Lesson 8 · 40 min
Kinetics Applications
Two classic problem types complete the module. A particle pushed along a fixed curved guide by a slotted arm, where the tangent angle \(\psi\) sets the direction of the guide's push. And central-force motion, where the transverse equation alone shows that \(r^2\dot\theta\) cannot change.
Learning objectives
- Find the forces of a slotted arm and of a smooth fixed guide on a particle moving along \(r = f(\theta)\), using the tangent angle \(\psi\).
- Show that a force with no transverse component keeps \(r^2\dot\theta\) constant, and use this for cords pulled through a hole and for orbits.
- Find the force needed to keep a central-force motion going.
- Choose rectangular, path or cylindrical components for a kinetics problem.
A particle on a fixed guide, pushed by an arm
A pin \(P\) of mass \(m\) moves in a smooth fixed groove \(r = f(\theta)\) in a horizontal plane, pushed along by a smooth slotted arm that turns about \(O\) (Figure 8.1). In the plane two forces act on it:
- The arm pushes perpendicular to itself: \(F\,\et\).
- The guide pushes perpendicular to the path. The tangent makes the angle \(\psi\) with \(\er\), so the normal to the path makes the angle \(\psi\) with \(\et\): \(\mathbf{N} = N\left(-\sin\psi\,\er + \cos\psi\,\et\right)\).
The weight and the vertical support cancel. The two in-plane equations of motion then give the two unknowns:
Arm and guide forces
\[ \begin{aligned} \textstyle\sum F_r = m a_r: &\quad -N\sin\psi = m a_r &&\Rightarrow\quad N = -\frac{m a_r}{\sin\psi} \\ \textstyle\sum F_\theta = m a_\theta: &\quad F + N\cos\psi = m a_\theta &&\Rightarrow\quad F = m a_\theta - N\cos\psi \end{aligned} \]with \(\tan\psi = r/(dr/d\theta)\) from Lesson 6. A negative \(N\) or \(F\) means that force points the other way: the other wall of the groove, or the other face of the slot.
Only the guide can supply the radial force, because the arm pushes across itself. That is why the radial equation gives \(N\) directly. If the plane is vertical, add the weight's components \(-mg\sin\theta\) and \(-mg\cos\theta\) to the two sums.
Example 8.1 — Forces in a spiral feeder
The part in the spiral feeder of Example 6.2 has a mass of \(0.4\ \text{kg}\): it rides in the smooth groove \(r = 0.05\theta\ \text{m}\) in a horizontal plane, pushed by a smooth arm turning at a constant \(5\ \text{rad/s}\). Find the forces of the arm and of the groove on the part when \(\theta = 2\pi\).
Show solution
Kinematics (Example 6.2): \(r = 0.3142\ \text{m}\), \(a_r = -7.854\ \text{m/s}^2\), \(a_\theta = 2.5\ \text{m/s}^2\), and \(\tan\psi = r/(dr/d\theta) = 2\pi\), so \(\psi = 80.96^\circ\).
Radial equation (only the groove has a radial component):
\[ -N\sin\psi = m a_r: \qquad N = -\frac{(0.4)(-7.854)}{\sin 80.96^\circ} = \frac{3.142}{0.9876} \approx 3.181\ \text{N} \]Transverse equation:
\[ F + N\cos\psi = m a_\theta: \qquad F = (0.4)(2.5) - (3.181)(0.1571) = 1.0 - 0.5 = 0.5\ \text{N} \]Both are positive. The groove pushes on the part with \(3.18\ \text{N}\), mostly toward \(O\) (it is the outer wall that pushes, supplying the centripetal force), with a small forward part \(N\cos\psi = 0.5\ \text{N}\). The arm supplies the other \(0.5\ \text{N}\) of the forward force \(m a_\theta = 1.0\ \text{N}\), which is all Coriolis: the part moves outward while it turns.
Central-force motion
Suppose the only force on a particle in the plane points along the line through \(O\): a cord pulled through a hole at \(O\), a spring anchored at \(O\), or the gravity of a planet at \(O\). Such a force is a central force. It has no transverse component, so
\[ \textstyle\sum F_\theta = m\left(r\ddot\theta + 2\dot r\dot\theta\right) = 0 \]Multiply by \(r/m\): \(r^2\ddot\theta + 2r\dot r\dot\theta = 0\). The left side is exactly \(\dfrac{d}{dt}\left(r^2\dot\theta\right)\), by the product rule. So \(r^2\dot\theta\) does not change:
Central-force motion
\[ r^2\dot\theta = h = \text{constant}, \qquad\text{equivalently}\qquad r\,v_\theta = h \]As \(r\) shrinks, the particle turns faster: \(\dot\theta = h/r^2\) and \(v_\theta = h/r\). The radial equation \(\sum F_r = m\left(\ddot r - r\dot\theta^2\right) = m\left(\ddot r - h^2/r^3\right)\) gives the force.
\(m r v_\theta\) is the particle's angular momentum about \(O\), so this is conservation of angular momentum, which you will meet again with impulse and momentum. It is also Kepler's second law: the line from the Sun to a planet sweeps out area at the rate \(\tfrac12 r^2\dot\theta = h/2\), the same at every point of the orbit.
Example 8.2 — Pulling a puck in
A \(0.3\ \text{kg}\) puck on a frictionless table moves on a circle of radius \(0.8\ \text{m}\) at \(\dot\theta = 2\ \text{rad/s}\), held by a cord through a hole at the center. The cord is then pulled down at a steady \(0.2\ \text{m/s}\). When \(r = 0.4\ \text{m}\), find \(\dot\theta\), the speed of the puck and the cord tension.
Show solution
The cord force is central, so \(r^2\dot\theta\) keeps its starting value:
\[ h = r_1^2\dot\theta_1 = (0.8)^2(2) = 1.28\ \text{m}^2/\text{s} \qquad\Rightarrow\qquad \dot\theta_2 = \frac{h}{r_2^2} = \frac{1.28}{(0.4)^2} = 8\ \text{rad/s} \]Halving the radius quadruples the turning rate. The velocity components are \(v_r = \dot r = -0.2\ \text{m/s}\) and \(v_\theta = r\dot\theta = (0.4)(8) = 3.2\ \text{m/s}\), so \(v = \sqrt{0.04 + 10.24} \approx 3.206\ \text{m/s}\), up from \(1.6\ \text{m/s}\).
Tension. The pull rate is steady, so \(\ddot r = 0\). The radial equation with the cord force \(-T\,\er\):
\[ -T = m\left(\ddot r - r\dot\theta^2\right) = 0.3\left(0 - (0.4)(8)^2\right) = -7.68 \qquad\Rightarrow\qquad T = 7.68\ \text{N} \]At the start it was \(T_1 = m r_1\dot\theta_1^2 = 0.96\ \text{N}\): eight times smaller, because \(T = m h^2/r^3\) here. The kinetic energy rises from \(0.384\ \text{J}\) to \(1.542\ \text{J}\); the hand pulling the cord does that work.
Example 8.3 — A satellite at apogee
A satellite moves on an elliptical orbit around the Earth. At perigee (closest point) it is \(7000\ \text{km}\) from the Earth's center, moving at \(8.185\ \text{km/s}\) perpendicular to the radial line. Find its speed at apogee, \(10\,000\ \text{km}\) from the center, where its velocity is again perpendicular to the radial line.
Show solution
Gravity points at the Earth's center, so it is a central force and \(r\,v_\theta\) is constant. At perigee and apogee \(v_r = 0\), so \(v = v_\theta\) there:
\[ r_p v_p = r_a v_a \qquad\Rightarrow\qquad v_a = \frac{r_p}{r_a}\,v_p = \frac{7000}{10\,000}(8.185) \approx 5.730\ \text{km/s} \]The satellite is fastest where it is closest. Its turning rate changes even more: \(\dot\theta = v/r\) is \(1.169 \times 10^{-3}\ \text{rad/s}\) at perigee and \(0.573 \times 10^{-3}\ \text{rad/s}\) at apogee, in the ratio \((r_a/r_p)^2 \approx 2.04\). (The shape of the orbit itself, an ellipse, also follows from the radial equation with Newton's law of gravitation; that is beyond this module.)
Choosing coordinates for a kinetics problem
| Use | When | Typical problems |
|---|---|---|
| Rectangular \(x, y, z\) | The forces act in fixed directions. | Projectiles, blocks on fixed inclines. |
| Path \(t, n\) | The path and its radius of curvature are known, and speed along it matters. | Cars on curves, roller coasters, banked turns. |
| Cylindrical \(r, \theta, z\) | The motion is driven or measured from a fixed point or axis, or the forces point along or across a line through it. | Rotating rods and arms, slotted guides and cams, cords through a hole, central forces, spiral paths. |
For a circle centered on \(O\), the path and polar components coincide (\(a_n = -a_r\), \(a_t = a_\theta\)), so either works. When a problem is stated in terms of an arm angle \(\theta\) and a distance \(r\) along it, cylindrical components are almost always the shortest route.
Check your understanding
Key takeaways
- For a particle in a smooth fixed guide pushed by a smooth arm: \(N = -m a_r/\sin\psi\) and \(F = m a_\theta - N\cos\psi\), with \(\tan\psi = r/(dr/d\theta)\).
- A central force has \(\sum F_\theta = 0\), which makes \(r^2\dot\theta = h\) constant: closer in means faster round.
- The radial equation \(\sum F_r = m(\ddot r - h^2/r^3)\) then gives the cord tension or the attracting force.
- Pick the components that match the forces and constraints; for arms, guides and central forces that means \(r\), \(\theta\), \(z\).
- This is the last lesson. Practice with the Practice Lab, test yourself with the Self-Check Quiz, and keep the Formula Sheet handy.