PHSX 220 Physics I (w/ Calculus) · Unit 1: Kinematics & Newton’s laws · on Exam 1
Ch 4.4–4.5 Uniform circular motion and relative motion
Study notes for Uniform circular motion and relative motion in PHSX 220 Physics I (w/ Calculus) at Montana State University: the big ideas, key formulas, a worked example, common mistakes and an exam tip. The same topic has endless practice questions and flashcards in Mathub.
Big ideas
- Moving in a circle at constant speed still means accelerating, because the velocity's direction changes. The centripetal acceleration points toward the center with magnitude $a_c = v^2/r$.
- Period $T$ is the time for one revolution: $v = 2\pi r/T$, so $a_c = 4\pi^2 r/T^2$. Angular speed $\omega = 2\pi/T$ and $v = r\omega$, $a_c = \omega^2 r$.
- If the speed also changes there is a tangential acceleration $a_t = dv/dt$; the total acceleration is the vector sum of $a_c$ (radial) and $a_t$ (tangential).
- Relative velocity: $\vec v_{A/C} = \vec v_{A/B} + \vec v_{B/C}$ (subscripts chain like fractions). A boat heading straight across a river is carried downstream; its ground velocity is the vector sum.
Key formulas
Circular motion
$$a_c = \frac{v^2}{r} = \omega^2 r = \frac{4\pi^2 r}{T^2},\qquad v = \frac{2\pi r}{T} = r\omega$$
Relative motion
$$\vec v_{\text{boat/ground}} = \vec v_{\text{boat/water}} + \vec v_{\text{water/ground}}$$
Worked example
A boat can move at 4.0 m/s in still water and aims straight across a river flowing at 3.0 m/s. The river is 120 m wide. How long does the crossing take, and how far downstream does the boat land?
Show the solution
Crossing uses only the boat's across-river velocity: $t = 120/4 = 30$ s. Downstream drift $= 3(30) = 90$ m. Ground speed $\sqrt{16+9} = 5$ m/s at $36.9^\circ$ downstream of straight across.
Common mistakes
- Drawing the centripetal acceleration tangent to the circle; it points to the center.
- Confusing period (seconds) with frequency (revolutions per second).
- Adding relative speeds as scalars when the velocities are not parallel.
Exam tipIn clicker questions about a car on a curve: constant speed does not mean zero acceleration. The net force must point toward the center.
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