Projectile motion, uniform circular motion and relative velocity

This week there are three classic applications of two-dimensional kinematics: projectile motion, uniform circular motion and relative velocity. All three are question types worth free points on the midterm. The basis of all of them is last week’s rule: $x$ and $y$ are independent, and the only thing they share is time.

Golden rules:

1. In projectile motion the horizontal acceleration is zero and the vertical acceleration is $-g$. The horizontal velocity is constant throughout the flight.
2. The range formula $R = v_0^2\sin2\theta/g$ is valid only if the landing level equals the launch level.
3. At the top, only $v_y = 0$. The speed is $v_0\cos\theta$, and the acceleration is still $g$.
4. An object moving in a circle at constant speed has an acceleration, toward the center and equal to $v^2/r$.
5. In relative velocity, chain the subscripts like links: $\vec v_{PA} = \vec v_{PB} + \vec v_{BA}$.

Projectile motion

No air resistance, $g$ constant. The launch point is the origin, up is $+$. The initial velocity is $v_0$, at an angle $\theta$ to the horizontal:

$v_{0x} = v_0\cos\theta \qquad v_{0y} = v_0\sin\theta$

$x = v_{0x}\,t \qquad y = v_{0y}\,t - \tfrac12gt^2 \qquad v_x = v_{0x} \qquad v_y = v_{0y} - gt$

The trajectory is a parabola. At any moment the speed is $\sqrt{v_x^2 + v_y^2}$ and the direction is $\tan^{-1}(v_y/v_x)$.

Ready results for a launch landing at the same level

$t_{\text{top}} = \frac{v_0\sin\theta}{g} \qquad H = \frac{(v_0\sin\theta)^2}{2g} \qquad T_{\text{flight}} = \frac{2v_0\sin\theta}{g} \qquad R = \frac{v_0^2\sin2\theta}{g}$

$R$ takes its largest value at $\theta = 45^\circ$. Complementary angles (such as $30^\circ$ and $60^\circ$) give the same range; the higher angle stays in the air longer.

Example 1. A soccer ball is kicked from level ground at 22.0 m/s, $40.0^\circ$ above the horizontal.

v0x = 22.0 cos 40.0° = 16.9 m/s       v0y = 22.0 sin 40.0° = 14.1 m/s
t_top = 14.1 / 9.80 = 1.44 s
H     = 14.1² / (2 × 9.80) = 10.2 m
T     = 2 × 1.44 = 2.89 s
R     = 22.0² sin 80.0° / 9.80 = 48.6 m
check: R = v0x × T = 16.9 × 2.89 = 48.6 m

Horizontal launch

An object leaves from height $h$ with horizontal $v_0$: $v_{0y} = 0$. The fall time is independent of the horizontal velocity, as if it had been dropped:

$t = \sqrt{\frac{2h}{g}} \qquad x = v_0\,t$

Example 2. A stone thrown horizontally at 8.0 m/s from a height of 12.0 m.

t  = √(2 × 12.0 / 9.80) = 1.56 s
x  = 8.0 × 1.56 = 12.5 m
vy = 9.80 × 1.56 = 15.3 m/s (downward),  vx = 8.0 m/s
v  = √(8.0² + 15.3²) = 17.3 m/s,  below the horizontal by tan⁻¹(15.3/8.0) = 62.4°

PITFALL · using the range formula in the wrong place. In a launch from a cliff, a building or a table, the landing point is below the launch point. $R = v_0^2\sin2\theta/g$ gives a wrong result here, and that wrong result is surely among the options. In this case write $y = -h$ and solve the quadratic equation, take the positive root, and then $x = v_{0x}t$.

Uniform circular motion

The object turns in a circle of radius $r$ at constant speed. The velocity is tangent and its direction changes continuously, so there is an acceleration. This acceleration is toward the center of the circle at every moment: centripetal acceleration.

$a_c = \frac{v^2}{r} \qquad v = \frac{2\pi r}{T} \qquad a_c = \frac{4\pi^2 r}{T^2}$

$T$ is the period (time for one revolution, s), $f = 1/T$ is the frequency (rev/s, Hz).

Example 3. A Ferris wheel of radius 12 m makes one turn in 40 s. What are the speed and acceleration of a passenger in a cabin?

v   = 2π(12) / 40 = 1.88 m/s
a_c = v² / r = 1.88² / 12 = 0.296 m/s²   (toward the center)
check: 4π²(12) / 40² = 0.296 m/s²

Tangential and radial acceleration

If the speed is changing too, the acceleration has two perpendicular components:

$a_r = \frac{v^2}{r}\ \ (\text{toward the center}) \qquad a_t = \frac{d|\vec v|}{dt}\ \ (\text{parallel to the velocity}) \qquad a = \sqrt{a_r^2 + a_t^2}$

When speeding up, $a_t$ is in the same direction as the velocity; when slowing down, in the opposite direction. In uniform circular motion $a_t = 0$.

Relative velocity

Velocity changes with the observer who measures it (reference frame). The velocity of particle $P$ relative to observer $A$ is the vector sum of its velocity relative to $B$ and the velocity of $B$ relative to $A$:

$\vec v_{PA} = \vec v_{PB} + \vec v_{BA} \qquad \vec v_{AB} = -\vec v_{BA}$

Think of the inner subscripts (B) as “canceling”. Observers moving at constant velocity measure the same acceleration.

Example 4. A river 60 m wide flows at 2.0 m/s. The boat moves at 4.0 m/s relative to the water.

Case A, the boat keeps its bow pointed straight across:

crossing time: only the perpendicular component  →  t = 60 / 4.0 = 15 s
drift (downstream): 2.0 × 15 = 30 m
speed relative to the ground: √(4.0² + 2.0²) = 4.47 m/s

Case B, the boat wants to reach the point directly opposite:

bow at an angle φ against the current:  4.0 sin φ = 2.0   →  φ = 30°
velocity relative to the ground (perpendicular): √(4.0² - 2.0²) = 3.46 m/s
t = 60 / 3.46 = 17.3 s   (longer than A)

PITFALL · which one is the hypotenuse in the river? In case A the boat’s velocity relative to the water is one leg, and the velocity relative to the ground is the hypotenuse. In case B the boat’s velocity relative to the water is the hypotenuse, and the velocity relative to the ground is a leg. First draw the vector triangle and look at which vector points straight across.

Common mistakes