slot that turns with the arm | \(N\,\et\) (plus a vertical part) | A smooth rod cannot push along itself: no \(\er\) component. |
| Spring along the arm, attached at \(O\) | \(-k(r - r_0)\,\er\) | \(r_0\) is the unstretched length; a stretched spring pulls toward \(O\). |
| Cord pulled through \(O\) | \(-T\,\er\) | A cord can only pull: \(T \ge 0\). |
| Smooth fixed guide \(r = f(\theta)\) | perpendicular to the path | Its direction comes from the tangent angle \(\psi\): Lesson 8. |
| Kinetic friction | \(\mu_k N\), opposite to the sliding velocity | Along the path, against the motion relative to the surface. |
Procedure for analysis
- Coordinates. Put \(O\) at the pivot or the center of the motion, and choose the direction of positive \(\theta\).
- Diagrams. Draw the free-body and kinetic diagrams at the instant of interest, with \(\er\) and \(\et\) at the particle.
- Kinematics. Find \(r, \dot r, \ddot r\) and \(\dot\theta, \ddot\theta\) (and \(\ddot z\)), from functions of time (Lesson 5) or from a path \(r = f(\theta)\) (Lesson 6). Then \(a_r\), \(a_\theta\), \(a_z\).
- Equations of motion. Sum the force components along \(\er\), \(\et\), \(\ez\) and set them equal to \(m a_r\), \(m a_\theta\), \(m a_z\).
- Solve and check. A negative answer means the force points opposite to the direction you assumed. A cord with \(T \lt 0\) or a contact force pulling the particle means the assumed motion cannot happen.
Worked examples
Example 7.1 — A collar on a spinning rod
A \(0.5\ \text{kg}\) collar slides freely on a smooth rod that a motor turns in a horizontal plane at a constant \(\dot\theta = 4\ \text{rad/s}\). At the instant when \(r = 0.6\ \text{m}\), the collar is sliding outward at \(\dot r = 0.8\ \text{m/s}\). Find (a) \(\ddot r\), and (b) the force the rod exerts on the collar.
Show solution
Forces. The rod is smooth, so it pushes only perpendicular to itself: a horizontal force \(N_\theta\,\et\) and a vertical force \(N_z\,\ez\). The weight is \(-mg\,\ez\). Nothing acts along the rod, so \(\sum F_r = 0\).
Kinematics. \(\dot\theta = 4\ \text{rad/s}\) and \(\ddot\theta = 0\); \(\ddot r\) is unknown; \(z\) is constant.
(a) Radial equation:
\[ \textstyle\sum F_r = m(\ddot r - r\dot\theta^2): \quad 0 = 0.5\left(\ddot r - (0.6)(4)^2\right) \quad\Rightarrow\quad \ddot r = 9.6\ \text{m/s}^2 \]With no force along the rod, nothing supplies the centripetal acceleration, so the collar accelerates outward along the rod: \(\ddot r = r\dot\theta^2\).
(b) Transverse and vertical equations:
\[ \begin{aligned} \textstyle\sum F_\theta &= m(r\ddot\theta + 2\dot r\dot\theta): & N_\theta &= 0.5\left(0 + 2(0.8)(4)\right) = 3.2\ \text{N} \\ \textstyle\sum F_z &= m\ddot z: & N_z - (0.5)(9.81) &= 0 \quad\Rightarrow\quad N_z = 4.905\ \text{N} \end{aligned} \]The rod pushes the collar forward (in the direction of rotation) with \(3.2\ \text{N}\), entirely because of the Coriolis term, and holds it up with \(4.905\ \text{N}\). The total force from the rod is \(\sqrt{3.2^2 + 4.905^2} \approx 5.857\ \text{N}\).
Example 7.2 — An arm turning in a vertical plane
The \(0.5\ \text{kg}\) collar of Figure 7.1 rides on a smooth arm that turns in a vertical plane at a constant \(\dot\theta = 3\ \text{rad/s}\). A cord through \(O\) pulls it in at a constant \(0.5\ \text{m/s}\). Find the cord tension \(T\) and the force \(N\) of the arm on the collar at the instant when \(\theta = 30^\circ\) and \(r = 0.8\ \text{m}\).
Show solution
Kinematics. \(\dot r = -0.5\ \text{m/s}\) and \(\ddot r = 0\) (pulled in at a constant rate); \(\dot\theta = 3\ \text{rad/s}\) and \(\ddot\theta = 0\).
\[ \begin{aligned} a_r &= \ddot r - r\dot\theta^2 = 0 - (0.8)(3)^2 = -7.2\ \text{m/s}^2 \\ a_\theta &= r\ddot\theta + 2\dot r\dot\theta = 0 + 2(-0.5)(3) = -3.0\ \text{m/s}^2 \end{aligned} \]Forces (Figure 7.1): the cord \(-T\,\er\), the arm \(N\,\et\), and the weight \(-mg\sin\theta\,\er - mg\cos\theta\,\et\) with \(mg = 4.905\ \text{N}\).
\[ \begin{aligned} \textstyle\sum F_r = m a_r: &\quad -T - 4.905\sin 30^\circ = 0.5(-7.2) &&\Rightarrow\quad T = 3.6 - 2.453 = 1.148\ \text{N} \\ \textstyle\sum F_\theta = m a_\theta: &\quad N - 4.905\cos 30^\circ = 0.5(-3.0) &&\Rightarrow\quad N = -1.5 + 4.248 = 2.748\ \text{N} \end{aligned} \]Both are positive, so the cord pulls toward \(O\) and the arm pushes in the \(+\et\) direction, as assumed. Note that \(m a_\theta \lt 0\): the Coriolis term asks for a backward push, but the arm must also hold the collar up against \(mg\cos\theta\). Check the cord: at the top (\(\theta = 90^\circ\)) the radial equation would give \(T = m r\dot\theta^2 - mg = 3.6 - 4.905 \lt 0\). A cord cannot push, so at this speed it would go slack near the top. Figure 7.3 shows where.
Example 7.3 — The force from a spiral slide
A \(25\ \text{kg}\) child rides the spiral slide of Example 5.2: \(r = 2\ \text{m}\), \(\dot\theta = 0.8\ \text{rad/s}\) and \(\dot z = -0.5\ \text{m/s}\), all constant. Find the total force the chute exerts on the child.
Show solution
From Example 5.2, \(\avec = -1.28\,\er\ \text{m/s}^2\). Let \(\mathbf{F}_c\) be the total chute force (normal force plus friction). The other force is the weight \(-mg\,\ez\).
\[ \begin{aligned} \textstyle\sum F_r &= m a_r: & F_{c,r} &= (25)(-1.28) = -32\ \text{N} \\ \textstyle\sum F_\theta &= m a_\theta: & F_{c,\theta} &= 0 \\ \textstyle\sum F_z &= m a_z: & F_{c,z} - (25)(9.81) &= 0 \quad\Rightarrow\quad F_{c,z} = 245.25\ \text{N} \end{aligned} \]So \(\mathbf{F}_c = (-32\,\er + 245.25\,\ez)\ \text{N}\), with magnitude \(\approx 247.3\ \text{N}\): the chute holds the child up and pushes inward with \(32\ \text{N}\), from the outer wall.
Going further. The child moves at constant speed, so the friction along the chute must balance the part of the weight along the path. The path direction is \(\mathbf{u}_t = (1.6\,\et - 0.5\,\ez)/1.676\), so the weight's component along it is \(mg(0.5/1.676) \approx 73.15\ \text{N}\), and friction is \(73.15\ \text{N}\) back up the chute. The rest of \(\mathbf{F}_c\) is the normal force, about \(236.3\ \text{N}\), so the friction coefficient is \(\mu_k \approx 73.15/236.3 \approx 0.31\).
Check your understanding
Key takeaways
- \(\sum F_r = m(\ddot r - r\dot\theta^2)\), \(\sum F_\theta = m(r\ddot\theta + 2\dot r\dot\theta)\), \(\sum F_z = m\ddot z\).
- Resolve every force along \(\er\), \(\et\), \(\ez\) at the particle's current position: in a vertical plane the weight has both \(r\) and \(\theta\) components.
- A smooth rod pushes only perpendicular to itself; a spring or a cord along the rod acts only along \(\er\).
- With no radial force, a collar on a spinning rod slides outward with \(\ddot r = r\dot\theta^2\), and the rod must push it sideways with \(2m\dot r\dot\theta\).
- Next, Lesson 8 handles particles pushed along curved guides and motion under a central force.